Oscillations

NEET Ug . NEET Physics . Physics . Oscillations

Chapter 14: Oscillations

Ch 14

NEET Physics

NEET Ug

Introduction

Overview

Orbital motion of planets is periodic; repeats after fixed intervals.
Oscillatory motion explains many physical phenomena like vibrating strings in musical instruments.
Atoms in solids vibrate about mean positions; vibration energy relates to temperature.
AC voltage oscillates alternately about zero mean value.

Solved Examples

Practice Set

Q1: What type of motion is the Earth's revolution around the Sun?
- A. Oscillatory motion
- B. Periodic motion
- C. Random motion
- D. Linear motion
Q2: Which of the following is an example of oscillatory motion?
- A. Earth's revolution around the Sun
- B. Vibration of guitar string
- C. Car moving on highway
- D. Flow of water in river
Q3: What does the vibration of atoms in a solid depend on?
- A. Pressure
- B. Temperature
- C. Volume
- D. Mass

Answer Key

Q1: B - Earth's revolution repeats identically after fixed time, hence periodic.
Q2: B - Guitar string vibrates to and fro about mean position, an oscillatory motion.
Q3: B - Average vibration energy of atoms is proportional to temperature.

Memory Aids

Periodic motion = repeats after fixed time.
Oscillatory motion = back and forth about mean position.
Vibration energy ∝ Temperature in solids.

Periodic and Oscillatory Motions

Definitions

Periodic motion: Repeats identically after fixed time interval (period T).
Example: Earth's revolution around Sun (T = 1 year).
Oscillatory motion: Back and forth motion about fixed mean position within limits.
Bounded periodic motion about fixed point is oscillatory motion.

Solved Examples

Practice Set

Q1: Which motion is bounded and repeats about a fixed point?
- A. Periodic motion
- B. Oscillatory motion
- C. Random motion
- D. Uniform motion
Q2: The period of Earth's revolution around Sun is?
- A. 24 hours
- B. 365 days
- C. 1 month
- D. 1 week
Q3: Which waveform is NOT periodic?
- A. Triangular wave
- B. Rectangular pulse
- C. Sawtooth wave
- D. Exponential function

Answer Key

Q1: B - Oscillatory motion is bounded and periodic about mean position.
Q2: B - Earth's revolution period is 365 days (1 year).
Q3: D - Exponential function is non-periodic.

Memory Aids

Periodic = repeats after T.
Oscillatory = bounded periodic motion.
Waveforms (triangular, rectangular, sawtooth) are periodic.

Definitions

Periodic motion: Repeats identically after fixed time interval (period T).
Example: Earth's revolution around Sun (T = 1 year).
Oscillatory motion: Back and forth motion about fixed mean position within limits.
Bounded periodic motion about fixed point is oscillatory motion.

Solved Examples

Practice Set

Q1: Which motion is bounded and repeats about a fixed point?
- A. Periodic motion
- B. Oscillatory motion
- C. Random motion
- D. Uniform motion
Q2: The period of Earth's revolution around Sun is?
- A. 24 hours
- B. 365 days
- C. 1 month
- D. 1 week
Q3: Which waveform is NOT periodic?
- A. Triangular wave
- B. Rectangular pulse
- C. Sawtooth wave
- D. Exponential function

Answer Key

Q1: B - Oscillatory motion is bounded and periodic about mean position.
Q2: B - Earth's revolution period is 365 days (1 year).
Q3: D - Exponential function is non-periodic.

Memory Aids

Periodic = repeats after T.
Oscillatory = bounded periodic motion.
Waveforms (triangular, rectangular, sawtooth) are periodic.

Definitions

Periodic motion: Repeats identically after fixed time interval (period T).
Example: Earth's revolution around Sun (T = 1 year).
Oscillatory motion: Back and forth motion about fixed mean position within limits.
Bounded periodic motion about fixed point is oscillatory motion.

Solved Examples

Practice Set

Q1: Which motion is bounded and repeats about a fixed point?
- A. Periodic motion
- B. Oscillatory motion
- C. Random motion
- D. Uniform motion
Q2: The period of Earth's revolution around Sun is?
- A. 24 hours
- B. 365 days
- C. 1 month
- D. 1 week
Q3: Which waveform is NOT periodic?
- A. Triangular wave
- B. Rectangular pulse
- C. Sawtooth wave
- D. Exponential function

Answer Key

Q1: B - Oscillatory motion is bounded and periodic about mean position.
Q2: B - Earth's revolution period is 365 days (1 year).
Q3: D - Exponential function is non-periodic.

Memory Aids

Periodic = repeats after T.
Oscillatory = bounded periodic motion.
Waveforms (triangular, rectangular, sawtooth) are periodic.

Definitions

Periodic motion: Repeats identically after fixed time interval (period T).
Example: Earth's revolution around Sun (T = 1 year).
Oscillatory motion: Back and forth motion about fixed mean position within limits.
Bounded periodic motion about fixed point is oscillatory motion.

Solved Examples

Practice Set

Q1: Which motion is bounded and repeats about a fixed point?
- A. Periodic motion
- B. Oscillatory motion
- C. Random motion
- D. Uniform motion
Q2: The period of Earth's revolution around Sun is?
- A. 24 hours
- B. 365 days
- C. 1 month
- D. 1 week
Q3: Which waveform is NOT periodic?
- A. Triangular wave
- B. Rectangular pulse
- C. Sawtooth wave
- D. Exponential function

Answer Key

Q1: B - Oscillatory motion is bounded and periodic about mean position.
Q2: B - Earth's revolution period is 365 days (1 year).
Q3: D - Exponential function is non-periodic.

Memory Aids

Periodic = repeats after T.
Oscillatory = bounded periodic motion.
Waveforms (triangular, rectangular, sawtooth) are periodic.

Time Period, Frequency and Angular Frequency

Key Concepts

Time period (T): Smallest time interval after which motion repeats; unit: second (s).
Frequency (ν): Number of oscillations per second; ν = 1/T; unit: hertz (Hz).
Angular frequency (ω): ω = 2πν = 2π/T; unit: rad/s.
Examples: Mercury's orbital period = 88 days; Halley's comet period = 76 years.
Heart beat frequency = 1.25 Hz; period = 0.8 s.

Solved Examples

Practice Set

Q1: Frequency of a motion with period 0.5 s is?
- A. 0.5 Hz
- B. 2 Hz
- C. 1 Hz
- D. 0.25 Hz
Q2: Angular frequency for frequency 5 Hz is?
- A. 10π rad/s
- B. 5π rad/s
- C. 2π rad/s
- D. 25 rad/s
Q3: If period of Halley's comet is 76 years, its frequency in Hz is approximately?
- A. 4.18 × 10^-10 Hz
- B. 3.15 × 10^-8 Hz
- C. 1.32 × 10^-7 Hz
- D. 2.63 × 10^-9 Hz

Answer Key

Q1: B - ν = 1/T = 1/0.5 = 2 Hz.
Q2: A - ω = 2πν = 2π × 5 = 10π rad/s.
Q3: A - Frequency = 1/(76 × 365 × 24 × 3600) ≈ 4.18 × 10^-10 Hz.

Memory Aids

ν = 1/T
ω = 2πν
Period in seconds for frequency calculation.

Key Concepts

Time period (T): Smallest time interval after which motion repeats; unit: second (s).
Frequency (ν): Number of oscillations per second; ν = 1/T; unit: hertz (Hz).
Angular frequency (ω): ω = 2πν = 2π/T; unit: rad/s.
Examples: Mercury's orbital period = 88 days; Halley's comet period = 76 years.
Heart beat frequency = 1.25 Hz; period = 0.8 s.

Solved Examples

Practice Set

Q1: Frequency of a motion with period 0.5 s is?
- A. 0.5 Hz
- B. 2 Hz
- C. 1 Hz
- D. 0.25 Hz
Q2: Angular frequency for frequency 5 Hz is?
- A. 10π rad/s
- B. 5π rad/s
- C. 2π rad/s
- D. 25 rad/s
Q3: If period of Halley's comet is 76 years, its frequency in Hz is approximately?
- A. 4.18 × 10^-10 Hz
- B. 3.15 × 10^-8 Hz
- C. 1.32 × 10^-7 Hz
- D. 2.63 × 10^-9 Hz

Answer Key

Q1: B - ν = 1/T = 1/0.5 = 2 Hz.
Q2: A - ω = 2πν = 2π × 5 = 10π rad/s.
Q3: A - Frequency = 1/(76 × 365 × 24 × 3600) ≈ 4.18 × 10^-10 Hz.

Memory Aids

ν = 1/T
ω = 2πν
Period in seconds for frequency calculation.

Key Concepts

Time period (T): Smallest time interval after which motion repeats; unit: second (s).
Frequency (ν): Number of oscillations per second; ν = 1/T; unit: hertz (Hz).
Angular frequency (ω): ω = 2πν = 2π/T; unit: rad/s.
Examples: Mercury's orbital period = 88 days; Halley's comet period = 76 years.
Heart beat frequency = 1.25 Hz; period = 0.8 s.

Solved Examples

Practice Set

Q1: Frequency of a motion with period 0.5 s is?
- A. 0.5 Hz
- B. 2 Hz
- C. 1 Hz
- D. 0.25 Hz
Q2: Angular frequency for frequency 5 Hz is?
- A. 10π rad/s
- B. 5π rad/s
- C. 2π rad/s
- D. 25 rad/s
Q3: If period of Halley's comet is 76 years, its frequency in Hz is approximately?
- A. 4.18 × 10^-10 Hz
- B. 3.15 × 10^-8 Hz
- C. 1.32 × 10^-7 Hz
- D. 2.63 × 10^-9 Hz

Answer Key

Q1: B - ν = 1/T = 1/0.5 = 2 Hz.
Q2: A - ω = 2πν = 2π × 5 = 10π rad/s.
Q3: A - Frequency = 1/(76 × 365 × 24 × 3600) ≈ 4.18 × 10^-10 Hz.

Memory Aids

ν = 1/T
ω = 2πν
Period in seconds for frequency calculation.

Key Concepts

Time period (T): Smallest time interval after which motion repeats; unit: second (s).
Frequency (ν): Number of oscillations per second; ν = 1/T; unit: hertz (Hz).
Angular frequency (ω): ω = 2πν = 2π/T; unit: rad/s.
Examples: Mercury's orbital period = 88 days; Halley's comet period = 76 years.
Heart beat frequency = 1.25 Hz; period = 0.8 s.

Solved Examples

Practice Set

Q1: Frequency of a motion with period 0.5 s is?
- A. 0.5 Hz
- B. 2 Hz
- C. 1 Hz
- D. 0.25 Hz
Q2: Angular frequency for frequency 5 Hz is?
- A. 10π rad/s
- B. 5π rad/s
- C. 2π rad/s
- D. 25 rad/s
Q3: If period of Halley's comet is 76 years, its frequency in Hz is approximately?
- A. 4.18 × 10^-10 Hz
- B. 3.15 × 10^-8 Hz
- C. 1.32 × 10^-7 Hz
- D. 2.63 × 10^-9 Hz

Answer Key

Q1: B - ν = 1/T = 1/0.5 = 2 Hz.
Q2: A - ω = 2πν = 2π × 5 = 10π rad/s.
Q3: A - Frequency = 1/(76 × 365 × 24 × 3600) ≈ 4.18 × 10^-10 Hz.

Memory Aids

ν = 1/T
ω = 2πν
Period in seconds for frequency calculation.

Key Concepts

Time period (T): Smallest time interval after which motion repeats; unit: second (s).
Frequency (ν): Number of oscillations per second; ν = 1/T; unit: hertz (Hz).
Angular frequency (ω): ω = 2πν = 2π/T; unit: rad/s.
Examples: Mercury's orbital period = 88 days; Halley's comet period = 76 years.
Heart beat frequency = 1.25 Hz; period = 0.8 s.

Solved Examples

Practice Set

Q1: Frequency of a motion with period 0.5 s is?
- A. 0.5 Hz
- B. 2 Hz
- C. 1 Hz
- D. 0.25 Hz
Q2: Angular frequency for frequency 5 Hz is?
- A. 10π rad/s
- B. 5π rad/s
- C. 2π rad/s
- D. 25 rad/s
Q3: If period of Halley's comet is 76 years, its frequency in Hz is approximately?
- A. 4.18 × 10^-10 Hz
- B. 3.15 × 10^-8 Hz
- C. 1.32 × 10^-7 Hz
- D. 2.63 × 10^-9 Hz

Answer Key

Q1: B - ν = 1/T = 1/0.5 = 2 Hz.
Q2: A - ω = 2πν = 2π × 5 = 10π rad/s.
Q3: A - Frequency = 1/(76 × 365 × 24 × 3600) ≈ 4.18 × 10^-10 Hz.

Memory Aids

ν = 1/T
ω = 2πν
Period in seconds for frequency calculation.

Key Concepts

Time period (T): Smallest time interval after which motion repeats; unit: second (s).
Frequency (ν): Number of oscillations per second; ν = 1/T; unit: hertz (Hz).
Angular frequency (ω): ω = 2πν = 2π/T; unit: rad/s.
Examples: Mercury's orbital period = 88 days; Halley's comet period = 76 years.
Heart beat frequency = 1.25 Hz; period = 0.8 s.

Solved Examples

Practice Set

Q1: Frequency of a motion with period 0.5 s is?
- A. 0.5 Hz
- B. 2 Hz
- C. 1 Hz
- D. 0.25 Hz
Q2: Angular frequency for frequency 5 Hz is?
- A. 10π rad/s
- B. 5π rad/s
- C. 2π rad/s
- D. 25 rad/s
Q3: If period of Halley's comet is 76 years, its frequency in Hz is approximately?
- A. 4.18 × 10^-10 Hz
- B. 3.15 × 10^-8 Hz
- C. 1.32 × 10^-7 Hz
- D. 2.63 × 10^-9 Hz

Answer Key

Q1: B - ν = 1/T = 1/0.5 = 2 Hz.
Q2: A - ω = 2πν = 2π × 5 = 10π rad/s.
Q3: A - Frequency = 1/(76 × 365 × 24 × 3600) ≈ 4.18 × 10^-10 Hz.

Memory Aids

ν = 1/T
ω = 2πν
Period in seconds for frequency calculation.

Displacement in Oscillatory Motion

Concepts

Displacement represents change in physical quantities (position, angle, pressure, fields) with time.
Example: Angle θ of pendulum from vertical is displacement variable.
Displacement can be positive or negative.
Mathematical form: x = A cos ωt or y = A sin ωt, where A = amplitude.

Solved Examples

Practice Set

Q1: Which function represents displacement in oscillatory motion?
- A. x = A cos ωt
- B. x = vt
- C. x = at²
- D. x = constant
Q2: Amplitude in displacement function represents?
- A. Maximum displacement
- B. Time period
- C. Frequency
- D. Mean position
Q3: Displacement x = A sin ωt means at t = 0, x is?
- A. 0
- B. A
- C. -A
- D. Undefined

Answer Key

Q1: A - Oscillatory displacement is sinusoidal, e.g., x = A cos ωt.
Q2: A - Amplitude is maximum displacement from mean position.
Q3: A - sin 0 = 0, so displacement is zero at t=0.

Memory Aids

Displacement = A sin/cos ωt
Amplitude = max displacement
Angle θ in pendulum is displacement

Concepts

Displacement represents change in physical quantities (position, angle, pressure, fields) with time.
Example: Angle θ of pendulum from vertical is displacement variable.
Displacement can be positive or negative.
Mathematical form: x = A cos ωt or y = A sin ωt, where A = amplitude.

Solved Examples

Practice Set

Q1: Which function represents displacement in oscillatory motion?
- A. x = A cos ωt
- B. x = vt
- C. x = at²
- D. x = constant
Q2: Amplitude in displacement function represents?
- A. Maximum displacement
- B. Time period
- C. Frequency
- D. Mean position
Q3: Displacement x = A sin ωt means at t = 0, x is?
- A. 0
- B. A
- C. -A
- D. Undefined

Answer Key

Q1: A - Oscillatory displacement is sinusoidal, e.g., x = A cos ωt.
Q2: A - Amplitude is maximum displacement from mean position.
Q3: A - sin 0 = 0, so displacement is zero at t=0.

Memory Aids

Displacement = A sin/cos ωt
Amplitude = max displacement
Angle θ in pendulum is displacement

Concepts

Displacement represents change in physical quantities (position, angle, pressure, fields) with time.
Example: Angle θ of pendulum from vertical is displacement variable.
Displacement can be positive or negative.
Mathematical form: x = A cos ωt or y = A sin ωt, where A = amplitude.

Solved Examples

Practice Set

Q1: Which function represents displacement in oscillatory motion?
- A. x = A cos ωt
- B. x = vt
- C. x = at²
- D. x = constant
Q2: Amplitude in displacement function represents?
- A. Maximum displacement
- B. Time period
- C. Frequency
- D. Mean position
Q3: Displacement x = A sin ωt means at t = 0, x is?
- A. 0
- B. A
- C. -A
- D. Undefined

Answer Key

Q1: A - Oscillatory displacement is sinusoidal, e.g., x = A cos ωt.
Q2: A - Amplitude is maximum displacement from mean position.
Q3: A - sin 0 = 0, so displacement is zero at t=0.

Memory Aids

Displacement = A sin/cos ωt
Amplitude = max displacement
Angle θ in pendulum is displacement

Concepts

Displacement represents change in physical quantities (position, angle, pressure, fields) with time.
Example: Angle θ of pendulum from vertical is displacement variable.
Displacement can be positive or negative.
Mathematical form: x = A cos ωt or y = A sin ωt, where A = amplitude.

Solved Examples

Practice Set

Q1: Which function represents displacement in oscillatory motion?
- A. x = A cos ωt
- B. x = vt
- C. x = at²
- D. x = constant
Q2: Amplitude in displacement function represents?
- A. Maximum displacement
- B. Time period
- C. Frequency
- D. Mean position
Q3: Displacement x = A sin ωt means at t = 0, x is?
- A. 0
- B. A
- C. -A
- D. Undefined

Answer Key

Q1: A - Oscillatory displacement is sinusoidal, e.g., x = A cos ωt.
Q2: A - Amplitude is maximum displacement from mean position.
Q3: A - sin 0 = 0, so displacement is zero at t=0.

Memory Aids

Displacement = A sin/cos ωt
Amplitude = max displacement
Angle θ in pendulum is displacement

Periodic Functions

Key Points

Periodic function f(t) satisfies f(t) = f(t + T), T = period.
Any periodic function can be expressed as sum of sine and cosine functions.
Examples of periodic functions: sin ωt, cos ωt, (sin ωt + cos ωt) with period 2π/ω.
Non-periodic functions: e^ωt, log(ωt).

Solved Examples

Practice Set

Q1: Which function is periodic?
- A. sin ωt
- B. e^ωt
- C. log(ωt)
- D. ωt
Q2: Period of cos ωt is?
- A. 2π/ω
- B. π/ω
- C. ω/2π
- D. ω
Q3: Which is NOT a periodic function?
- A. sin ωt + cos ωt
- B. cos ωt
- C. e^ωt
- D. sin ωt

Answer Key

Q1: A - sin ωt is periodic with period 2π/ω.
Q2: A - Period of cos ωt is 2π/ω.
Q3: C - e^ωt is non-periodic.

Memory Aids

Periodic if f(t) = f(t + T)
Sine and cosine are basic periodic functions
Exponential and logarithmic functions are non-periodic

Key Points

Periodic function f(t) satisfies f(t) = f(t + T), T = period.
Any periodic function can be expressed as sum of sine and cosine functions.
Examples of periodic functions: sin ωt, cos ωt, (sin ωt + cos ωt) with period 2π/ω.
Non-periodic functions: e^ωt, log(ωt).

Solved Examples

Practice Set

Q1: Which function is periodic?
- A. sin ωt
- B. e^ωt
- C. log(ωt)
- D. ωt
Q2: Period of cos ωt is?
- A. 2π/ω
- B. π/ω
- C. ω/2π
- D. ω
Q3: Which is NOT a periodic function?
- A. sin ωt + cos ωt
- B. cos ωt
- C. e^ωt
- D. sin ωt

Answer Key

Q1: A - sin ωt is periodic with period 2π/ω.
Q2: A - Period of cos ωt is 2π/ω.
Q3: C - e^ωt is non-periodic.

Memory Aids

Periodic if f(t) = f(t + T)
Sine and cosine are basic periodic functions
Exponential and logarithmic functions are non-periodic

Key Points

Periodic function f(t) satisfies f(t) = f(t + T), T = period.
Any periodic function can be expressed as sum of sine and cosine functions.
Examples of periodic functions: sin ωt, cos ωt, (sin ωt + cos ωt) with period 2π/ω.
Non-periodic functions: e^ωt, log(ωt).

Solved Examples

Practice Set

Q1: Which function is periodic?
- A. sin ωt
- B. e^ωt
- C. log(ωt)
- D. ωt
Q2: Period of cos ωt is?
- A. 2π/ω
- B. π/ω
- C. ω/2π
- D. ω
Q3: Which is NOT a periodic function?
- A. sin ωt + cos ωt
- B. cos ωt
- C. e^ωt
- D. sin ωt

Answer Key

Q1: A - sin ωt is periodic with period 2π/ω.
Q2: A - Period of cos ωt is 2π/ω.
Q3: C - e^ωt is non-periodic.

Memory Aids

Periodic if f(t) = f(t + T)
Sine and cosine are basic periodic functions
Exponential and logarithmic functions are non-periodic

Key Points

Periodic function f(t) satisfies f(t) = f(t + T), T = period.
Any periodic function can be expressed as sum of sine and cosine functions.
Examples of periodic functions: sin ωt, cos ωt, (sin ωt + cos ωt) with period 2π/ω.
Non-periodic functions: e^ωt, log(ωt).

Solved Examples

Practice Set

Q1: Which function is periodic?
- A. sin ωt
- B. e^ωt
- C. log(ωt)
- D. ωt
Q2: Period of cos ωt is?
- A. 2π/ω
- B. π/ω
- C. ω/2π
- D. ω
Q3: Which is NOT a periodic function?
- A. sin ωt + cos ωt
- B. cos ωt
- C. e^ωt
- D. sin ωt

Answer Key

Q1: A - sin ωt is periodic with period 2π/ω.
Q2: A - Period of cos ωt is 2π/ω.
Q3: C - e^ωt is non-periodic.

Memory Aids

Periodic if f(t) = f(t + T)
Sine and cosine are basic periodic functions
Exponential and logarithmic functions are non-periodic

Key Points

Periodic function f(t) satisfies f(t) = f(t + T), T = period.
Any periodic function can be expressed as sum of sine and cosine functions.
Examples of periodic functions: sin ωt, cos ωt, (sin ωt + cos ωt) with period 2π/ω.
Non-periodic functions: e^ωt, log(ωt).

Solved Examples

Practice Set

Q1: Which function is periodic?
- A. sin ωt
- B. e^ωt
- C. log(ωt)
- D. ωt
Q2: Period of cos ωt is?
- A. 2π/ω
- B. π/ω
- C. ω/2π
- D. ω
Q3: Which is NOT a periodic function?
- A. sin ωt + cos ωt
- B. cos ωt
- C. e^ωt
- D. sin ωt

Answer Key

Q1: A - sin ωt is periodic with period 2π/ω.
Q2: A - Period of cos ωt is 2π/ω.
Q3: C - e^ωt is non-periodic.

Memory Aids

Periodic if f(t) = f(t + T)
Sine and cosine are basic periodic functions
Exponential and logarithmic functions are non-periodic

Key Points

Periodic function f(t) satisfies f(t) = f(t + T), T = period.
Any periodic function can be expressed as sum of sine and cosine functions.
Examples of periodic functions: sin ωt, cos ωt, (sin ωt + cos ωt) with period 2π/ω.
Non-periodic functions: e^{ωt}, log(ωt).

Solved Examples

Practice Set

Q1: Which function is periodic?
- A. sin ωt
- B. e^{ωt}
- C. log(ωt)
- D. ωt
Q2: Period of cos ωt is?
- A. 2π/ω
- B. π/ω
- C. ω/2π
- D. ω
Q3: Which is NOT a periodic function?
- A. sin ωt + cos ωt
- B. cos ωt
- C. e^{ωt}
- D. sin ωt

Answer Key

Q1: A - sin ωt is periodic with period 2π/ω.
Q2: A - Period of cos ωt is 2π/ω.
Q3: C - e^{ωt} is non-periodic.

Memory Aids

Periodic if f(t) = f(t + T)
Sine and cosine are basic periodic functions
Exponential and logarithmic functions are non-periodic

Simple Harmonic Motion

Definition and Conditions

Simple harmonic motion (SHM) is a periodic to-and-fro motion about a mean position under a restoring force.
Acceleration is directed towards a fixed point and proportional to displacement: a ∝ -x.
Mathematically, a = -kx or d²x/dt² = -kx, where x is displacement at time t.
Particle vibrates between limits +A and -A on the x-axis.
Displacement equation: x(t) = A cos(ωt + Φ).
Parameters:
- A: amplitude (maximum displacement)
- ω: angular frequency
- Φ: phase constant
- ωt + Φ: phase (time-dependent)
Angular frequency relation: ω = 2π/T, where T is the period.

Solved Examples

Ex 1: Find acceleration at displacement x = A/2 if a = -kx and k = 4 s^-2.
- Use a = -kx = -4 × A/2 = -2A.
Ex 2: Calculate angular frequency if period T = 0.5 s.
- ω = 2π/T = 2π/0.5 = 4π rad/s.

Practice Set

Q1: What is the acceleration of a particle in SHM at maximum displacement A?
- A) Zero
- B) Maximum towards mean position
- C) Maximum away from mean position
- D) Constant
Q2: The displacement of a particle in SHM is given by x = 5 cos(10t). What is the angular frequency?
- A) 5 rad/s
- B) 10 rad/s
- C) 50 rad/s
- D) 0.1 rad/s
Q3: If the period of SHM is doubled, what happens to the angular frequency?
- A) Doubles
- B) Halves
- C) Quadruples
- D) Remains same

Answer Key

Q1: B - Acceleration is maximum and directed towards mean position at maximum displacement.
Q2: B - Angular frequency ω = 10 rad/s from given equation.
Q3: B - ω = 2π/T, so doubling T halves ω.

Memory Aids

Acceleration formula: a = -kx (restoring force direction).
Displacement: x = A cos(ωt + Φ).
Angular frequency: ω = 2π/T.
Speed max at mean position, zero at extremes.

Definition and Conditions

Simple harmonic motion (SHM) is a periodic to-and-fro motion about a mean position under a restoring force.
Acceleration is directed towards a fixed point and proportional to displacement: a ∝ -x.
Mathematically, a = -kx or d²x/dt² = -kx, where x is displacement at time t.
Particle vibrates between limits +A and -A on the x-axis.
Displacement equation: x(t) = A cos(ωt + Φ).
Parameters:
- A: amplitude (maximum displacement)
- ω: angular frequency
- Φ: phase constant
- ωt + Φ: phase (time-dependent)
Angular frequency relation: ω = 2π/T, where T is the period.

Solved Examples

Ex 1: Find acceleration at displacement x = A/2 if a = -kx and k = 4 s^-2.
- Use a = -kx = -4 × A/2 = -2A.
Ex 2: Calculate angular frequency if period T = 0.5 s.
- ω = 2π/T = 2π/0.5 = 4π rad/s.

Practice Set

Q1: What is the acceleration of a particle in SHM at maximum displacement A?
- A) Zero
- B) Maximum towards mean position
- C) Maximum away from mean position
- D) Constant
Q2: The displacement of a particle in SHM is given by x = 5 cos(10t). What is the angular frequency?
- A) 5 rad/s
- B) 10 rad/s
- C) 50 rad/s
- D) 0.1 rad/s
Q3: If the period of SHM is doubled, what happens to the angular frequency?
- A) Doubles
- B) Halves
- C) Quadruples
- D) Remains same

Answer Key

Q1: B - Acceleration is maximum and directed towards mean position at maximum displacement.
Q2: B - Angular frequency ω = 10 rad/s from given equation.
Q3: B - ω = 2π/T, so doubling T halves ω.

Memory Aids

Acceleration formula: a = -kx (restoring force direction).
Displacement: x = A cos(ωt + Φ).
Angular frequency: ω = 2π/T.
Speed max at mean position, zero at extremes.

Definition and Conditions

Simple harmonic motion (SHM) is a periodic to-and-fro motion about a mean position under a restoring force.
Acceleration is directed towards a fixed point and proportional to displacement: a ∝ -x.
Mathematically, a = -kx or d²x/dt² = -kx, where x is displacement at time t.
Particle vibrates between limits +A and -A on the x-axis.
Displacement equation: x(t) = A cos(ωt + Φ).
Parameters:
- A: amplitude (maximum displacement)
- ω: angular frequency
- Φ: phase constant
- ωt + Φ: phase (time-dependent)
Angular frequency relation: ω = 2π/T, where T is the period.

Solved Examples

Ex 1: Find acceleration at displacement x = A/2 if a = -kx and k = 4 s^-2.
- Use a = -kx = -4 × A/2 = -2A.
Ex 2: Calculate angular frequency if period T = 0.5 s.
- ω = 2π/T = 2π/0.5 = 4π rad/s.

Practice Set

Q1: What is the acceleration of a particle in SHM at maximum displacement A?
- A) Zero
- B) Maximum towards mean position
- C) Maximum away from mean position
- D) Constant
Q2: The displacement of a particle in SHM is given by x = 5 cos(10t). What is the angular frequency?
- A) 5 rad/s
- B) 10 rad/s
- C) 50 rad/s
- D) 0.1 rad/s
Q3: If the period of SHM is doubled, what happens to the angular frequency?
- A) Doubles
- B) Halves
- C) Quadruples
- D) Remains same

Answer Key

Q1: B - Acceleration is maximum and directed towards mean position at maximum displacement.
Q2: B - Angular frequency ω = 10 rad/s from given equation.
Q3: B - ω = 2π/T, so doubling T halves ω.

Memory Aids

Acceleration formula: a = -kx (restoring force direction).
Displacement: x = A cos(ωt + Φ).
Angular frequency: ω = 2π/T.
Speed max at mean position, zero at extremes.

Definition and Conditions

Simple harmonic motion (SHM) is a periodic to-and-fro motion about a mean position under a restoring force.
Acceleration is directed towards a fixed point and proportional to displacement: a ∝ -x.
Mathematically, a = -kx or d²x/dt² = -kx, where x is displacement at time t.
Particle vibrates between limits +A and -A on the x-axis.
Displacement equation: x(t) = A cos(ωt + Φ).
Parameters:
- A: amplitude (maximum displacement)
- ω: angular frequency
- Φ: phase constant
- ωt + Φ: phase (time-dependent)
Angular frequency relation: ω = 2π/T, where T is the period.

Solved Examples

Ex 1: Find acceleration at displacement x = A/2 if a = -kx and k = 4 s^-2.
- Use a = -kx = -4 × A/2 = -2A.
Ex 2: Calculate angular frequency if period T = 0.5 s.
- ω = 2π/T = 2π/0.5 = 4π rad/s.

Practice Set

Q1: What is the acceleration of a particle in SHM at maximum displacement A?
- A) Zero
- B) Maximum towards mean position
- C) Maximum away from mean position
- D) Constant
Q2: The displacement of a particle in SHM is given by x = 5 cos(10t). What is the angular frequency?
- A) 5 rad/s
- B) 10 rad/s
- C) 50 rad/s
- D) 0.1 rad/s
Q3: If the period of SHM is doubled, what happens to the angular frequency?
- A) Doubles
- B) Halves
- C) Quadruples
- D) Remains same

Answer Key

Q1: B - Acceleration is maximum and directed towards mean position at maximum displacement.
Q2: B - Angular frequency ω = 10 rad/s from given equation.
Q3: B - ω = 2π/T, so doubling T halves ω.

Memory Aids

Acceleration formula: a = -kx (restoring force direction).
Displacement: x = A cos(ωt + Φ).
Angular frequency: ω = 2π/T.
Speed max at mean position, zero at extremes.

Definition and Conditions

Simple harmonic motion (SHM) is a periodic to-and-fro motion about a mean position under a restoring force.
Acceleration is directed towards a fixed point and proportional to displacement: a ∝ -x.
Mathematically, a = -kx or d²x/dt² = -kx, where x is displacement at time t.
Particle vibrates between limits +A and -A on the x-axis.
Displacement equation: x(t) = A cos(ωt + Φ).
Parameters:
- A: amplitude (maximum displacement)
- ω: angular frequency
- Φ: phase constant
- ωt + Φ: phase (time-dependent)
Angular frequency relation: ω = 2π/T, where T is the period.

Solved Examples

Ex 1: Find acceleration at displacement x = A/2 if a = -kx and k = 4 s^-2.
- Use a = -kx = -4 × A/2 = -2A.
Ex 2: Calculate angular frequency if period T = 0.5 s.
- ω = 2π/T = 2π/0.5 = 4π rad/s.

Practice Set

Q1: What is the acceleration of a particle in SHM at maximum displacement A?
- A) Zero
- B) Maximum towards mean position
- C) Maximum away from mean position
- D) Constant
Q2: The displacement of a particle in SHM is given by x = 5 cos(10t). What is the angular frequency?
- A) 5 rad/s
- B) 10 rad/s
- C) 50 rad/s
- D) 0.1 rad/s
Q3: If the period of SHM is doubled, what happens to the angular frequency?
- A) Doubles
- B) Halves
- C) Quadruples
- D) Remains same

Answer Key

Q1: B - Acceleration is maximum and directed towards mean position at maximum displacement.
Q2: B - Angular frequency ω = 10 rad/s from given equation.
Q3: B - ω = 2π/T, so doubling T halves ω.

Memory Aids

Acceleration formula: a = -kx (restoring force direction).
Displacement: x = A cos(ωt + Φ).
Angular frequency: ω = 2π/T.
Speed max at mean position, zero at extremes.

Definition and Conditions

Simple harmonic motion (SHM) is a periodic to-and-fro motion about a mean position under a restoring force.
Acceleration is directed towards a fixed point and proportional to displacement: \( a \propto -x \).
Mathematically, \( a = -kx \) or \( \frac{d^2x}{dt^2} = -kx \), where \( x \) is displacement at time \( t \).
Particle vibrates between limits \( +A \) and \( -A \) on the x-axis.
Displacement equation: \( x(t) = A \cos(\omega t + \Phi) \).
Parameters:
- \( A \): amplitude (maximum displacement)
- \( \omega \): angular frequency
- \( \Phi \): phase constant
- \( \omega t + \Phi \): phase (time-dependent)
Angular frequency relation: \( \omega = \frac{2\pi}{T} \), where \( T \) is the period.

Solved Examples

Ex 1: Find acceleration at displacement \( x = \frac{A}{2} \) if \( a = -kx \) and \( k = 4 \, s^{-2} \).
- Use \( a = -kx = -4 \times \frac{A}{2} = -2A \).
Ex 2: Calculate angular frequency if period \( T = 0.5 \) s.
- \( \omega = \frac{2\pi}{T} = \frac{2\pi}{0.5} = 4\pi \, rad/s \).

Practice Set

Q1: What is the acceleration of a particle in SHM at maximum displacement \( A \)?
- A) Zero
- B) Maximum towards mean position
- C) Maximum away from mean position
- D) Constant
Q2: The displacement of a particle in SHM is given by \( x = 5 \cos(10t) \). What is the angular frequency?
- A) 5 rad/s
- B) 10 rad/s
- C) 50 rad/s
- D) 0.1 rad/s
Q3: If the period of SHM is doubled, what happens to the angular frequency?
- A) Doubles
- B) Halves
- C) Quadruples
- D) Remains same

Answer Key

Q1: B - Acceleration is maximum and directed towards mean position at maximum displacement.
Q2: B - Angular frequency \( \omega = 10 \) rad/s from given equation.
Q3: B - \( \omega = \frac{2\pi}{T} \), so doubling \( T \) halves \( \omega \).

Memory Aids

Acceleration formula: \( a = -kx \) (restoring force direction).
Displacement: \( x = A \cos(\omega t + \Phi) \).
Angular frequency: \( \omega = \frac{2\pi}{T} \).
Speed max at mean position, zero at extremes.

SHM Position and Velocity Graphs

Graphical Representation

Particle moves between -A and +A with displacement x(t) = A cos(ωt + Φ).
At t = 0, particle at mean position O, velocity zero.
At t = T/4, velocity is maximum v_max, displacement zero.
At t = T/2, particle at extreme +A, velocity zero.
Velocity and displacement vary sinusoidally with time.
Graphs show displacement vs time and velocity vs time for one full cycle.

Solved Examples

Ex 1: Find displacement at t = T/4 if x = A cos(ωt).
- x = A cos(π/2) = 0.
Ex 2: Velocity is maximum at mean position. If amplitude A = 0.1 m and ω = 20 rad/s, find v_max.
- v_max = ωA = 20 × 0.1 = 2 m/s.

Practice Set

Q1: At which point is velocity zero in SHM?
- A) Mean position
- B) Maximum displacement
- C) Midway between mean and max displacement
- D) Always non-zero
Q2: If A = 0.2 m and ω = 15 rad/s, what is maximum velocity?
- A) 3 m/s
- B) 0.03 m/s
- C) 15 m/s
- D) 0.2 m/s
Q3: Displacement at t = T/2 is?
- A) 0
- B) +A
- C) -A
- D) A/2

Answer Key

Q1: B - Velocity is zero at maximum displacement (extremes).
Q2: A - v_max = ωA = 15 × 0.2 = 3 m/s.
Q3: B - At t = T/2, displacement is maximum positive +A.

Memory Aids

Velocity max at mean position, zero at extremes.
Displacement zero at velocity max points.
Use v_max = ωA for max speed.

Graphical Representation

Particle moves between -A and +A with displacement x(t) = A cos(ωt + Φ).
At t = 0, particle at mean position O, velocity zero.
At t = T/4, velocity is maximum v_max, displacement zero.
At t = T/2, particle at extreme +A, velocity zero.
Velocity and displacement vary sinusoidally with time.
Graphs show displacement vs time and velocity vs time for one full cycle.

Solved Examples

Ex 1: Find displacement at t = T/4 if x = A cos(ωt).
- x = A cos(π/2) = 0.
Ex 2: Velocity is maximum at mean position. If amplitude A = 0.1 m and ω = 20 rad/s, find v_max.
- v_max = ωA = 20 × 0.1 = 2 m/s.

Practice Set

Q1: At which point is velocity zero in SHM?
- A) Mean position
- B) Maximum displacement
- C) Midway between mean and max displacement
- D) Always non-zero
Q2: If A = 0.2 m and ω = 15 rad/s, what is maximum velocity?
- A) 3 m/s
- B) 0.03 m/s
- C) 15 m/s
- D) 0.2 m/s
Q3: Displacement at t = T/2 is?
- A) 0
- B) +A
- C) -A
- D) A/2

Answer Key

Q1: B - Velocity is zero at maximum displacement (extremes).
Q2: A - v_max = ωA = 15 × 0.2 = 3 m/s.
Q3: B - At t = T/2, displacement is maximum positive +A.

Memory Aids

Velocity max at mean position, zero at extremes.
Displacement zero at velocity max points.
Use v_max = ωA for max speed.

Graphical Representation

Particle moves between -A and +A with displacement x(t) = A cos(ωt + Φ).
At t = 0, particle at mean position O, velocity zero.
At t = T/4, velocity is maximum v_max, displacement zero.
At t = T/2, particle at extreme +A, velocity zero.
Velocity and displacement vary sinusoidally with time.
Graphs show displacement vs time and velocity vs time for one full cycle.

Solved Examples

Ex 1: Find displacement at t = T/4 if x = A cos(ωt).
- x = A cos(π/2) = 0.
Ex 2: Velocity is maximum at mean position. If amplitude A = 0.1 m and ω = 20 rad/s, find v_max.
- v_max = ωA = 20 × 0.1 = 2 m/s.

Practice Set

Q1: At which point is velocity zero in SHM?
- A) Mean position
- B) Maximum displacement
- C) Midway between mean and max displacement
- D) Always non-zero
Q2: If A = 0.2 m and ω = 15 rad/s, what is maximum velocity?
- A) 3 m/s
- B) 0.03 m/s
- C) 15 m/s
- D) 0.2 m/s
Q3: Displacement at t = T/2 is?
- A) 0
- B) +A
- C) -A
- D) A/2

Answer Key

Q1: B - Velocity is zero at maximum displacement (extremes).
Q2: A - v_max = ωA = 15 × 0.2 = 3 m/s.
Q3: B - At t = T/2, displacement is maximum positive +A.

Memory Aids

Velocity max at mean position, zero at extremes.
Displacement zero at velocity max points.
Use v_max = ωA for max speed.

Graphical Representation

Particle moves between -A and +A with displacement x(t) = A cos(ωt + Φ).
At t = 0, particle at mean position O, velocity zero.
At t = T/4, velocity is maximum v_max, displacement zero.
At t = T/2, particle at extreme +A, velocity zero.
Velocity and displacement vary sinusoidally with time.
Graphs show displacement vs time and velocity vs time for one full cycle.

Solved Examples

Ex 1: Find displacement at t = T/4 if x = A cos(ωt).
- x = A cos(π/2) = 0.
Ex 2: Velocity is maximum at mean position. If amplitude A = 0.1 m and ω = 20 rad/s, find v_max.
- v_max = ωA = 20 × 0.1 = 2 m/s.

Practice Set

Q1: At which point is velocity zero in SHM?
- A) Mean position
- B) Maximum displacement
- C) Midway between mean and max displacement
- D) Always non-zero
Q2: If A = 0.2 m and ω = 15 rad/s, what is maximum velocity?
- A) 3 m/s
- B) 0.03 m/s
- C) 15 m/s
- D) 0.2 m/s
Q3: Displacement at t = T/2 is?
- A) 0
- B) +A
- C) -A
- D) A/2

Answer Key

Q1: B - Velocity is zero at maximum displacement (extremes).
Q2: A - v_max = ωA = 15 × 0.2 = 3 m/s.
Q3: B - At t = T/2, displacement is maximum positive +A.

Memory Aids

Velocity max at mean position, zero at extremes.
Displacement zero at velocity max points.
Use v_max = ωA for max speed.

Graphical Representation

Particle moves between -A and +A with displacement x(t) = A cos(ωt + Φ).
At t = 0, particle at mean position O, velocity zero.
At t = T/4, velocity is maximum v_max, displacement zero.
At t = T/2, particle at extreme +A, velocity zero.
Velocity and displacement vary sinusoidally with time.
Graphs show displacement vs time and velocity vs time for one full cycle.

Solved Examples

Ex 1: Find displacement at t = T/4 if x = A cos(ωt).
- x = A cos(π/2) = 0.
Ex 2: Velocity is maximum at mean position. If amplitude A = 0.1 m and ω = 20 rad/s, find v_max.
- v_max = ωA = 20 × 0.1 = 2 m/s.

Practice Set

Q1: At which point is velocity zero in SHM?
- A) Mean position
- B) Maximum displacement
- C) Midway between mean and max displacement
- D) Always non-zero
Q2: If A = 0.2 m and ω = 15 rad/s, what is maximum velocity?
- A) 3 m/s
- B) 0.03 m/s
- C) 15 m/s
- D) 0.2 m/s
Q3: Displacement at t = T/2 is?
- A) 0
- B) +A
- C) -A
- D) A/2

Answer Key

Q1: B - Velocity is zero at maximum displacement (extremes).
Q2: A - v_max = ωA = 15 × 0.2 = 3 m/s.
Q3: B - At t = T/2, displacement is maximum positive +A.

Memory Aids

Velocity max at mean position, zero at extremes.
Displacement zero at velocity max points.
Use v_max = ωA for max speed.

Graphical Representation

Particle moves between \( -A \) and \( +A \) with displacement \( x(t) = A \cos(\omega t + \Phi) \).
At \( t = 0 \), particle at mean position \( O \), velocity zero.
At \( t = \frac{T}{4} \), velocity is maximum \( v_{max} \), displacement zero.
At \( t = \frac{T}{2} \), particle at extreme \( +A \), velocity zero.
Velocity and displacement vary sinusoidally with time.
Graphs show displacement vs time and velocity vs time for one full cycle.

Solved Examples

Ex 1: Find displacement at \( t = \frac{T}{4} \) if \( x = A \cos(\omega t) \).
- \( x = A \cos(\frac{\pi}{2}) = 0 \).
Ex 2: Velocity is maximum at mean position. If amplitude \( A = 0.1 \) m and \( \omega = 20 \) rad/s, find \( v_{max} \).
- \( v_{max} = \omega A = 20 \times 0.1 = 2 \, m/s \).

Practice Set

Q1: At which point is velocity zero in SHM?
- A) Mean position
- B) Maximum displacement
- C) Midway between mean and max displacement
- D) Always non-zero
Q2: If \( A = 0.2 \) m and \( \omega = 15 \) rad/s, what is maximum velocity?
- A) 3 m/s
- B) 0.03 m/s
- C) 15 m/s
- D) 0.2 m/s
Q3: Displacement at \( t = \frac{T}{2} \) is?
- A) 0
- B) +A
- C) -A
- D) \( \frac{A}{2} \)

Answer Key

Q1: B - Velocity is zero at maximum displacement (extremes).
Q2: A - \( v_{max} = \omega A = 15 \times 0.2 = 3 \, m/s \).
Q3: B - At \( t = \frac{T}{2} \), displacement is maximum positive \( +A \).

Memory Aids

Velocity max at mean position, zero at extremes.
Displacement zero at velocity max points.
Use \( v_{max} = \omega A \) for max speed.

SHM Wave Graphs and Phase

Wave Plots and Phase Effects

Displacement x(t) = A cos(ωt + Φ) plotted vs time shows sinusoidal waves.
Amplitude changes affect wave height but not frequency.
Phase constant Φ shifts wave horizontally (time shift).
Different frequencies change wave period and number of cycles in given time.
Wave superposition occurs when two waves combine, adding displacements.

Solved Examples

Ex 1: Compare two waves with Φ = 0 and Φ = -π/4.
- Phase shift causes horizontal displacement of wave.
Ex 2: Find frequency if period T = 0.25 s.
- Frequency f = 1/T = 4 Hz.

Practice Set

Q1: What does phase constant Φ affect in SHM?
- A) Amplitude
- B) Frequency
- C) Horizontal shift of wave
- D) Vertical shift of wave
Q2: Two waves with different frequencies have?
- A) Same period
- B) Different periods
- C) Same phase
- D) Same amplitude
Q3: Superposition of two waves results in?
- A) Multiplication of amplitudes
- B) Addition of displacements
- C) No change
- D) Change in frequency

Answer Key

Q1: C - Phase constant shifts wave horizontally (time shift).
Q2: B - Different frequencies mean different periods.
Q3: B - Superposition adds displacements of waves.

Memory Aids

Phase constant Φ shifts wave left/right.
Frequency f = 1/T changes wave cycles.
Superposition = sum of wave displacements.

Wave Plots and Phase Effects

Displacement x(t) = A cos(ωt + Φ) plotted vs time shows sinusoidal waves.
Amplitude changes affect wave height but not frequency.
Phase constant Φ shifts wave horizontally (time shift).
Different frequencies change wave period and number of cycles in given time.
Wave superposition occurs when two waves combine, adding displacements.

Solved Examples

Ex 1: Compare two waves with Φ = 0 and Φ = -π/4.
- Phase shift causes horizontal displacement of wave.
Ex 2: Find frequency if period T = 0.25 s.
- Frequency f = 1/T = 4 Hz.

Practice Set

Q1: What does phase constant Φ affect in SHM?
- A) Amplitude
- B) Frequency
- C) Horizontal shift of wave
- D) Vertical shift of wave
Q2: Two waves with different frequencies have?
- A) Same period
- B) Different periods
- C) Same phase
- D) Same amplitude
Q3: Superposition of two waves results in?
- A) Multiplication of amplitudes
- B) Addition of displacements
- C) No change
- D) Change in frequency

Answer Key

Q1: C - Phase constant shifts wave horizontally (time shift).
Q2: B - Different frequencies mean different periods.
Q3: B - Superposition adds displacements of waves.

Memory Aids

Phase constant Φ shifts wave left/right.
Frequency f = 1/T changes wave cycles.
Superposition = sum of wave displacements.

Wave Plots and Phase Effects

Displacement x(t) = A cos(ωt + Φ) plotted vs time shows sinusoidal waves.
Amplitude changes affect wave height but not frequency.
Phase constant Φ shifts wave horizontally (time shift).
Different frequencies change wave period and number of cycles in given time.
Wave superposition occurs when two waves combine, adding displacements.

Solved Examples

Ex 1: Compare two waves with Φ = 0 and Φ = -π/4.
- Phase shift causes horizontal displacement of wave.
Ex 2: Find frequency if period T = 0.25 s.
- Frequency f = 1/T = 4 Hz.

Practice Set

Q1: What does phase constant Φ affect in SHM?
- A) Amplitude
- B) Frequency
- C) Horizontal shift of wave
- D) Vertical shift of wave
Q2: Two waves with different frequencies have?
- A) Same period
- B) Different periods
- C) Same phase
- D) Same amplitude
Q3: Superposition of two waves results in?
- A) Multiplication of amplitudes
- B) Addition of displacements
- C) No change
- D) Change in frequency

Answer Key

Q1: C - Phase constant shifts wave horizontally (time shift).
Q2: B - Different frequencies mean different periods.
Q3: B - Superposition adds displacements of waves.

Memory Aids

Phase constant Φ shifts wave left/right.
Frequency f = 1/T changes wave cycles.
Superposition = sum of wave displacements.

Wave Plots and Phase Effects

Displacement x(t) = A cos(ωt + Φ) plotted vs time shows sinusoidal waves.
Amplitude changes affect wave height but not frequency.
Phase constant Φ shifts wave horizontally (time shift).
Different frequencies change wave period and number of cycles in given time.
Wave superposition occurs when two waves combine, adding displacements.

Solved Examples

Ex 1: Compare two waves with Φ = 0 and Φ = -π/4.
- Phase shift causes horizontal displacement of wave.
Ex 2: Find frequency if period T = 0.25 s.
- Frequency f = 1/T = 4 Hz.

Practice Set

Q1: What does phase constant Φ affect in SHM?
- A) Amplitude
- B) Frequency
- C) Horizontal shift of wave
- D) Vertical shift of wave
Q2: Two waves with different frequencies have?
- A) Same period
- B) Different periods
- C) Same phase
- D) Same amplitude
Q3: Superposition of two waves results in?
- A) Multiplication of amplitudes
- B) Addition of displacements
- C) No change
- D) Change in frequency

Answer Key

Q1: C - Phase constant shifts wave horizontally (time shift).
Q2: B - Different frequencies mean different periods.
Q3: B - Superposition adds displacements of waves.

Memory Aids

Phase constant Φ shifts wave left/right.
Frequency f = 1/T changes wave cycles.
Superposition = sum of wave displacements.

Wave Plots and Phase Effects

Displacement x(t) = A cos(ωt + Φ) plotted vs time shows sinusoidal waves.
Amplitude changes affect wave height but not frequency.
Phase constant Φ shifts wave horizontally (time shift).
Different frequencies change wave period and number of cycles in given time.
Wave superposition occurs when two waves combine, adding displacements.

Solved Examples

Ex 1: Compare two waves with Φ = 0 and Φ = -π/4.
- Phase shift causes horizontal displacement of wave.
Ex 2: Find frequency if period T = 0.25 s.
- Frequency f = 1/T = 4 Hz.

Practice Set

Q1: What does phase constant Φ affect in SHM?
- A) Amplitude
- B) Frequency
- C) Horizontal shift of wave
- D) Vertical shift of wave
Q2: Two waves with different frequencies have?
- A) Same period
- B) Different periods
- C) Same phase
- D) Same amplitude
Q3: Superposition of two waves results in?
- A) Multiplication of amplitudes
- B) Addition of displacements
- C) No change
- D) Change in frequency

Answer Key

Q1: C - Phase constant shifts wave horizontally (time shift).
Q2: B - Different frequencies mean different periods.
Q3: B - Superposition adds displacements of waves.

Memory Aids

Phase constant Φ shifts wave left/right.
Frequency f = 1/T changes wave cycles.
Superposition = sum of wave displacements.

Wave Plots and Phase Effects

Displacement x(t) = A cos(ωt + Φ) plotted vs time shows sinusoidal waves.
Amplitude changes affect wave height but not frequency.
Phase constant Φ shifts wave horizontally (time shift).
Different frequencies change wave period and number of cycles in given time.
Wave superposition occurs when two waves combine, adding displacements.

Solved Examples

Ex 1: Compare two waves with Φ = 0 and Φ = -π/4.
- Phase shift causes horizontal displacement of wave.
Ex 2: Find frequency if period T = 0.25 s.
- Frequency f = 1/T = 4 Hz.

Practice Set

Q1: What does phase constant Φ affect in SHM?
- A) Amplitude
- B) Frequency
- C) Horizontal shift of wave
- D) Vertical shift of wave
Q2: Two waves with different frequencies have?
- A) Same period
- B) Different periods
- C) Same phase
- D) Same amplitude
Q3: Superposition of two waves results in?
- A) Multiplication of amplitudes
- B) Addition of displacements
- C) No change
- D) Change in frequency

Answer Key

Q1: C - Phase constant shifts wave horizontally (time shift).
Q2: B - Different frequencies mean different periods.
Q3: B - Superposition adds displacements of waves.

Memory Aids

Phase constant Φ shifts wave left/right.
Frequency f = 1/T changes wave cycles.
Superposition = sum of wave displacements.

Wave Plots and Phase Effects

Displacement x(t) = A cos(ωt + Φ) plotted vs time shows sinusoidal waves.
Amplitude changes affect wave height but not frequency.
Phase constant Φ shifts wave horizontally (time shift).
Different frequencies change wave period and number of cycles in given time.
Wave superposition occurs when two waves combine, adding displacements.

Solved Examples

Ex 1: Compare two waves with Φ = 0 and Φ = -π/4.
- Phase shift causes horizontal displacement of wave.
Ex 2: Find frequency if period T = 0.25 s.
- Frequency f = 1/T = 4 Hz.

Practice Set

Q1: What does phase constant Φ affect in SHM?
- A) Amplitude
- B) Frequency
- C) Horizontal shift of wave
- D) Vertical shift of wave
Q2: Two waves with different frequencies have?
- A) Same period
- B) Different periods
- C) Same phase
- D) Same amplitude
Q3: Superposition of two waves results in?
- A) Multiplication of amplitudes
- B) Addition of displacements
- C) No change
- D) Change in frequency

Answer Key

Q1: C - Phase constant shifts wave horizontally (time shift).
Q2: B - Different frequencies mean different periods.
Q3: B - Superposition adds displacements of waves.

Memory Aids

Phase constant Φ shifts wave left/right.
Frequency f = 1/T changes wave cycles.
Superposition = sum of wave displacements.

Wave Plots and Phase Effects

Displacement x(t) = A cos(ωt + Φ) plotted vs time shows sinusoidal waves.
Amplitude changes affect wave height but not frequency.
Phase constant Φ shifts wave horizontally (time shift).
Different frequencies change wave period and number of cycles in given time.
Wave superposition occurs when two waves combine, adding displacements.

Solved Examples

Ex 1: Compare two waves with Φ = 0 and Φ = -π/4.
- Phase shift causes horizontal displacement of wave.
Ex 2: Find frequency if period T = 0.25 s.
- Frequency f = 1/T = 4 Hz.

Practice Set

Q1: What does phase constant Φ affect in SHM?
- A) Amplitude
- B) Frequency
- C) Horizontal shift of wave
- D) Vertical shift of wave
Q2: Two waves with different frequencies have?
- A) Same period
- B) Different periods
- C) Same phase
- D) Same amplitude
Q3: Superposition of two waves results in?
- A) Multiplication of amplitudes
- B) Addition of displacements
- C) No change
- D) Change in frequency

Answer Key

Q1: C - Phase constant shifts wave horizontally (time shift).
Q2: B - Different frequencies mean different periods.
Q3: B - Superposition adds displacements of waves.

Memory Aids

Phase constant Φ shifts wave left/right.
Frequency f = 1/T changes wave cycles.
Superposition = sum of wave displacements.

Wave Plots and Phase Effects

Displacement x(t) = A cos(ωt + Φ) plotted vs time shows sinusoidal waves.
Amplitude changes affect wave height but not frequency.
Phase constant Φ shifts wave horizontally (time shift).
Different frequencies change wave period and number of cycles in given time.
Wave superposition occurs when two waves combine, adding displacements.

Solved Examples

Ex 1: Compare two waves with Φ = 0 and Φ = -π/4.
- Phase shift causes horizontal displacement of wave.
Ex 2: Find frequency if period T = 0.25 s.
- Frequency f = 1/T = 4 Hz.

Practice Set

Q1: What does phase constant Φ affect in SHM?
- A) Amplitude
- B) Frequency
- C) Horizontal shift of wave
- D) Vertical shift of wave
Q2: Two waves with different frequencies have?
- A) Same period
- B) Different periods
- C) Same phase
- D) Same amplitude
Q3: Superposition of two waves results in?
- A) Multiplication of amplitudes
- B) Addition of displacements
- C) No change
- D) Change in frequency

Answer Key

Q1: C - Phase constant shifts wave horizontally (time shift).
Q2: B - Different frequencies mean different periods.
Q3: B - Superposition adds displacements of waves.

Memory Aids

Phase constant Φ shifts wave left/right.
Frequency f = 1/T changes wave cycles.
Superposition = sum of wave displacements.

Wave Plots and Phase Effects

Displacement x(t) = A cos(ωt + Φ) plotted vs time shows sinusoidal waves.
Amplitude changes affect wave height but not frequency.
Phase constant Φ shifts wave horizontally (time shift).
Different frequencies change wave period and number of cycles in given time.
Wave superposition occurs when two waves combine, adding displacements.

Solved Examples

Ex 1: Compare two waves with Φ = 0 and Φ = -π/4.
- Phase shift causes horizontal displacement of wave.
Ex 2: Find frequency if period T = 0.25 s.
- Frequency f = 1/T = 4 Hz.

Practice Set

Q1: What does phase constant Φ affect in SHM?
- A) Amplitude
- B) Frequency
- C) Horizontal shift of wave
- D) Vertical shift of wave
Q2: Two waves with different frequencies have?
- A) Same period
- B) Different periods
- C) Same phase
- D) Same amplitude
Q3: Superposition of two waves results in?
- A) Multiplication of amplitudes
- B) Addition of displacements
- C) No change
- D) Change in frequency

Answer Key

Q1: C - Phase constant shifts wave horizontally (time shift).
Q2: B - Different frequencies mean different periods.
Q3: B - Superposition adds displacements of waves.

Memory Aids

Phase constant Φ shifts wave left/right.
Frequency f = 1/T changes wave cycles.
Superposition = sum of wave displacements.

Wave Plots and Phase Effects

Displacement x(t) = A cos(ωt + Φ) plotted vs time shows sinusoidal waves.
Amplitude changes affect wave height but not frequency.
Phase constant Φ shifts wave horizontally (time shift).
Different frequencies change wave period and number of cycles in given time.
Wave superposition occurs when two waves combine, adding displacements.

Solved Examples

Ex 1: Compare two waves with Φ = 0 and Φ = -π/4.
- Phase shift causes horizontal displacement of wave.
Ex 2: Find frequency if period T = 0.25 s.
- Frequency f = 1/T = 4 Hz.

Practice Set

Q1: What does phase constant Φ affect in SHM?
- A) Amplitude
- B) Frequency
- C) Horizontal shift of wave
- D) Vertical shift of wave
Q2: Two waves with different frequencies have?
- A) Same period
- B) Different periods
- C) Same phase
- D) Same amplitude
Q3: Superposition of two waves results in?
- A) Multiplication of amplitudes
- B) Addition of displacements
- C) No change
- D) Change in frequency

Answer Key

Q1: C - Phase constant shifts wave horizontally (time shift).
Q2: B - Different frequencies mean different periods.
Q3: B - Superposition adds displacements of waves.

Memory Aids

Phase constant Φ shifts wave left/right.
Frequency f = 1/T changes wave cycles.
Superposition = sum of wave displacements.

Wave Plots and Phase Effects

Displacement x(t) = A cos(ωt + Φ) plotted vs time shows sinusoidal waves.
Amplitude changes affect wave height but not frequency.
Phase constant Φ shifts wave horizontally (time shift).
Different frequencies change wave period and number of cycles in given time.
Wave superposition occurs when two waves combine, adding displacements.

Solved Examples

Ex 1: Compare two waves with Φ = 0 and Φ = -π/4.
- Phase shift causes horizontal displacement of wave.
Ex 2: Find frequency if period T = 0.25 s.
- Frequency f = 1/T = 4 Hz.

Practice Set

Q1: What does phase constant Φ affect in SHM?
- A) Amplitude
- B) Frequency
- C) Horizontal shift of wave
- D) Vertical shift of wave
Q2: Two waves with different frequencies have?
- A) Same period
- B) Different periods
- C) Same phase
- D) Same amplitude
Q3: Superposition of two waves results in?
- A) Multiplication of amplitudes
- B) Addition of displacements
- C) No change
- D) Change in frequency

Answer Key

Q1: C - Phase constant shifts wave horizontally (time shift).
Q2: B - Different frequencies mean different periods.
Q3: B - Superposition adds displacements of waves.

Memory Aids

Phase constant Φ shifts wave left/right.
Frequency f = 1/T changes wave cycles.
Superposition = sum of wave displacements.

Wave Plots and Phase Effects

Displacement x(t) = A cos(ωt + Φ) plotted vs time shows sinusoidal waves.
Amplitude changes affect wave height but not frequency.
Phase constant Φ shifts wave horizontally (time shift).
Different frequencies change wave period and number of cycles in given time.
Wave superposition occurs when two waves combine, adding displacements.

Solved Examples

Ex 1: Compare two waves with Φ = 0 and Φ = -π/4.
- Phase shift causes horizontal displacement of wave.
Ex 2: Find frequency if period T = 0.25 s.
- Frequency f = 1/T = 4 Hz.

Practice Set

Q1: What does phase constant Φ affect in SHM?
- A) Amplitude
- B) Frequency
- C) Horizontal shift of wave
- D) Vertical shift of wave
Q2: Two waves with different frequencies have?
- A) Same period
- B) Different periods
- C) Same phase
- D) Same amplitude
Q3: Superposition of two waves results in?
- A) Multiplication of amplitudes
- B) Addition of displacements
- C) No change
- D) Change in frequency

Answer Key

Q1: C - Phase constant shifts wave horizontally (time shift).
Q2: B - Different frequencies mean different periods.
Q3: B - Superposition adds displacements of waves.

Memory Aids

Phase constant Φ shifts wave left/right.
Frequency f = 1/T changes wave cycles.
Superposition = sum of wave displacements.

Wave Plots and Phase Effects

Displacement x(t) = A cos(ωt + Φ) plotted vs time shows sinusoidal waves.
Amplitude changes affect wave height but not frequency.
Phase constant Φ shifts wave horizontally (time shift).
Different frequencies change wave period and number of cycles in given time.
Wave superposition occurs when two waves combine, adding displacements.

Solved Examples

Ex 1: Compare two waves with Φ = 0 and Φ = -π/4.
- Phase shift causes horizontal displacement of wave.
Ex 2: Find frequency if period T = 0.25 s.
- Frequency f = 1/T = 4 Hz.

Practice Set

Q1: What does phase constant Φ affect in SHM?
- A) Amplitude
- B) Frequency
- C) Horizontal shift of wave
- D) Vertical shift of wave
Q2: Two waves with different frequencies have?
- A) Same period
- B) Different periods
- C) Same phase
- D) Same amplitude
Q3: Superposition of two waves results in?
- A) Multiplication of amplitudes
- B) Addition of displacements
- C) No change
- D) Change in frequency

Answer Key

Q1: C - Phase constant shifts wave horizontally (time shift).
Q2: B - Different frequencies mean different periods.
Q3: B - Superposition adds displacements of waves.

Memory Aids

Phase constant Φ shifts wave left/right.
Frequency f = 1/T changes wave cycles.
Superposition = sum of wave displacements.

Wave Plots and Phase Effects

Displacement x(t) = A cos(ωt + Φ) plotted vs time shows sinusoidal waves.
Amplitude changes affect wave height but not frequency.
Phase constant Φ shifts wave horizontally (time shift).
Different frequencies change wave period and number of cycles in given time.
Wave superposition occurs when two waves combine, adding displacements.

Solved Examples

Ex 1: Compare two waves with Φ = 0 and Φ = -π/4.
- Phase shift causes horizontal displacement of wave.
Ex 2: Find frequency if period T = 0.25 s.
- Frequency f = 1/T = 4 Hz.

Practice Set

Q1: What does phase constant Φ affect in SHM?
- A) Amplitude
- B) Frequency
- C) Horizontal shift of wave
- D) Vertical shift of wave
Q2: Two waves with different frequencies have?
- A) Same period
- B) Different periods
- C) Same phase
- D) Same amplitude
Q3: Superposition of two waves results in?
- A) Multiplication of amplitudes
- B) Addition of displacements
- C) No change
- D) Change in frequency

Answer Key

Q1: C - Phase constant shifts wave horizontally (time shift).
Q2: B - Different frequencies mean different periods.
Q3: B - Superposition adds displacements of waves.

Memory Aids

Phase constant Φ shifts wave left/right.
Frequency f = 1/T changes wave cycles.
Superposition = sum of wave displacements.

Wave Plots and Phase Effects

Displacement x(t) = A cos(ωt + Φ) plotted vs time shows sinusoidal waves.
Amplitude changes affect wave height but not frequency.
Phase constant Φ shifts wave horizontally (time shift).
Different frequencies change wave period and number of cycles in given time.
Wave superposition occurs when two waves combine, adding displacements.

Solved Examples

Ex 1: Compare two waves with Φ = 0 and Φ = -π/4.
- Phase shift causes horizontal displacement of wave.
Ex 2: Find frequency if period T = 0.25 s.
- Frequency f = 1/T = 4 Hz.

Practice Set

Q1: What does phase constant Φ affect in SHM?
- A) Amplitude
- B) Frequency
- C) Horizontal shift of wave
- D) Vertical shift of wave
Q2: Two waves with different frequencies have?
- A) Same period
- B) Different periods
- C) Same phase
- D) Same amplitude
Q3: Superposition of two waves results in?
- A) Multiplication of amplitudes
- B) Addition of displacements
- C) No change
- D) Change in frequency

Answer Key

Q1: C - Phase constant shifts wave horizontally (time shift).
Q2: B - Different frequencies mean different periods.
Q3: B - Superposition adds displacements of waves.

Memory Aids

Phase constant Φ shifts wave left/right.
Frequency f = 1/T changes wave cycles.
Superposition = sum of wave displacements.

Wave Plots and Phase Effects

Displacement x(t) = A cos(ωt + Φ) plotted vs time shows sinusoidal waves.
Amplitude changes affect wave height but not frequency.
Phase constant Φ shifts wave horizontally (time shift).
Different frequencies change wave period and number of cycles in given time.
Wave superposition occurs when two waves combine, adding displacements.

Solved Examples

Ex 1: Compare two waves with Φ = 0 and Φ = -π/4.
- Phase shift causes horizontal displacement of wave.
Ex 2: Find frequency if period T = 0.25 s.
- Frequency f = 1/T = 4 Hz.

Practice Set

Q1: What does phase constant Φ affect in SHM?
- A) Amplitude
- B) Frequency
- C) Horizontal shift of wave
- D) Vertical shift of wave
Q2: Two waves with different frequencies have?
- A) Same period
- B) Different periods
- C) Same phase
- D) Same amplitude
Q3: Superposition of two waves results in?
- A) Multiplication of amplitudes
- B) Addition of displacements
- C) No change
- D) Change in frequency

Answer Key

Q1: C - Phase constant shifts wave horizontally (time shift).
Q2: B - Different frequencies mean different periods.
Q3: B - Superposition adds displacements of waves.

Memory Aids

Phase constant Φ shifts wave left/right.
Frequency f = 1/T changes wave cycles.
Superposition = sum of wave displacements.

Wave Plots and Phase Effects

Displacement x(t) = A cos(ωt + Φ) plotted vs time shows sinusoidal waves.
Amplitude changes affect wave height but not frequency.
Phase constant Φ shifts wave horizontally (time shift).
Different frequencies change wave period and number of cycles in given time.
Wave superposition occurs when two waves combine, adding displacements.

Solved Examples

Ex 1: Compare two waves with Φ = 0 and Φ = -π/4.
- Phase shift causes horizontal displacement of wave.
Ex 2: Find frequency if period T = 0.25 s.
- Frequency f = 1/T = 4 Hz.

Practice Set

Q1: What does phase constant Φ affect in SHM?
- A) Amplitude
- B) Frequency
- C) Horizontal shift of wave
- D) Vertical shift of wave
Q2: Two waves with different frequencies have?
- A) Same period
- B) Different periods
- C) Same phase
- D) Same amplitude
Q3: Superposition of two waves results in?
- A) Multiplication of amplitudes
- B) Addition of displacements
- C) No change
- D) Change in frequency

Answer Key

Q1: C - Phase constant shifts wave horizontally (time shift).
Q2: B - Different frequencies mean different periods.
Q3: B - Superposition adds displacements of waves.

Memory Aids

Phase constant Φ shifts wave left/right.
Frequency f = 1/T changes wave cycles.
Superposition = sum of wave displacements.

Wave Plots and Phase Effects

Displacement x(t) = A cos(ωt + Φ) plotted vs time shows sinusoidal waves.
Amplitude changes affect wave height but not frequency.
Phase constant Φ shifts wave horizontally (time shift).
Different frequencies change wave period and number of cycles in given time.
Wave superposition occurs when two waves combine, adding displacements.

Solved Examples

Ex 1: Compare two waves with Φ = 0 and Φ = -π/4.
- Phase shift causes horizontal displacement of wave.
Ex 2: Find frequency if period T = 0.25 s.
- Frequency f = 1/T = 4 Hz.

Practice Set

Q1: What does phase constant Φ affect in SHM?
- A) Amplitude
- B) Frequency
- C) Horizontal shift of wave
- D) Vertical shift of wave
Q2: Two waves with different frequencies have?
- A) Same period
- B) Different periods
- C) Same phase
- D) Same amplitude
Q3: Superposition of two waves results in?
- A) Multiplication of amplitudes
- B) Addition of displacements
- C) No change
- D) Change in frequency

Answer Key

Q1: C - Phase constant shifts wave horizontally (time shift).
Q2: B - Different frequencies mean different periods.
Q3: B - Superposition adds displacements of waves.

Memory Aids

Phase constant Φ shifts wave left/right.
Frequency f = 1/T changes wave cycles.
Superposition = sum of wave displacements.

Wave Plots and Phase Effects

Displacement x(t) = A cos(ωt + Φ) plotted vs time shows sinusoidal waves.
Amplitude changes affect wave height but not frequency.
Phase constant Φ shifts wave horizontally (time shift).
Different frequencies change wave period and number of cycles in given time.
Wave superposition occurs when two waves combine, adding displacements.

Solved Examples

Ex 1: Compare two waves with Φ = 0 and Φ = -π/4.
- Phase shift causes horizontal displacement of wave.
Ex 2: Find frequency if period T = 0.25 s.
- Frequency f = 1/T = 4 Hz.

Practice Set

Q1: What does phase constant Φ affect in SHM?
- A) Amplitude
- B) Frequency
- C) Horizontal shift of wave
- D) Vertical shift of wave
Q2: Two waves with different frequencies have?
- A) Same period
- B) Different periods
- C) Same phase
- D) Same amplitude
Q3: Superposition of two waves results in?
- A) Multiplication of amplitudes
- B) Addition of displacements
- C) No change
- D) Change in frequency

Answer Key

Q1: C - Phase constant shifts wave horizontally (time shift).
Q2: B - Different frequencies mean different periods.
Q3: B - Superposition adds displacements of waves.

Memory Aids

Phase constant Φ shifts wave left/right.
Frequency f = 1/T changes wave cycles.
Superposition = sum of wave displacements.

Wave Plots and Phase Effects

Displacement \( x(t) = A \cos(\omega t + \Phi) \) plotted vs time shows sinusoidal waves.
Amplitude changes affect wave height but not frequency.
Phase constant \( \Phi \) shifts wave horizontally (time shift).
Different frequencies change wave period and number of cycles in given time.
Wave superposition occurs when two waves combine, adding displacements.

Solved Examples

Ex 1: Compare two waves with \( \Phi = 0 \) and \( \Phi = -\frac{\pi}{4} \).
- Phase shift causes horizontal displacement of wave.
Ex 2: Find frequency if period \( T = 0.25 \) s.
- Frequency \( f = \frac{1}{T} = 4 \) Hz.

Practice Set

Q1: What does phase constant \( \Phi \) affect in SHM?
- A) Amplitude
- B) Frequency
- C) Horizontal shift of wave
- D) Vertical shift of wave
Q2: Two waves with different frequencies have?
- A) Same period
- B) Different periods
- C) Same phase
- D) Same amplitude
Q3: Superposition of two waves results in?
- A) Multiplication of amplitudes
- B) Addition of displacements
- C) No change
- D) Change in frequency

Answer Key

Q1: C - Phase constant shifts wave horizontally (time shift).
Q2: B - Different frequencies mean different periods.
Q3: B - Superposition adds displacements of waves.

Memory Aids

Phase constant \( \Phi \) shifts wave left/right.
Frequency \( f = \frac{1}{T} \) changes wave cycles.
Superposition = sum of wave displacements.

Simple Harmonic Motion and Uniform Circular Motion

Key Concepts

Projection of uniform circular motion on a diameter follows simple harmonic motion (SHM).
Particle moves on circle radius A with angular speed ω anticlockwise.
At time t=0, particle at point X on x-axis; at time t, angle θ = ωt.
Coordinates of particle at point P:
- x = A cos θ = A cos ωt
- y = A sin θ = A sin ωt
Foot of perpendicular from P on x-axis (point M) executes SHM with amplitude A, angular frequency ω.
Foot of perpendicular from P on y-axis (point N) also executes SHM with same amplitude and frequency but phase difference π/2.

Solved Examples

Ex 1: Find x-coordinate of particle at t = 2 s, given A = 5 cm, ω = 3 rad/s.
- Use x = A cos ωt = 5 cos (3 × 2) = 5 cos 6
- Calculate cos 6 and multiply by 5 cm.
Ex 2: Find y-coordinate at t = 1 s for same values.
- Use y = A sin ωt = 5 sin (3 × 1) = 5 sin 3
- Calculate sin 3 and multiply by 5 cm.

Practice Set

Q1: The projection of uniform circular motion on a diameter is:
- A) Uniform motion
- B) Simple harmonic motion
- C) Random motion
- D) Accelerated motion
Q2: For a particle in uniform circular motion with radius 10 cm and angular speed 2 rad/s, what is the x-coordinate at t = π/4 s?
- A) 0 cm
- B) 7.07 cm
- C) 10 cm
- D) 5 cm
Q3: Two SHMs along x and y axes have amplitude 4 cm, angular frequency 5 rad/s, and phase difference π/2. What is the y-coordinate at t = 0.2 s?
- A) 4 cm
- B) 3.42 cm
- C) 2.94 cm
- D) 0 cm

Answer Key

Q1: B - Projection of uniform circular motion on a diameter is simple harmonic motion.
Q2: B - x = 10 cos (2 × π/4) = 10 cos (π/2) = 0, correct option is 0 cm, so option A is correct. (Correction: Option A is correct, not B.)
Q3: B - y = 4 sin (5 × 0.2) = 4 sin 1 = 4 × 0.84 = 3.36 ≈ 3.42 cm.

Memory Aids

SHM projection formula: x = A cos ωt, y = A sin ωt.
Phase difference between x and y SHM is π/2.
Amplitude = radius of circle.
Angular speed ω relates angle and time: θ = ωt.

Key Concepts

Projection of uniform circular motion on a diameter follows simple harmonic motion (SHM).
Particle moves on circle radius A with angular speed ω anticlockwise.
At time t=0, particle at point X on x-axis; at time t, angle θ = ωt.
Coordinates of particle at point P:
- x = A cos θ = A cos ωt
- y = A sin θ = A sin ωt
Foot of perpendicular from P on x-axis (point M) executes SHM with amplitude A, angular frequency ω.
Foot of perpendicular from P on y-axis (point N) also executes SHM with same amplitude and frequency but phase difference π/2.

Solved Examples

Ex 1: Find x-coordinate of particle at t = 2 s, given A = 5 cm, ω = 3 rad/s.
- Use x = A cos ωt = 5 cos (3 × 2) = 5 cos 6
- Calculate cos 6 and multiply by 5 cm.
Ex 2: Find y-coordinate at t = 1 s for same values.
- Use y = A sin ωt = 5 sin (3 × 1) = 5 sin 3
- Calculate sin 3 and multiply by 5 cm.

Practice Set

Q1: The projection of uniform circular motion on a diameter is:
- A) Uniform motion
- B) Simple harmonic motion
- C) Random motion
- D) Accelerated motion
Q2: For a particle in uniform circular motion with radius 10 cm and angular speed 2 rad/s, what is the x-coordinate at t = π/4 s?
- A) 0 cm
- B) 7.07 cm
- C) 10 cm
- D) 5 cm
Q3: Two SHMs along x and y axes have amplitude 4 cm, angular frequency 5 rad/s, and phase difference π/2. What is the y-coordinate at t = 0.2 s?
- A) 4 cm
- B) 3.42 cm
- C) 2.94 cm
- D) 0 cm

Answer Key

Q1: B - Projection of uniform circular motion on a diameter is simple harmonic motion.
Q2: B - x = 10 cos (2 × π/4) = 10 cos (π/2) = 0, correct option is 0 cm, so option A is correct. (Correction: Option A is correct, not B.)
Q3: B - y = 4 sin (5 × 0.2) = 4 sin 1 = 4 × 0.84 = 3.36 ≈ 3.42 cm.

Memory Aids

SHM projection formula: x = A cos ωt, y = A sin ωt.
Phase difference between x and y SHM is π/2.
Amplitude = radius of circle.
Angular speed ω relates angle and time: θ = ωt.

Key Concepts

Projection of uniform circular motion on a diameter follows simple harmonic motion (SHM).
Particle moves on circle radius A with angular speed ω anticlockwise.
At time t=0, particle at point X on x-axis; at time t, angle θ = ωt.
Coordinates of particle at point P:
- x = A cos θ = A cos ωt
- y = A sin θ = A sin ωt
Foot of perpendicular from P on x-axis (point M) executes SHM with amplitude A, angular frequency ω.
Foot of perpendicular from P on y-axis (point N) also executes SHM with same amplitude and frequency but phase difference π/2.

Solved Examples

Ex 1: Find x-coordinate of particle at t = 2 s, given A = 5 cm, ω = 3 rad/s.
- Use x = A cos ωt = 5 cos (3 × 2) = 5 cos 6
- Calculate cos 6 and multiply by 5 cm.
Ex 2: Find y-coordinate at t = 1 s for same values.
- Use y = A sin ωt = 5 sin (3 × 1) = 5 sin 3
- Calculate sin 3 and multiply by 5 cm.

Practice Set

Q1: The projection of uniform circular motion on a diameter is:
- A) Uniform motion
- B) Simple harmonic motion
- C) Random motion
- D) Accelerated motion
Q2: For a particle in uniform circular motion with radius 10 cm and angular speed 2 rad/s, what is the x-coordinate at t = π/4 s?
- A) 0 cm
- B) 7.07 cm
- C) 10 cm
- D) 5 cm
Q3: Two SHMs along x and y axes have amplitude 4 cm, angular frequency 5 rad/s, and phase difference π/2. What is the y-coordinate at t = 0.2 s?
- A) 4 cm
- B) 3.42 cm
- C) 2.94 cm
- D) 0 cm

Answer Key

Q1: B - Projection of uniform circular motion on a diameter is simple harmonic motion.
Q2: B - x = 10 cos (2 × π/4) = 10 cos (π/2) = 0, correct option is 0 cm, so option A is correct. (Correction: Option A is correct, not B.)
Q3: B - y = 4 sin (5 × 0.2) = 4 sin 1 = 4 × 0.84 = 3.36 ≈ 3.42 cm.

Memory Aids

SHM projection formula: x = A cos ωt, y = A sin ωt.
Phase difference between x and y SHM is π/2.
Amplitude = radius of circle.
Angular speed ω relates angle and time: θ = ωt.

Key Concepts

Projection of uniform circular motion on a diameter follows simple harmonic motion (SHM).
Particle moves on circle radius A with angular speed ω anticlockwise.
At time t=0, particle at point X on x-axis; at time t, angle θ = ωt.
Coordinates of particle at point P:
- x = A cos θ = A cos ωt
- y = A sin θ = A sin ωt
Foot of perpendicular from P on x-axis (point M) executes SHM with amplitude A, angular frequency ω.
Foot of perpendicular from P on y-axis (point N) also executes SHM with same amplitude and frequency but phase difference π/2.

Solved Examples

Ex 1: Find x-coordinate of particle at t = 2 s, given A = 5 cm, ω = 3 rad/s.
- Use x = A cos ωt = 5 cos (3 × 2) = 5 cos 6
- Calculate cos 6 and multiply by 5 cm.
Ex 2: Find y-coordinate at t = 1 s for same values.
- Use y = A sin ωt = 5 sin (3 × 1) = 5 sin 3
- Calculate sin 3 and multiply by 5 cm.

Practice Set

Q1: The projection of uniform circular motion on a diameter is:
- A) Uniform motion
- B) Simple harmonic motion
- C) Random motion
- D) Accelerated motion
Q2: For a particle in uniform circular motion with radius 10 cm and angular speed 2 rad/s, what is the x-coordinate at t = π/4 s?
- A) 0 cm
- B) 7.07 cm
- C) 10 cm
- D) 5 cm
Q3: Two SHMs along x and y axes have amplitude 4 cm, angular frequency 5 rad/s, and phase difference π/2. What is the y-coordinate at t = 0.2 s?
- A) 4 cm
- B) 3.42 cm
- C) 2.94 cm
- D) 0 cm

Answer Key

Q1: B - Projection of uniform circular motion on a diameter is simple harmonic motion.
Q2: B - x = 10 cos (2 × π/4) = 10 cos (π/2) = 0, correct option is 0 cm, so option A is correct. (Correction: Option A is correct, not B.)
Q3: B - y = 4 sin (5 × 0.2) = 4 sin 1 = 4 × 0.84 = 3.36 ≈ 3.42 cm.

Memory Aids

SHM projection formula: x = A cos ωt, y = A sin ωt.
Phase difference between x and y SHM is π/2.
Amplitude = radius of circle.
Angular speed ω relates angle and time: θ = ωt.

Key Concepts

Projection of uniform circular motion on a diameter follows simple harmonic motion (SHM).
Particle moves on circle radius A with angular speed \( \omega \) anticlockwise.
At time \( t=0 \), particle at point X on x-axis; at time \( t \), angle \( \theta = \omega t \).
Coordinates of particle at point P:
- \( x = A \cos \theta = A \cos \omega t \)
- \( y = A \sin \theta = A \sin \omega t \)
Foot of perpendicular from P on x-axis (point M) executes SHM with amplitude A, angular frequency \( \omega \).
Foot of perpendicular from P on y-axis (point N) also executes SHM with same amplitude and frequency but phase difference \( \pi/2 \).

Solved Examples

Ex 1: Find x-coordinate of particle at \( t = 2 \text{ s} \), given \( A = 5 \text{ cm} \), \( \omega = 3 \text{ rad/s} \).
- Use \( x = A \cos \omega t = 5 \cos (3 \times 2) = 5 \cos 6 \)
- Calculate \( \cos 6 \) and multiply by 5 cm.
Ex 2: Find y-coordinate at \( t = 1 \text{ s} \) for same values.
- Use \( y = A \sin \omega t = 5 \sin (3 \times 1) = 5 \sin 3 \)
- Calculate \( \sin 3 \) and multiply by 5 cm.

Practice Set

Q1: The projection of uniform circular motion on a diameter is:
- A) Uniform motion
- B) Simple harmonic motion
- C) Random motion
- D) Accelerated motion
Q2: For a particle in uniform circular motion with radius 10 cm and angular speed 2 rad/s, what is the x-coordinate at \( t = \pi/4 \) s?
- A) 0 cm
- B) 7.07 cm
- C) 10 cm
- D) 5 cm
Q3: Two SHMs along x and y axes have amplitude 4 cm, angular frequency 5 rad/s, and phase difference \( \pi/2 \). What is the y-coordinate at \( t = 0.2 \) s?
- A) 4 cm
- B) 3.42 cm
- C) 2.94 cm
- D) 0 cm

Answer Key

Q1: B - Projection of uniform circular motion on a diameter is simple harmonic motion.
Q2: B - \( x = 10 \cos (2 \times \pi/4) = 10 \cos (\pi/2) = 0 \), correct option is 0 cm, so option A is correct. (Correction: Option A is correct, not B.)
Q3: B - \( y = 4 \sin (5 \times 0.2) = 4 \sin 1 = 4 \times 0.84 = 3.36 \approx 3.42 \) cm.

Memory Aids

SHM projection formula: \( x = A \cos \omega t \), \( y = A \sin \omega t \).
Phase difference between x and y SHM is \( \pi/2 \).
Amplitude = radius of circle.
Angular speed \( \omega \) relates angle and time: \( \theta = \omega t \).

Velocity and Acceleration in Simple Harmonic Motion

Key Concepts

Displacement in SHM: x(t) = A cos(ωt + φ), where A is amplitude, ω angular frequency, φ phase angle.
Velocity: Rate of change of displacement, v(t) = dx/dt = -ωA sin(ωt + φ) = -ω√(A² - x²).
Velocity is maximum at mean position (x=0) with v_max = ωA, zero at extremes (x=±A).
Acceleration: Rate of change of velocity, a(t) = dv/dt = -ω²A cos(ωt + φ) = -ω²x(t).
Acceleration is zero at mean position, maximum magnitude at extremes (a_max = ω²A) directed opposite to displacement.
Displacement, velocity, and acceleration have same period T but differ in phase.

Solved Examples

Ex 1: Find velocity at displacement x = A/2.
- Use v = -ω√(A² - x²) = -ω√(A² - (A/2)²) = -ωA√(1 - 1/4) = -ωA√3/2.
Ex 2: Find acceleration at displacement x = 0.
- Use a = -ω²x = 0.

Practice Set

Q1: What is the velocity of a particle in SHM at maximum displacement?
- A) ωA
- B) 0
- C) -ωA
- D) ω²A
Q2: The acceleration of a particle in SHM is zero at:
- A) Maximum displacement
- B) Mean position
- C) Half amplitude
- D) None of these
Q3: If amplitude A = 0.1 m and angular frequency ω = 10 rad/s, what is the maximum velocity?
- A) 1 m/s
- B) 10 m/s
- C) 0.1 m/s
- D) 100 m/s

Answer Key

Q1: B - Velocity is zero at maximum displacement.
Q2: B - Acceleration is zero at mean position where displacement is zero.
Q3: A - v_max = ωA = 10 × 0.1 = 1 m/s.

Memory Aids

Velocity formula: v = -ω√(A² - x²), max at center.
Acceleration formula: a = -ω²x, max at extremes.
Velocity zero at extremes, acceleration zero at center.

Key Concepts

Displacement in SHM: x(t) = A cos(ωt + φ), where A is amplitude, ω angular frequency, φ phase angle.
Velocity: Rate of change of displacement, v(t) = dx/dt = -ωA sin(ωt + φ) = -ω√(A² - x²).
Velocity is maximum at mean position (x=0) with v_max = ωA, zero at extremes (x=±A).
Acceleration: Rate of change of velocity, a(t) = dv/dt = -ω²A cos(ωt + φ) = -ω²x(t).
Acceleration is zero at mean position, maximum magnitude at extremes (a_max = ω²A) directed opposite to displacement.
Displacement, velocity, and acceleration have same period T but differ in phase.

Solved Examples

Ex 1: Find velocity at displacement x = A/2.
- Use v = -ω√(A² - x²) = -ω√(A² - (A/2)²) = -ωA√(1 - 1/4) = -ωA√3/2.
Ex 2: Find acceleration at displacement x = 0.
- Use a = -ω²x = 0.

Practice Set

Q1: What is the velocity of a particle in SHM at maximum displacement?
- A) ωA
- B) 0
- C) -ωA
- D) ω²A
Q2: The acceleration of a particle in SHM is zero at:
- A) Maximum displacement
- B) Mean position
- C) Half amplitude
- D) None of these
Q3: If amplitude A = 0.1 m and angular frequency ω = 10 rad/s, what is the maximum velocity?
- A) 1 m/s
- B) 10 m/s
- C) 0.1 m/s
- D) 100 m/s

Answer Key

Q1: B - Velocity is zero at maximum displacement.
Q2: B - Acceleration is zero at mean position where displacement is zero.
Q3: A - v_max = ωA = 10 × 0.1 = 1 m/s.

Memory Aids

Velocity formula: v = -ω√(A² - x²), max at center.
Acceleration formula: a = -ω²x, max at extremes.
Velocity zero at extremes, acceleration zero at center.

Key Concepts

Displacement in SHM: x(t) = A cos(ωt + φ), where A is amplitude, ω angular frequency, φ phase angle.
Velocity: Rate of change of displacement, v(t) = dx/dt = -ωA sin(ωt + φ) = -ω√(A^2 - x^2).
Velocity is maximum at mean position (x=0) with v_max = ωA, zero at extremes (x=±A).
Acceleration: Rate of change of velocity, a(t) = dv/dt = -ω²A cos(ωt + φ) = -ω²x(t).
Acceleration is zero at mean position, maximum magnitude at extremes (a_max = ω²A) directed opposite to displacement.
Displacement, velocity, and acceleration have same period T but differ in phase.

Solved Examples

Ex 1: Find velocity at displacement x = A/2.
- Use v = -ω√(A^2 - x^2) = -ω√(A^2 - (A/2)^2) = -ωA√(1 - 1/4) = -ωA√3/2.
Ex 2: Find acceleration at displacement x = 0.
- Use a = -ω^2x = 0.

Practice Set

Q1: What is the velocity of a particle in SHM at maximum displacement?
- A) ωA
- B) 0
- C) -ωA
- D) ω²A
Q2: The acceleration of a particle in SHM is zero at:
- A) Maximum displacement
- B) Mean position
- C) Half amplitude
- D) None of these
Q3: If amplitude A = 0.1 m and angular frequency ω = 10 rad/s, what is the maximum velocity?
- A) 1 m/s
- B) 10 m/s
- C) 0.1 m/s
- D) 100 m/s

Answer Key

Q1: B - Velocity is zero at maximum displacement.
Q2: B - Acceleration is zero at mean position where displacement is zero.
Q3: A - v_max = ωA = 10 × 0.1 = 1 m/s.

Memory Aids

Velocity formula: v = -ω√(A^2 - x^2), max at center.
Acceleration formula: a = -ω²x, max at extremes.
Velocity zero at extremes, acceleration zero at center.

Key Concepts

Displacement in SHM: x(t) = A cos(ωt + φ), where A is amplitude, ω angular frequency, φ phase angle.
Velocity: Rate of change of displacement, v(t) = dx/dt = -ωA sin(ωt + φ) = -ω√(A² - x²).
Velocity is maximum at mean position (x=0) with v_max = ωA, zero at extremes (x=±A).
Acceleration: Rate of change of velocity, a(t) = dv/dt = -ω²A cos(ωt + φ) = -ω²x(t).
Acceleration is zero at mean position, maximum magnitude at extremes (a_max = ω²A) directed opposite to displacement.
Displacement, velocity, and acceleration have same period T but differ in phase.

Solved Examples

Ex 1: Find velocity at displacement x = A/2.
- Use v = -ω√(A² - x²) = -ω√(A² - (A/2)²) = -ωA√(1 - 1/4) = -ωA√3/2.
Ex 2: Find acceleration at displacement x = 0.
- Use a = -ω²x = 0.

Practice Set

Q1: What is the velocity of a particle in SHM at maximum displacement?
- A) ωA
- B) 0
- C) -ωA
- D) ω²A
Q2: The acceleration of a particle in SHM is zero at:
- A) Maximum displacement
- B) Mean position
- C) Half amplitude
- D) None of these
Q3: If amplitude A = 0.1 m and angular frequency ω = 10 rad/s, what is the maximum velocity?
- A) 1 m/s
- B) 10 m/s
- C) 0.1 m/s
- D) 100 m/s

Answer Key

Q1: B - Velocity is zero at maximum displacement.
Q2: B - Acceleration is zero at mean position where displacement is zero.
Q3: A - v_max = ωA = 10 × 0.1 = 1 m/s.

Memory Aids

Velocity formula: v = -ω√(A² - x²), max at center.
Acceleration formula: a = -ω²x, max at extremes.
Velocity zero at extremes, acceleration zero at center.

Key Concepts

Displacement in SHM: \( x(t) = A \cos(\omega t + \phi) \), where \( A \) is amplitude, \( \omega \) angular frequency, \( \phi \) phase angle.
Velocity: Rate of change of displacement, \( v(t) = \frac{dx}{dt} = -\omega A \sin(\omega t + \phi) = -\omega \sqrt{A^2 - x^2} \).
Velocity is maximum at mean position (\( x=0 \)) with \( v_{max} = \omega A \), zero at extremes (\( x=\pm A \)).
Acceleration: Rate of change of velocity, \( a(t) = \frac{dv}{dt} = -\omega^2 A \cos(\omega t + \phi) = -\omega^2 x(t) \).
Acceleration is zero at mean position, maximum magnitude at extremes (\( a_{max} = \omega^2 A \)) directed opposite to displacement.
Displacement, velocity, and acceleration have same period \( T \) but differ in phase.

Solved Examples

Ex 1: Find velocity at displacement \( x = \frac{A}{2} \).
- Use \( v = -\omega \sqrt{A^2 - x^2} = -\omega \sqrt{A^2 - \left(\frac{A}{2}\right)^2} = -\omega A \sqrt{1 - \frac{1}{4}} = -\omega A \frac{\sqrt{3}}{2} \).
Ex 2: Find acceleration at displacement \( x = 0 \).
- Use \( a = -\omega^2 x = 0 \).

Practice Set

Q1: What is the velocity of a particle in SHM at maximum displacement?
- A) \( \omega A \)
- B) 0
- C) \( -\omega A \)
- D) \( \omega^2 A \)
Q2: The acceleration of a particle in SHM is zero at:
- A) Maximum displacement
- B) Mean position
- C) Half amplitude
- D) None of these
Q3: If amplitude \( A = 0.1 \) m and angular frequency \( \omega = 10 \) rad/s, what is the maximum velocity?
- A) 1 m/s
- B) 10 m/s
- C) 0.1 m/s
- D) 100 m/s

Answer Key

Q1: B - Velocity is zero at maximum displacement.
Q2: B - Acceleration is zero at mean position where displacement is zero.
Q3: A - \( v_{max} = \omega A = 10 \times 0.1 = 1 \) m/s.

Memory Aids

Velocity formula: \( v = -\omega \sqrt{A^2 - x^2} \), max at center.
Acceleration formula: \( a = -\omega^2 x \), max at extremes.
Velocity zero at extremes, acceleration zero at center.

SHM Graphs

Key Concepts

Displacement vs Time: Sinusoidal wave between +A and -A, period T.
Velocity vs Time: Sine wave, max at zero displacement, zero at extremes.
Acceleration vs Time: Cosine wave, zero at mean, max magnitude at extremes opposite displacement.
All have same period but differ in phase.

Solved Examples

Ex 1: Identify phase difference between displacement and velocity.
- Velocity leads displacement by π/2 (90° phase difference).
Ex 2: At what time in period is velocity zero?
- At maximum displacement points t = T/4, 3T/4.

Practice Set

Q1: At which point is velocity maximum in SHM?
- A) Maximum displacement
- B) Mean position
- C) Half amplitude
- D) None
Q2: Acceleration is maximum when displacement is:
- A) Zero
- B) Maximum
- C) Half amplitude
- D) None
Q3: Which graph leads displacement by 90°?
- A) Displacement
- B) Velocity
- C) Acceleration
- D) None

Answer Key

Q1: B - Velocity is max at mean position where displacement is zero.
Q2: B - Acceleration max at maximum displacement.
Q3: B - Velocity leads displacement by 90°.

Memory Aids

Displacement: cosine wave, velocity: sine wave (90° ahead).
Acceleration opposite displacement, same frequency.
Velocity zero at extremes, acceleration zero at mean.

Key Concepts

Displacement vs Time: Sinusoidal wave between +A and -A, period \( T \).
Velocity vs Time: Sine wave, max at zero displacement, zero at extremes.
Acceleration vs Time: Cosine wave, zero at mean, max magnitude at extremes opposite displacement.
All have same period but differ in phase.

Solved Examples

Ex 1: Identify phase difference between displacement and velocity.
- Velocity leads displacement by \( \frac{\pi}{2} \) (90° phase difference).
Ex 2: At what time in period is velocity zero?
- At maximum displacement points \( t = \frac{T}{4}, \frac{3T}{4} \).

Practice Set

Q1: At which point is velocity maximum in SHM?
- A) Maximum displacement
- B) Mean position
- C) Half amplitude
- D) None
Q2: Acceleration is maximum when displacement is:
- A) Zero
- B) Maximum
- C) Half amplitude
- D) None
Q3: Which graph leads displacement by 90°?
- A) Displacement
- B) Velocity
- C) Acceleration
- D) None

Answer Key

Q1: B - Velocity is max at mean position where displacement is zero.
Q2: B - Acceleration max at maximum displacement.
Q3: B - Velocity leads displacement by 90°.

Memory Aids

Displacement: cosine wave, velocity: sine wave (90° ahead).
Acceleration opposite displacement, same frequency.
Velocity zero at extremes, acceleration zero at mean.

Force Law for Simple Harmonic Motion

Key Concepts

Acceleration in S.H.M.: a = -ω²x
Force from Newton's 2nd law: F = ma = -mω²x = -kx
Spring constant: k = mω², Angular frequency: ω = √(k/m)
Force law for S.H.M.: Restoring force is proportional and opposite to displacement

Solved Examples

Ex 1: Find force on mass m = 2kg at displacement x = 0.1m with k = 50N/m.
- Use F = -kx = -50 × 0.1 = -5N

Practice Set

Q1: What is the force on a particle in S.H.M. at displacement x?
- A) kx
- B) -kx
- C) mω²x
- D) Zero
Q2: If ω = 10rad/s and m = 0.5kg, what is k?
- A) 5 N/m
- B) 50 N/m
- C) 100 N/m
- D) 0.5 N/m
Q3: For m = 1kg, k = 25N/m, find force at x = 0.2m.
- A) -5 N
- B) 5 N
- C) -10 N
- D) 10 N

Answer Key

Q1: B) -kx - Force is restoring, opposite to displacement.
Q2: C) 100 N/m - k = mω² = 0.5 × 10² = 50 (Correction: Actually 50 N/m, so option B is correct.)
Q3: A) -5 N - F = -kx = -25 × 0.2 = -5 N

Memory Aids

Force in S.H.M. always F = -kx
Angular frequency ω = √(k/m)
Acceleration a = -ω²x

Key Concepts

Acceleration in S.H.M.: a = -ω²x
Force from Newton's 2nd law: F = ma = -mω²x = -kx
Spring constant: k = mω², Angular frequency: ω = √(k/m)
Force law for S.H.M.: Restoring force is proportional and opposite to displacement

Solved Examples

Ex 1: Find force on mass m = 2kg at displacement x = 0.1m with k = 50N/m.
- Use F = -kx = -50 × 0.1 = -5N

Practice Set

Q1: What is the force on a particle in S.H.M. at displacement x?
- A) kx
- B) -kx
- C) mω²x
- D) Zero
Q2: If ω = 10rad/s and m = 0.5kg, what is k?
- A) 5 N/m
- B) 50 N/m
- C) 100 N/m
- D) 0.5 N/m
Q3: For m = 1kg, k = 25N/m, find force at x = 0.2m.
- A) -5 N
- B) 5 N
- C) -10 N
- D) 10 N

Answer Key

Q1: B) -kx - Force is restoring, opposite to displacement.
Q2: C) 100 N/m - k = mω² = 0.5 × 10² = 50 (Correction: Actually 50 N/m, so option B is correct.)
Q3: A) -5 N - F = -kx = -25 × 0.2 = -5 N

Memory Aids

Force in S.H.M. always F = -kx
Angular frequency ω = √(k/m)
Acceleration a = -ω²x

Key Concepts

Acceleration in S.H.M.: a = -ω²x
Force from Newton's 2nd law: F = ma = -mω²x = -kx
Spring constant: k = mω², Angular frequency: ω = √(k/m)
Force law for S.H.M.: Restoring force is proportional and opposite to displacement

Solved Examples

Ex 1: Find force on mass m = 2kg at displacement x = 0.1m with k = 50N/m.
- Use F = -kx = -50 × 0.1 = -5N

Practice Set

Q1: What is the force on a particle in S.H.M. at displacement x?
- A) kx
- B) -kx
- C) mω²x
- D) Zero
Q2: If ω = 10rad/s and m = 0.5kg, what is k?
- A) 5 N/m
- B) 50 N/m
- C) 100 N/m
- D) 0.5 N/m
Q3: For m = 1kg, k = 25N/m, find force at x = 0.2m.
- A) -5 N
- B) 5 N
- C) -10 N
- D) 10 N

Answer Key

Q1: B) -kx - Force is restoring, opposite to displacement.
Q2: C) 100 N/m - k = mω² = 0.5 × 10² = 50 (Correction: Actually 50 N/m, so option B is correct.)
Q3: A) -5 N - F = -kx = -25 × 0.2 = -5 N

Memory Aids

Force in S.H.M. always F = -kx
Angular frequency ω = √(k/m)
Acceleration a = -ω²x

Key Concepts

Acceleration in S.H.M.: a = -ω²x
Force from Newton's 2nd law: F = ma = -mω²x = -kx
Spring constant: k = mω², Angular frequency: ω = √(k/m)
Force law for S.H.M.: Restoring force is proportional and opposite to displacement

Solved Examples

Ex 1: Find force on mass m = 2kg at displacement x = 0.1m with k = 50N/m.
- Use F = -kx = -50 × 0.1 = -5N

Practice Set

Q1: What is the force on a particle in S.H.M. at displacement x?
- A) kx
- B) -kx
- C) mω²x
- D) Zero
Q2: If ω = 10rad/s and m = 0.5kg, what is k?
- A) 5 N/m
- B) 50 N/m
- C) 100 N/m
- D) 0.5 N/m
Q3: For m = 1kg, k = 25N/m, find force at x = 0.2m.
- A) -5 N
- B) 5 N
- C) -10 N
- D) 10 N

Answer Key

Q1: B) -kx - Force is restoring, opposite to displacement.
Q2: C) 100 N/m - k = mω² = 0.5 × 10² = 50 (Correction: Actually 50 N/m, so option B is correct.)
Q3: A) -5 N - F = -kx = -25 × 0.2 = -5 N

Memory Aids

Force in S.H.M. always F = -kx
Angular frequency ω = √(k/m)
Acceleration a = -ω²x

Key Concepts

Acceleration in S.H.M.: \( a = -\omega^2 x \)
Force from Newton's 2nd law: \( F = ma = -m\omega^2 x = -kx \)
Spring constant: \( k = m\omega^2 \), Angular frequency: \( \omega = \sqrt{\frac{k}{m}} \)
Force law for S.H.M.: Restoring force is proportional and opposite to displacement

Solved Examples

Ex 1: Find force on mass \( m = 2\,kg \) at displacement \( x = 0.1\,m \) with \( k = 50\,N/m \).
- Use \( F = -kx = -50 \times 0.1 = -5\,N \)

Practice Set

Q1: What is the force on a particle in S.H.M. at displacement \( x \)?
- A) \( kx \)
- B) \( -kx \)
- C) \( m\omega^2 x \)
- D) Zero
Q2: If \( \omega = 10\,rad/s \) and \( m = 0.5\,kg \), what is \( k \)?
- A) 5 N/m
- B) 50 N/m
- C) 100 N/m
- D) 0.5 N/m
Q3: For \( m = 1\,kg \), \( k = 25\,N/m \), find force at \( x = 0.2\,m \).
- A) -5 N
- B) 5 N
- C) -10 N
- D) 10 N

Answer Key

Q1: B) \( -kx \) - Force is restoring, opposite to displacement.
Q2: C) 100 N/m - \( k = m\omega^2 = 0.5 \times 10^2 = 50 \) (Correction: Actually 50 N/m, so option B is correct.)
Q3: A) -5 N - \( F = -kx = -25 \times 0.2 = -5 \) N

Memory Aids

Force in S.H.M. always \( F = -kx \)
Angular frequency \( \omega = \sqrt{k/m} \)
Acceleration \( a = -\omega^2 x \)

Oscillations Due to Spring

Key Concepts

Massless spring on smooth surface, one end fixed, other attached to mass m
Displacement x causes restoring force F = -kx
Force constant k is restoring force per unit displacement
Force direction opposite to displacement

Solved Examples

Ex 1: Calculate restoring force for k = 40N/m, x = 0.05m.
- F = -kx = -40 × 0.05 = -2N

Practice Set

Q1: What is the restoring force when spring is stretched by x?
- A) kx
- B) -kx
- C) Zero
- D) mω²x
Q2: If k = 100N/m, m = 0.25kg, find angular frequency ω.
- A) 20 rad/s
- B) 40 rad/s
- C) 10 rad/s
- D) 5 rad/s
Q3: For k = 50N/m, m = 2kg, find time period T.
- A) 1.26 s
- B) 2.81 s
- C) 0.89 s
- D) 3.14 s

Answer Key

Q1: B) -kx - Restoring force opposes displacement.
Q2: A) 20 rad/s - ω = √(k/m) = √(100/0.25) = 20
Q3: A) 1.26 s - T = 2π√(m/k) = 2π√(2/50) ≈ 1.26

Memory Aids

Restoring force formula: F = -kx
Angular frequency: ω = √(k/m)
Time period: T = 2π√(m/k)

Key Concepts

Massless spring on smooth surface, one end fixed, other attached to mass m
Displacement x causes restoring force F = -kx
Force constant k is restoring force per unit displacement
Force direction opposite to displacement

Solved Examples

Ex 1: Calculate restoring force for k = 40N/m, x = 0.05m.
- F = -kx = -40 × 0.05 = -2N

Practice Set

Q1: What is the restoring force when spring is stretched by x?
- A) kx
- B) -kx
- C) Zero
- D) mω²x
Q2: If k = 100N/m, m = 0.25kg, find angular frequency ω.
- A) 20 rad/s
- B) 40 rad/s
- C) 10 rad/s
- D) 5 rad/s
Q3: For k = 50N/m, m = 2kg, find time period T.
- A) 1.26 s
- B) 2.81 s
- C) 0.89 s
- D) 3.14 s

Answer Key

Q1: B) -kx - Restoring force opposes displacement.
Q2: A) 20 rad/s - ω = √(k/m) = √(100/0.25) = 20
Q3: A) 1.26 s - T = 2π√(m/k) = 2π√(2/50) ≈ 1.26

Memory Aids

Restoring force formula: F = -kx
Angular frequency: ω = √(k/m)
Time period: T = 2π√(m/k)

Key Concepts

Massless spring on smooth surface, one end fixed, other attached to mass m
Displacement x causes restoring force F = -kx
Force constant k is restoring force per unit displacement
Force direction opposite to displacement

Solved Examples

Ex 1: Calculate restoring force for k = 40N/m, x = 0.05m.
- F = -kx = -40 × 0.05 = -2N

Practice Set

Q1: What is the restoring force when spring is stretched by x?
- A) kx
- B) -kx
- C) Zero
- D) mω²x
Q2: If k = 100N/m, m = 0.25kg, find angular frequency ω.
- A) 20 rad/s
- B) 40 rad/s
- C) 10 rad/s
- D) 5 rad/s
Q3: For k = 50N/m, m = 2kg, find time period T.
- A) 1.26 s
- B) 2.81 s
- C) 0.89 s
- D) 3.14 s

Answer Key

Q1: B) -kx - Restoring force opposes displacement.
Q2: A) 20 rad/s - ω = √(k/m) = √(100/0.25) = 20
Q3: A) 1.26 s - T = 2π√(m/k) = 2π√(2/50) ≈ 1.26

Memory Aids

Restoring force formula: F = -kx
Angular frequency: ω = √(k/m)
Time period: T = 2π√(m/k)

Key Concepts

Massless spring on smooth surface, one end fixed, other attached to mass m
Displacement x causes restoring force F = -kx
Force constant k is restoring force per unit displacement
Force direction opposite to displacement

Solved Examples

Ex 1: Calculate restoring force for k = 40N/m, x = 0.05m.
- F = -kx = -40 × 0.05 = -2N

Practice Set

Q1: What is the restoring force when spring is stretched by x?
- A) kx
- B) -kx
- C) Zero
- D) mω²x
Q2: If k = 100N/m, m = 0.25kg, find angular frequency ω.
- A) 20 rad/s
- B) 40 rad/s
- C) 10 rad/s
- D) 5 rad/s
Q3: For k = 50N/m, m = 2kg, find time period T.
- A) 1.26 s
- B) 2.81 s
- C) 0.89 s
- D) 3.14 s

Answer Key

Q1: B) -kx - Restoring force opposes displacement.
Q2: A) 20 rad/s - ω = √(k/m) = √(100/0.25) = 20
Q3: A) 1.26 s - T = 2π√(m/k) = 2π√(2/50) ≈ 1.26

Memory Aids

Restoring force formula: F = -kx
Angular frequency: ω = √(k/m)
Time period: T = 2π√(m/k)

Key Concepts

Massless spring on smooth surface, one end fixed, other attached to mass m
Displacement x causes restoring force F = -kx
Force constant k is restoring force per unit displacement
Force direction opposite to displacement

Solved Examples

Ex 1: Calculate restoring force for k = 40N/m, x = 0.05m.
- F = -kx = -40 × 0.05 = -2N

Practice Set

Q1: What is the restoring force when spring is stretched by x?
- A) kx
- B) -kx
- C) Zero
- D) mω²x
Q2: If k = 100N/m, m = 0.25kg, find angular frequency ω.
- A) 20 rad/s
- B) 40 rad/s
- C) 10 rad/s
- D) 5 rad/s
Q3: For k = 50N/m, m = 2kg, find time period T.
- A) 1.26 s
- B) 2.81 s
- C) 0.89 s
- D) 3.14 s

Answer Key

Q1: B) -kx - Restoring force opposes displacement.
Q2: A) 20 rad/s - ω = √(k/m) = √(100/0.25) = 20
Q3: A) 1.26 s - T = 2π√(m/k) = 2π√(2/50) ≈ 1.26

Memory Aids

Restoring force formula: F = -kx
Angular frequency: ω = √(k/m)
Time period: T = 2π√(m/k)

Key Concepts

Massless spring on smooth surface, one end fixed, other attached to mass m
Displacement x causes restoring force F = -kx
Force constant k is restoring force per unit displacement
Force direction opposite to displacement

Solved Examples

Ex 1: Calculate restoring force for k = 40N/m, x = 0.05m.
- F = -kx = -40 × 0.05 = -2N

Practice Set

Q1: What is the restoring force when spring is stretched by x?
- A) kx
- B) -kx
- C) Zero
- D) mω²x
Q2: If k = 100N/m, m = 0.25kg, find angular frequency ω.
- A) 20 rad/s
- B) 40 rad/s
- C) 10 rad/s
- D) 5 rad/s
Q3: For k = 50N/m, m = 2kg, find time period T.
- A) 1.26 s
- B) 2.81 s
- C) 0.89 s
- D) 3.14 s

Answer Key

Q1: B) -kx - Restoring force opposes displacement.
Q2: A) 20 rad/s - ω = √(k/m) = √(100/0.25) = 20
Q3: A) 1.26 s - T = 2π√(m/k) = 2π√(2/50) ≈ 1.26

Memory Aids

Restoring force formula: F = -kx
Angular frequency: ω = √(k/m)
Time period: T = 2π√(m/k)

Key Concepts

Massless spring on smooth surface, one end fixed, other attached to mass m
Displacement x causes restoring force F = -kx
Force constant k is restoring force per unit displacement
Force direction opposite to displacement

Solved Examples

Ex 1: Calculate restoring force for k = 40N/m, x = 0.05m.
- F = -kx = -40 × 0.05 = -2N

Practice Set

Q1: What is the restoring force when spring is stretched by x?
- A) kx
- B) -kx
- C) Zero
- D) mω²x
Q2: If k = 100N/m, m = 0.25kg, find angular frequency ω.
- A) 20 rad/s
- B) 40 rad/s
- C) 10 rad/s
- D) 5 rad/s
Q3: For k = 50N/m, m = 2kg, find time period T.
- A) 1.26 s
- B) 2.81 s
- C) 0.89 s
- D) 3.14 s

Answer Key

Q1: B) -kx - Restoring force opposes displacement.
Q2: A) 20 rad/s - ω = √(k/m) = √(100/0.25) = 20
Q3: A) 1.26 s - T = 2π√(m/k) = 2π√(2/50) ≈ 1.26

Memory Aids

Restoring force formula: F = -kx
Angular frequency: ω = √(k/m)
Time period: T = 2π√(m/k)

Key Concepts

Massless spring on smooth surface, one end fixed, other attached to mass m
Displacement x causes restoring force F = -kx
Force constant k is restoring force per unit displacement
Force direction opposite to displacement

Solved Examples

Ex 1: Calculate restoring force for k = 40N/m, x = 0.05m.
- F = -kx = -40 × 0.05 = -2N

Practice Set

Q1: What is the restoring force when spring is stretched by x?
- A) kx
- B) -kx
- C) Zero
- D) mω²x
Q2: If k = 100N/m, m = 0.25kg, find angular frequency ω.
- A) 20 rad/s
- B) 40 rad/s
- C) 10 rad/s
- D) 5 rad/s
Q3: For k = 50N/m, m = 2kg, find time period T.
- A) 1.26 s
- B) 2.81 s
- C) 0.89 s
- D) 3.14 s

Answer Key

Q1: B) -kx - Restoring force opposes displacement.
Q2: A) 20 rad/s - ω = √(k/m) = √(100/0.25) = 20
Q3: A) 1.26 s - T = 2π√(m/k) = 2π√(2/50) ≈ 1.26

Memory Aids

Restoring force formula: F = -kx
Angular frequency: ω = √(k/m)
Time period: T = 2π√(m/k)

Key Concepts

Massless spring on smooth surface, one end fixed, other attached to mass m
Displacement x causes restoring force F = -kx
Force constant k is restoring force per unit displacement
Force direction opposite to displacement

Solved Examples

Ex 1: Calculate restoring force for k = 40N/m, x = 0.05m.
- F = -kx = -40 × 0.05 = -2N

Practice Set

Q1: What is the restoring force when spring is stretched by x?
- A) kx
- B) -kx
- C) Zero
- D) mω²x
Q2: If k = 100N/m, m = 0.25kg, find angular frequency ω.
- A) 20 rad/s
- B) 40 rad/s
- C) 10 rad/s
- D) 5 rad/s
Q3: For k = 50N/m, m = 2kg, find time period T.
- A) 1.26 s
- B) 2.81 s
- C) 0.89 s
- D) 3.14 s

Answer Key

Q1: B) -kx - Restoring force opposes displacement.
Q2: A) 20 rad/s - ω = √(k/m) = √(100/0.25) = 20
Q3: A) 1.26 s - T = 2π√(m/k) = 2π√(2/50) ≈ 1.26

Memory Aids

Restoring force formula: F = -kx
Angular frequency: ω = √(k/m)
Time period: T = 2π√(m/k)

Key Concepts

Massless spring on smooth surface, one end fixed, other attached to mass m
Displacement x causes restoring force F = -kx
Force constant k is restoring force per unit displacement
Force direction opposite to displacement

Solved Examples

Ex 1: Calculate restoring force for k = 40N/m, x = 0.05m.
- F = -kx = -40 × 0.05 = -2N

Practice Set

Q1: What is the restoring force when spring is stretched by x?
- A) kx
- B) -kx
- C) Zero
- D) mω²x
Q2: If k = 100N/m, m = 0.25kg, find angular frequency ω.
- A) 20 rad/s
- B) 40 rad/s
- C) 10 rad/s
- D) 5 rad/s
Q3: For k = 50N/m, m = 2kg, find time period T.
- A) 1.26 s
- B) 2.81 s
- C) 0.89 s
- D) 3.14 s

Answer Key

Q1: B) -kx - Restoring force opposes displacement.
Q2: A) 20 rad/s - ω = √(k/m) = √(100/0.25) = 20
Q3: A) 1.26 s - T = 2π√(m/k) = 2π√(2/50) ≈ 1.26

Memory Aids

Restoring force formula: F = -kx
Angular frequency: ω = √(k/m)
Time period: T = 2π√(m/k)

Key Concepts

Massless spring on smooth surface, one end fixed, other attached to mass \( m \)
Displacement \( x \) causes restoring force \( F = -kx \)
Force constant \( k \) is restoring force per unit displacement
Force direction opposite to displacement

Solved Examples

Ex 1: Calculate restoring force for \( k = 40\,N/m \), \( x = 0.05\,m \).
- \( F = -kx = -40 \times 0.05 = -2\,N \)

Practice Set

Q1: What is the restoring force when spring is stretched by \( x \)?
- A) \( kx \)
- B) \( -kx \)
- C) Zero
- D) \( m\omega^2 x \)
Q2: If \( k = 100\,N/m \), \( m = 0.25\,kg \), find angular frequency \( \omega \).
- A) 20 rad/s
- B) 40 rad/s
- C) 10 rad/s
- D) 5 rad/s
Q3: For \( k = 50\,N/m \), \( m = 2\,kg \), find time period \( T \).
- A) 1.26 s
- B) 2.81 s
- C) 0.89 s
- D) 3.14 s

Answer Key

Q1: B) \( -kx \) - Restoring force opposes displacement.
Q2: A) 20 rad/s - \( \omega = \sqrt{k/m} = \sqrt{100/0.25} = 20 \)
Q3: A) 1.26 s - \( T = 2\pi \sqrt{m/k} = 2\pi \sqrt{2/50} \approx 1.26 \)

Memory Aids

Restoring force formula: \( F = -kx \)
Angular frequency: \( \omega = \sqrt{k/m} \)
Time period: \( T = 2\pi \sqrt{m/k} \)

Time Period of SHM for Combination of Springs

Key Concepts

Series combination: 1/k_eq = 1/k₁ + 1/k₂
Time period for series: T = 2π√(m(k₁ + k₂)/(k₁k₂))
Parallel combination: k_eq = k₁ + k₂
Time period for parallel: T = 2π√(m/(k₁ + k₂))

Solved Examples

Ex 1: For series springs with k₁ = 100N/m, k₂ = 200N/m, m = 0.5kg, find T.
- k_eq = 1/(1/100 + 1/200) = 66.67N/m
- T = 2π√(0.5/66.67) = 1.72s
Ex 2: For parallel springs with k₁ = 100N/m, k₂ = 200N/m, m = 0.5kg, find T.
- k_eq = 100 + 200 = 300N/m
- T = 2π√(0.5/300) = 0.81s

Practice Set

Q1: For series springs k₁ = 50N/m, k₂ = 50N/m, m = 1kg, find T.
- A) 1.26 s
- B) 2.51 s
- C) 0.89 s
- D) 3.14 s
Q2: For parallel springs k₁ = 30N/m, k₂ = 70N/m, m = 0.5kg, find T.
- A) 1.77 s
- B) 1.26 s
- C) 0.89 s
- D) 0.56 s
Q3: Which combination results in higher frequency?
- A) Series
- B) Parallel
- C) Both same
- D) None

Answer Key

Q1: A) 1.26 s - Series k_eq = 25, T = 2π√(1/25) = 1.26
Q2: D) 0.56 s - Parallel k_eq = 100, T = 2π√(0.5/100) = 0.56
Q3: B) Parallel - Parallel springs have higher k_eq, higher frequency

Memory Aids

Series: 1/k_eq = 1/k₁ + 1/k₂
Parallel: k_eq = k₁ + k₂
Time period T = 2π√(m/k_eq)

Key Concepts

Series combination: 1/k_eq = 1/k₁ + 1/k₂
Time period for series: T = 2π√(m(k₁ + k₂)/(k₁k₂))
Parallel combination: k_eq = k₁ + k₂
Time period for parallel: T = 2π√(m/(k₁ + k₂))

Solved Examples

Ex 1: For series springs with k₁ = 100N/m, k₂ = 200N/m, m = 0.5kg, find T.
- k_eq = 1/(1/100 + 1/200) = 66.67N/m
- T = 2π√(0.5/66.67) = 1.72s
Ex 2: For parallel springs with k₁ = 100N/m, k₂ = 200N/m, m = 0.5kg, find T.
- k_eq = 100 + 200 = 300N/m
- T = 2π√(0.5/300) = 0.81s

Practice Set

Q1: For series springs k₁ = 50N/m, k₂ = 50N/m, m = 1kg, find T.
- A) 1.26 s
- B) 2.51 s
- C) 0.89 s
- D) 3.14 s
Q2: For parallel springs k₁ = 30N/m, k₂ = 70N/m, m = 0.5kg, find T.
- A) 1.77 s
- B) 1.26 s
- C) 0.89 s
- D) 0.56 s
Q3: Which combination results in higher frequency?
- A) Series
- B) Parallel
- C) Both same
- D) None

Answer Key

Q1: A) 1.26 s - Series k_eq = 25, T = 2π√(1/25) = 1.26
Q2: D) 0.56 s - Parallel k_eq = 100, T = 2π√(0.5/100) = 0.56
Q3: B) Parallel - Parallel springs have higher k_eq, higher frequency

Memory Aids

Series: 1/k_eq = 1/k₁ + 1/k₂
Parallel: k_eq = k₁ + k₂
Time period T = 2π√(m/k_eq)

Key Concepts

Series combination: 1/k_eq = 1/k₁ + 1/k₂
Time period for series: T = 2π√(m(k₁ + k₂)/(k₁k₂))
Parallel combination: k_eq = k₁ + k₂
Time period for parallel: T = 2π√(m/(k₁ + k₂))

Solved Examples

Ex 1: For series springs with k₁ = 100N/m, k₂ = 200N/m, m = 0.5kg, find T.
- k_eq = 1/(1/100 + 1/200) = 66.67N/m
- T = 2π√(0.5/66.67) = 1.72s
Ex 2: For parallel springs with k₁ = 100N/m, k₂ = 200N/m, m = 0.5kg, find T.
- k_eq = 100 + 200 = 300N/m
- T = 2π√(0.5/300) = 0.81s

Practice Set

Q1: For series springs k₁ = 50N/m, k₂ = 50N/m, m = 1kg, find T.
- A) 1.26 s
- B) 2.51 s
- C) 0.89 s
- D) 3.14 s
Q2: For parallel springs k₁ = 30N/m, k₂ = 70N/m, m = 0.5kg, find T.
- A) 1.77 s
- B) 1.26 s
- C) 0.89 s
- D) 0.56 s
Q3: Which combination results in higher frequency?
- A) Series
- B) Parallel
- C) Both same
- D) None

Answer Key

Q1: A) 1.26 s - Series k_eq = 25, T = 2π√(1/25) = 1.26
Q2: D) 0.56 s - Parallel k_eq = 100, T = 2π√(0.5/100) = 0.56
Q3: B) Parallel - Parallel springs have higher k_eq, higher frequency

Memory Aids

Series: 1/k_eq = 1/k₁ + 1/k₂
Parallel: k_eq = k₁ + k₂
Time period T = 2π√(m/k_eq)

Key Concepts

Series combination: 1/k_eq = 1/k₁ + 1/k₂
Time period for series: T = 2π√(m(k₁ + k₂)/(k₁k₂))
Parallel combination: k_eq = k₁ + k₂
Time period for parallel: T = 2π√(m/(k₁ + k₂))

Solved Examples

Ex 1: For series springs with k₁ = 100N/m, k₂ = 200N/m, m = 0.5kg, find T.
- k_eq = 1/(1/100 + 1/200) = 66.67N/m
- T = 2π√(0.5/66.67) = 1.72s
Ex 2: For parallel springs with k₁ = 100N/m, k₂ = 200N/m, m = 0.5kg, find T.
- k_eq = 100 + 200 = 300N/m
- T = 2π√(0.5/300) = 0.81s

Practice Set

Q1: For series springs k₁ = 50N/m, k₂ = 50N/m, m = 1kg, find T.
- A) 1.26 s
- B) 2.51 s
- C) 0.89 s
- D) 3.14 s
Q2: For parallel springs k₁ = 30N/m, k₂ = 70N/m, m = 0.5kg, find T.
- A) 1.77 s
- B) 1.26 s
- C) 0.89 s
- D) 0.56 s
Q3: Which combination results in higher frequency?
- A) Series
- B) Parallel
- C) Both same
- D) None

Answer Key

Q1: A) 1.26 s - Series k_eq = 25, T = 2π√(1/25) = 1.26
Q2: D) 0.56 s - Parallel k_eq = 100, T = 2π√(0.5/100) = 0.56
Q3: B) Parallel - Parallel springs have higher k_eq, higher frequency

Memory Aids

Series: 1/k_eq = 1/k₁ + 1/k₂
Parallel: k_eq = k₁ + k₂
Time period T = 2π√(m/k_eq)

Key Concepts

Series combination: 1/k_eq = 1/k₁ + 1/k₂
Time period for series: T = 2π√(m(k₁ + k₂)/(k₁k₂))
Parallel combination: k_eq = k₁ + k₂
Time period for parallel: T = 2π√(m/(k₁ + k₂))

Solved Examples

Ex 1: For series springs with k₁ = 100N/m, k₂ = 200N/m, m = 0.5kg, find T.
- k_eq = 1/(1/100 + 1/200) = 66.67N/m
- T = 2π√(0.5/66.67) = 1.72s
Ex 2: For parallel springs with k₁ = 100N/m, k₂ = 200N/m, m = 0.5kg, find T.
- k_eq = 100 + 200 = 300N/m
- T = 2π√(0.5/300) = 0.81s

Practice Set

Q1: For series springs k₁ = 50N/m, k₂ = 50N/m, m = 1kg, find T.
- A) 1.26 s
- B) 2.51 s
- C) 0.89 s
- D) 3.14 s
Q2: For parallel springs k₁ = 30N/m, k₂ = 70N/m, m = 0.5kg, find T.
- A) 1.77 s
- B) 1.26 s
- C) 0.89 s
- D) 0.56 s
Q3: Which combination results in higher frequency?
- A) Series
- B) Parallel
- C) Both same
- D) None

Answer Key

Q1: A) 1.26 s - Series k_eq = 25, T = 2π√(1/25) = 1.26
Q2: D) 0.56 s - Parallel k_eq = 100, T = 2π√(0.5/100) = 0.56
Q3: B) Parallel - Parallel springs have higher k_eq, higher frequency

Memory Aids

Series: 1/k_eq = 1/k₁ + 1/k₂
Parallel: k_eq = k₁ + k₂
Time period T = 2π√(m/k_eq)

Key Concepts

Series combination: 1/k_eq = 1/k₁ + 1/k₂
Time period for series: T = 2π√(m(k₁ + k₂)/(k₁k₂))
Parallel combination: k_eq = k₁ + k₂
Time period for parallel: T = 2π√(m/(k₁ + k₂))

Solved Examples

Ex 1: For series springs with k₁ = 100N/m, k₂ = 200N/m, m = 0.5kg, find T.
- k_eq = 1/(1/100 + 1/200) = 66.67N/m
- T = 2π√(0.5/66.67) = 1.72s
Ex 2: For parallel springs with k₁ = 100N/m, k₂ = 200N/m, m = 0.5kg, find T.
- k_eq = 100 + 200 = 300N/m
- T = 2π√(0.5/300) = 0.81s

Practice Set

Q1: For series springs k₁ = 50N/m, k₂ = 50N/m, m = 1kg, find T.
- A) 1.26 s
- B) 2.51 s
- C) 0.89 s
- D) 3.14 s
Q2: For parallel springs k₁ = 30N/m, k₂ = 70N/m, m = 0.5kg, find T.
- A) 1.77 s
- B) 1.26 s
- C) 0.89 s
- D) 0.56 s
Q3: Which combination results in higher frequency?
- A) Series
- B) Parallel
- C) Both same
- D) None

Answer Key

Q1: A) 1.26 s - Series k_eq = 25, T = 2π√(1/25) = 1.26
Q2: D) 0.56 s - Parallel k_eq = 100, T = 2π√(0.5/100) = 0.56
Q3: B) Parallel - Parallel springs have higher k_eq, higher frequency

Memory Aids

Series: 1/k_eq = 1/k₁ + 1/k₂
Parallel: k_eq = k₁ + k₂
Time period T = 2π√(m/k_eq)

Key Concepts

Series combination: 1/k_eq = 1/k₁ + 1/k₂
Time period for series: T = 2π√(m(k₁ + k₂)/(k₁k₂))
Parallel combination: k_eq = k₁ + k₂
Time period for parallel: T = 2π√(m/(k₁ + k₂))

Solved Examples

Ex 1: For series springs with k₁ = 100N/m, k₂ = 200N/m, m = 0.5kg, find T.
- k_eq = 1/(1/100 + 1/200) = 66.67N/m
- T = 2π√(0.5/66.67) = 1.72s
Ex 2: For parallel springs with k₁ = 100N/m, k₂ = 200N/m, m = 0.5kg, find T.
- k_eq = 100 + 200 = 300N/m
- T = 2π√(0.5/300) = 0.81s

Practice Set

Q1: For series springs k₁ = 50N/m, k₂ = 50N/m, m = 1kg, find T.
- A) 1.26 s
- B) 2.51 s
- C) 0.89 s
- D) 3.14 s
Q2: For parallel springs k₁ = 30N/m, k₂ = 70N/m, m = 0.5kg, find T.
- A) 1.77 s
- B) 1.26 s
- C) 0.89 s
- D) 0.56 s
Q3: Which combination results in higher frequency?
- A) Series
- B) Parallel
- C) Both same
- D) None

Answer Key

Q1: A) 1.26 s - Series k_eq = 25, T = 2π√(1/25) = 1.26
Q2: D) 0.56 s - Parallel k_eq = 100, T = 2π√(0.5/100) = 0.56
Q3: B) Parallel - Parallel springs have higher k_eq, higher frequency

Memory Aids

Series: 1/k_eq = 1/k₁ + 1/k₂
Parallel: k_eq = k₁ + k₂
Time period T = 2π√(m/k_eq)

Key Concepts

Series combination: 1/k_eq = 1/k₁ + 1/k₂
Time period for series: T = 2π√(m(k₁ + k₂)/(k₁k₂))
Parallel combination: k_eq = k₁ + k₂
Time period for parallel: T = 2π√(m/(k₁ + k₂))

Solved Examples

Ex 1: For series springs with k₁ = 100N/m, k₂ = 200N/m, m = 0.5kg, find T.
- k_eq = 1/(1/100 + 1/200) = 66.67N/m
- T = 2π√(0.5/66.67) = 1.72s
Ex 2: For parallel springs with k₁ = 100N/m, k₂ = 200N/m, m = 0.5kg, find T.
- k_eq = 100 + 200 = 300N/m
- T = 2π√(0.5/300) = 0.81s

Practice Set

Q1: For series springs k₁ = 50N/m, k₂ = 50N/m, m = 1kg, find T.
- A) 1.26 s
- B) 2.51 s
- C) 0.89 s
- D) 3.14 s
Q2: For parallel springs k₁ = 30N/m, k₂ = 70N/m, m = 0.5kg, find T.
- A) 1.77 s
- B) 1.26 s
- C) 0.89 s
- D) 0.56 s
Q3: Which combination results in higher frequency?
- A) Series
- B) Parallel
- C) Both same
- D) None

Answer Key

Q1: A) 1.26 s - Series k_eq = 25, T = 2π√(1/25) = 1.26
Q2: D) 0.56 s - Parallel k_eq = 100, T = 2π√(0.5/100) = 0.56
Q3: B) Parallel - Parallel springs have higher k_eq, higher frequency

Memory Aids

Series: 1/k_eq = 1/k₁ + 1/k₂
Parallel: k_eq = k₁ + k₂
Time period T = 2π√(m/k_eq)

Key Concepts

Series combination: 1/k_eq = 1/k₁ + 1/k₂
Time period for series: T = 2π√(m(k₁ + k₂)/(k₁k₂))
Parallel combination: k_eq = k₁ + k₂
Time period for parallel: T = 2π√(m/(k₁ + k₂))

Solved Examples

Ex 1: For series springs with k₁ = 100N/m, k₂ = 200N/m, m = 0.5kg, find T.
- k_eq = 1/(1/100 + 1/200) = 66.67N/m
- T = 2π√(0.5/66.67) = 1.72s
Ex 2: For parallel springs with k₁ = 100N/m, k₂ = 200N/m, m = 0.5kg, find T.
- k_eq = 100 + 200 = 300N/m
- T = 2π√(0.5/300) = 0.81s

Practice Set

Q1: For series springs k₁ = 50N/m, k₂ = 50N/m, m = 1kg, find T.
- A) 1.26 s
- B) 2.51 s
- C) 0.89 s
- D) 3.14 s
Q2: For parallel springs k₁ = 30N/m, k₂ = 70N/m, m = 0.5kg, find T.
- A) 1.77 s
- B) 1.26 s
- C) 0.89 s
- D) 0.56 s
Q3: Which combination results in higher frequency?
- A) Series
- B) Parallel
- C) Both same
- D) None

Answer Key

Q1: A) 1.26 s - Series k_eq = 25, T = 2π√(1/25) = 1.26
Q2: D) 0.56 s - Parallel k_eq = 100, T = 2π√(0.5/100) = 0.56
Q3: B) Parallel - Parallel springs have higher k_eq, higher frequency

Memory Aids

Series: 1/k_eq = 1/k₁ + 1/k₂
Parallel: k_eq = k₁ + k₂
Time period T = 2π√(m/k_eq)

Key Concepts

Series combination: \( \frac{1}{k_{eq}} = \frac{1}{k_1} + \frac{1}{k_2} \)
Time period for series: \( T = 2\pi \sqrt{\frac{m(k_1 + k_2)}{k_1 k_2}} \)
Parallel combination: \( k_{eq} = k_1 + k_2 \)
Time period for parallel: \( T = 2\pi \sqrt{\frac{m}{k_1 + k_2}} \)

Solved Examples

Ex 1: For series springs with \( k_1 = 100\,N/m \), \( k_2 = 200\,N/m \), \( m = 0.5\,kg \), find \( T \).
- \( k_{eq} = \frac{1}{\frac{1}{100} + \frac{1}{200}} = 66.67\,N/m \)
- \( T = 2\pi \sqrt{\frac{0.5}{66.67}} = 1.72\,s \)
Ex 2: For parallel springs with \( k_1 = 100\,N/m \), \( k_2 = 200\,N/m \), \( m = 0.5\,kg \), find \( T \).
- \( k_{eq} = 100 + 200 = 300\,N/m \)
- \( T = 2\pi \sqrt{\frac{0.5}{300}} = 0.81\,s \)

Practice Set

Q1: For series springs \( k_1 = 50\,N/m \), \( k_2 = 50\,N/m \), \( m = 1\,kg \), find \( T \).
- A) 1.26 s
- B) 2.51 s
- C) 0.89 s
- D) 3.14 s
Q2: For parallel springs \( k_1 = 30\,N/m \), \( k_2 = 70\,N/m \), \( m = 0.5\,kg \), find \( T \).
- A) 1.77 s
- B) 1.26 s
- C) 0.89 s
- D) 0.56 s
Q3: Which combination results in higher frequency?
- A) Series
- B) Parallel
- C) Both same
- D) None

Answer Key

Q1: A) 1.26 s - Series \( k_{eq} = 25 \), \( T = 2\pi \sqrt{1/25} = 1.26 \)
Q2: D) 0.56 s - Parallel \( k_{eq} = 100 \), \( T = 2\pi \sqrt{0.5/100} = 0.56 \)
Q3: B) Parallel - Parallel springs have higher \( k_{eq} \), higher frequency

Memory Aids

Series: \( \frac{1}{k_{eq}} = \frac{1}{k_1} + \frac{1}{k_2} \)
Parallel: \( k_{eq} = k_1 + k_2 \)
Time period \( T = 2\pi \sqrt{m/k_{eq}} \)

Energy in Simple Harmonic Motion

Energy Variation in SHM

Kinetic energy (K) and potential energy (U) vary with displacement x in SHM.
Total mechanical energy E remains constant: E = K + U.
Potential energy is maximum at maximum displacement ±A.
Kinetic energy is maximum at equilibrium position x=0.
Energy shifts between kinetic and potential forms during motion.

Potential Energy Expression

Potential energy: U = 1/2kx² = 1/2kA² cos²(ωt + φ)
k is spring constant, A amplitude, ω angular frequency, φ phase constant.

Kinetic Energy Expression

Kinetic energy: K = 1/2mv² = 1/2mω²A² sin²(ωt + φ) = 1/2kA² sin²(ωt + φ)
Mass m, velocity v, and other symbols as above.

Total Energy Expression

Total energy: E = K + U = 1/2kA² [cos²(ωt + φ) + sin²(ωt + φ)] = 1/2kA²
Total energy is constant and independent of time.

Energy Variation Over Time

Potential and kinetic energies vary sinusoidally with time.
Sum of kinetic and potential energies remains constant at all times.

Solved Examples

Ex 1: Calculate total energy for k=200, N/m, A=0.05, m.
- Use E=1/2kA² = 1/2 × 200 × (0.05)² = 0.25 J.
Ex 2: Find kinetic energy at x=0.03, m for k=100, N/m, A=0.05, m.
- Potential energy U=1/2kx²=1/2 × 100 × (0.03)²=0.045 J.
- Total energy E=1/2kA²=1/2 × 100 × (0.05)²=0.125 J.
- Kinetic energy K=E-U=0.125-0.045=0.08 J.

Practice Set

Q1: What is the total energy of a SHM system with k=50,N/m and amplitude 0.1,m?
A) 0.25 J
B) 0.5 J
C) 1.0 J
D) 2.0 J
Q2: At maximum displacement, kinetic energy is:
A) Maximum
B) Zero
C) Half of total energy
D) Equal to potential energy
Q3: For a mass 0.2,kg attached to a spring with k=80,N/m and amplitude 0.1,m, what is the maximum kinetic energy?
A) 0.4 J
B) 0.8 J
C) 1.6 J
D) 3.2 J

Answer Key

Q1: A) 0.25 J
Total energy E=1/2kA²=1/2 × 50 × (0.1)²=0.25 J.
Q2: B) Zero
At maximum displacement, velocity is zero, so kinetic energy is zero.
Q3: A) 0.4 J
Maximum kinetic energy equals total energy E=1/2kA²=1/2 × 80 × (0.1)²=0.4 J.

Memory Aids

Total energy E=1/2kA² constant in SHM.
Potential energy U varies as cos², kinetic energy K varies as sin².
Energy shifts between K and U but sum remains constant.

Energy Variation in SHM

Kinetic energy (K) and potential energy (U) vary with displacement x in SHM.
Total mechanical energy E remains constant: E = K + U.
Potential energy is maximum at maximum displacement ±A.
Kinetic energy is maximum at equilibrium position x=0.
Energy shifts between kinetic and potential forms during motion.

Potential Energy Expression

Potential energy: U = 1/2kx² = 1/2kA² cos²(ωt + φ)
k is spring constant, A amplitude, ω angular frequency, φ phase constant.

Kinetic Energy Expression

Kinetic energy: K = 1/2mv² = 1/2mω²A² sin²(ωt + φ) = 1/2kA² sin²(ωt + φ)
Mass m, velocity v, and other symbols as above.

Total Energy Expression

Total energy: E = K + U = 1/2kA² [cos²(ωt + φ) + sin²(ωt + φ)] = 1/2kA²
Total energy is constant and independent of time.

Energy Variation Over Time

Potential and kinetic energies vary sinusoidally with time.
Sum of kinetic and potential energies remains constant at all times.

Solved Examples

Ex 1: Calculate total energy for k=200, N/m, A=0.05, m.
- Use E=1/2kA² = 1/2 × 200 × (0.05)² = 0.25 J.
Ex 2: Find kinetic energy at x=0.03, m for k=100, N/m, A=0.05, m.
- Potential energy U=1/2kx²=1/2 × 100 × (0.03)²=0.045 J.
- Total energy E=1/2kA²=1/2 × 100 × (0.05)²=0.125 J.
- Kinetic energy K=E-U=0.125-0.045=0.08 J.

Practice Set

Q1: What is the total energy of a SHM system with k=50,N/m and amplitude 0.1,m?
A) 0.25 J
B) 0.5 J
C) 1.0 J
D) 2.0 J
Q2: At maximum displacement, kinetic energy is:
A) Maximum
B) Zero
C) Half of total energy
D) Equal to potential energy
Q3: For a mass 0.2,kg attached to a spring with k=80,N/m and amplitude 0.1,m, what is the maximum kinetic energy?
A) 0.4 J
B) 0.8 J
C) 1.6 J
D) 3.2 J

Answer Key

Q1: A) 0.25 J
Total energy E=1/2kA²=1/2 × 50 × (0.1)²=0.25 J.
Q2: B) Zero
At maximum displacement, velocity is zero, so kinetic energy is zero.
Q3: A) 0.4 J
Maximum kinetic energy equals total energy E=1/2kA²=1/2 × 80 × (0.1)²=0.4 J.

Memory Aids

Total energy E=1/2kA² constant in SHM.
Potential energy U varies as cos², kinetic energy K varies as sin².
Energy shifts between K and U but sum remains constant.

Energy Variation in SHM

Kinetic energy (K) and potential energy (U) vary with displacement x in SHM.
Total mechanical energy E remains constant: E = K + U.
Potential energy is maximum at maximum displacement ±A.
Kinetic energy is maximum at equilibrium position x=0.
Energy shifts between kinetic and potential forms during motion.

Potential Energy Expression

Potential energy: U = 1/2kx² = 1/2kA² cos²(ωt + φ)
k is spring constant, A amplitude, ω angular frequency, φ phase constant.

Kinetic Energy Expression

Kinetic energy: K = 1/2mv² = 1/2mω²A² sin²(ωt + φ) = 1/2kA² sin²(ωt + φ)
Mass m, velocity v, and other symbols as above.

Total Energy Expression

Total energy: E = K + U = 1/2kA² [cos²(ωt + φ) + sin²(ωt + φ)] = 1/2kA²
Total energy is constant and independent of time.

Energy Variation Over Time

Potential and kinetic energies vary sinusoidally with time.
Sum of kinetic and potential energies remains constant at all times.

Solved Examples

Ex 1: Calculate total energy for k=200, N/m, A=0.05, m.
- Use E=1/2kA² = 1/2 × 200 × (0.05)² = 0.25 J.
Ex 2: Find kinetic energy at x=0.03, m for k=100, N/m, A=0.05, m.
- Potential energy U=1/2kx²=1/2 × 100 × (0.03)²=0.045 J.
- Total energy E=1/2kA²=1/2 × 100 × (0.05)²=0.125 J.
- Kinetic energy K=E-U=0.125-0.045=0.08 J.

Practice Set

Q1: What is the total energy of a SHM system with k=50,N/m and amplitude 0.1,m?
A) 0.25 J
B) 0.5 J
C) 1.0 J
D) 2.0 J
Q2: At maximum displacement, kinetic energy is:
A) Maximum
B) Zero
C) Half of total energy
D) Equal to potential energy
Q3: For a mass 0.2,kg attached to a spring with k=80,N/m and amplitude 0.1,m, what is the maximum kinetic energy?
A) 0.4 J
B) 0.8 J
C) 1.6 J
D) 3.2 J

Answer Key

Q1: A) 0.25 J
Total energy E=1/2kA²=1/2 × 50 × (0.1)²=0.25 J.
Q2: B) Zero
At maximum displacement, velocity is zero, so kinetic energy is zero.
Q3: A) 0.4 J
Maximum kinetic energy equals total energy E=1/2kA²=1/2 × 80 × (0.1)²=0.4 J.

Memory Aids

Total energy E=1/2kA² constant in SHM.
Potential energy U varies as cos², kinetic energy K varies as sin².
Energy shifts between K and U but sum remains constant.

Energy Variation in SHM

Kinetic energy (K) and potential energy (U) vary with displacement x in SHM.
Total mechanical energy E remains constant: E = K + U.
Potential energy is maximum at maximum displacement ±A.
Kinetic energy is maximum at equilibrium position x=0.
Energy shifts between kinetic and potential forms during motion.

Potential Energy Expression

Potential energy: U = 1/2kx² = 1/2kA² cos²(ωt + φ)
k is spring constant, A amplitude, ω angular frequency, φ phase constant.

Kinetic Energy Expression

Kinetic energy: K = 1/2mv² = 1/2mω²A² sin²(ωt + φ) = 1/2kA² sin²(ωt + φ)
Mass m, velocity v, and other symbols as above.

Total Energy Expression

Total energy: E = K + U = 1/2kA² [cos²(ωt + φ) + sin²(ωt + φ)] = 1/2kA²
Total energy is constant and independent of time.

Energy Variation Over Time

Potential and kinetic energies vary sinusoidally with time.
Sum of kinetic and potential energies remains constant at all times.

Solved Examples

Ex 1: Calculate total energy for k=200, N/m, A=0.05, m.
- Use E=1/2kA² = 1/2 × 200 × (0.05)² = 0.25 J.
Ex 2: Find kinetic energy at x=0.03, m for k=100, N/m, A=0.05, m.
- Potential energy U=1/2kx²=1/2 × 100 × (0.03)²=0.045 J.
- Total energy E=1/2kA²=1/2 × 100 × (0.05)²=0.125 J.
- Kinetic energy K=E-U=0.125-0.045=0.08 J.

Practice Set

Q1: What is the total energy of a SHM system with k=50,N/m and amplitude 0.1,m?
A) 0.25 J
B) 0.5 J
C) 1.0 J
D) 2.0 J
Q2: At maximum displacement, kinetic energy is:
A) Maximum
B) Zero
C) Half of total energy
D) Equal to potential energy
Q3: For a mass 0.2,kg attached to a spring with k=80,N/m and amplitude 0.1,m, what is the maximum kinetic energy?
A) 0.4 J
B) 0.8 J
C) 1.6 J
D) 3.2 J

Answer Key

Q1: A) 0.25 J
Total energy E=1/2kA²=1/2 × 50 × (0.1)²=0.25 J.
Q2: B) Zero
At maximum displacement, velocity is zero, so kinetic energy is zero.
Q3: A) 0.4 J
Maximum kinetic energy equals total energy E=1/2kA²=1/2 × 80 × (0.1)²=0.4 J.

Memory Aids

Total energy E=1/2kA² constant in SHM.
Potential energy U varies as cos², kinetic energy K varies as sin².
Energy shifts between K and U but sum remains constant.

Energy Variation in SHM

Kinetic energy (K) and potential energy (U) vary with displacement x in SHM.
Total mechanical energy E remains constant: E = K + U.
Potential energy is maximum at maximum displacement ±A.
Kinetic energy is maximum at equilibrium position x=0.
Energy shifts between kinetic and potential forms during motion.

Potential Energy Expression

Potential energy: U = 1/2kx² = 1/2kA² cos²(ωt + φ)
k is spring constant, A amplitude, ω angular frequency, φ phase constant.

Kinetic Energy Expression

Kinetic energy: K = 1/2mv² = 1/2mω²A² sin²(ωt + φ) = 1/2kA² sin²(ωt + φ)
Mass m, velocity v, and other symbols as above.

Total Energy Expression

Total energy: E = K + U = 1/2kA² [cos²(ωt + φ) + sin²(ωt + φ)] = 1/2kA²
Total energy is constant and independent of time.

Energy Variation Over Time

Potential and kinetic energies vary sinusoidally with time.
Sum of kinetic and potential energies remains constant at all times.

Solved Examples

Ex 1: Calculate total energy for k=200, N/m, A=0.05, m.
- Use E=1/2kA² = 1/2 × 200 × (0.05)² = 0.25 J.
Ex 2: Find kinetic energy at x=0.03, m for k=100, N/m, A=0.05, m.
- Potential energy U=1/2kx²=1/2 × 100 × (0.03)²=0.045 J.
- Total energy E=1/2kA²=1/2 × 100 × (0.05)²=0.125 J.
- Kinetic energy K=E-U=0.125-0.045=0.08 J.

Practice Set

Q1: What is the total energy of a SHM system with k=50,N/m and amplitude 0.1,m?
A) 0.25 J
B) 0.5 J
C) 1.0 J
D) 2.0 J
Q2: At maximum displacement, kinetic energy is:
A) Maximum
B) Zero
C) Half of total energy
D) Equal to potential energy
Q3: For a mass 0.2,kg attached to a spring with k=80,N/m and amplitude 0.1,m, what is the maximum kinetic energy?
A) 0.4 J
B) 0.8 J
C) 1.6 J
D) 3.2 J

Answer Key

Q1: A) 0.25 J
Total energy E=1/2kA²=1/2 × 50 × (0.1)²=0.25 J.
Q2: B) Zero
At maximum displacement, velocity is zero, so kinetic energy is zero.
Q3: A) 0.4 J
Maximum kinetic energy equals total energy E=1/2kA²=1/2 × 80 × (0.1)²=0.4 J.

Memory Aids

Total energy E=1/2kA² constant in SHM.
Potential energy U varies as cos², kinetic energy K varies as sin².
Energy shifts between K and U but sum remains constant.

Energy Variation in SHM

Kinetic energy (K) and potential energy (U) vary with displacement x in SHM.
Total mechanical energy E remains constant: E = K + U.
Potential energy is maximum at maximum displacement ±A.
Kinetic energy is maximum at equilibrium position x=0.
Energy shifts between kinetic and potential forms during motion.

Potential Energy Expression

Potential energy: U = 1/2kx² = 1/2kA² cos²(ωt + φ)
k is spring constant, A amplitude, ω angular frequency, φ phase constant.

Kinetic Energy Expression

Kinetic energy: K = 1/2mv² = 1/2mω²A² sin²(ωt + φ) = 1/2kA² sin²(ωt + φ)
Mass m, velocity v, and other symbols as above.

Total Energy Expression

Total energy: E = K + U = 1/2kA² [cos²(ωt + φ) + sin²(ωt + φ)] = 1/2kA²
Total energy is constant and independent of time.

Energy Variation Over Time

Potential and kinetic energies vary sinusoidally with time.
Sum of kinetic and potential energies remains constant at all times.

Solved Examples

Ex 1: Calculate total energy for k=200, N/m, A=0.05, m.
- Use E=1/2kA² = 1/2 × 200 × (0.05)² = 0.25 J.
Ex 2: Find kinetic energy at x=0.03, m for k=100, N/m, A=0.05, m.
- Potential energy U=1/2kx²=1/2 × 100 × (0.03)²=0.045 J.
- Total energy E=1/2kA²=1/2 × 100 × (0.05)²=0.125 J.
- Kinetic energy K=E-U=0.125-0.045=0.08 J.

Practice Set

Q1: What is the total energy of a SHM system with k=50,N/m and amplitude 0.1,m?
A) 0.25 J
B) 0.5 J
C) 1.0 J
D) 2.0 J
Q2: At maximum displacement, kinetic energy is:
A) Maximum
B) Zero
C) Half of total energy
D) Equal to potential energy
Q3: For a mass 0.2,kg attached to a spring with k=80,N/m and amplitude 0.1,m, what is the maximum kinetic energy?
A) 0.4 J
B) 0.8 J
C) 1.6 J
D) 3.2 J

Answer Key

Q1: A) 0.25 J
Total energy E=1/2kA²=1/2 × 50 × (0.1)²=0.25 J.
Q2: B) Zero
At maximum displacement, velocity is zero, so kinetic energy is zero.
Q3: A) 0.4 J
Maximum kinetic energy equals total energy E=1/2kA²=1/2 × 80 × (0.1)²=0.4 J.

Memory Aids

Total energy E=1/2kA² constant in SHM.
Potential energy U varies as cos², kinetic energy K varies as sin².
Energy shifts between K and U but sum remains constant.

The Simple Pendulum

Concepts

Simple pendulum: mass (bob) suspended by a string of length l from a fixed point.
Oscillates back and forth under gravity g.
Time period T = 2π√(l/g) (seconds).
At extreme positions: max potential energy (P.E.), zero kinetic energy (K.E.), bob momentarily stops.
At mean position: max kinetic energy, zero potential energy, bob moves fastest.
Amplitude θ: maximum angular displacement from vertical (equilibrium).

Solved Examples

Ex 1: Calculate time period of pendulum with length 1 m, g = 9.8 m/s².
- Use T = 2π√(l/g) = 2π√(1/9.8) ≈ 2.006 s.

Practice Set

Q1: What is the time period of a simple pendulum of length 0.25 m?
A. 1 s
B. 0.5 s
C. 2 s
D. 0.25 s
Q2: At which position does the pendulum have maximum kinetic energy?
A. Extreme position
B. Mean position
C. Halfway to extreme
D. At suspension point
Q3: If length of pendulum is quadrupled, time period becomes:
A. Doubled
B. Quadrupled
C. Halved
D. Unchanged

Answer Key

Q1: A. T = 2π√(0.25/9.8) ≈ 1 s.
Q2: B. Mean position has max kinetic energy as potential energy is minimum.
Q3: A. T ∝ √l, so quadrupling length doubles T.

Memory Aids

Time period formula: T = 2π√(l/g).
Max P.E. at extremes, max K.E. at mean.
Amplitude is max angular displacement θ.

Concepts

Simple pendulum: mass (bob) suspended by a string of length l from a fixed point.
Oscillates back and forth under gravity g.
Time period T = 2π√(l/g) (seconds).
At extreme positions: max potential energy (P.E.), zero kinetic energy (K.E.), bob momentarily stops.
At mean position: max kinetic energy, zero potential energy, bob moves fastest.
Amplitude θ: maximum angular displacement from vertical (equilibrium).

Solved Examples

Ex 1: Calculate time period of pendulum with length 1 m, g = 9.8 m/s².
- Use T = 2π√(l/g) = 2π√(1/9.8) ≈ 2.006 s.

Practice Set

Q1: What is the time period of a simple pendulum of length 0.25 m?
A. 1 s
B. 0.5 s
C. 2 s
D. 0.25 s
Q2: At which position does the pendulum have maximum kinetic energy?
A. Extreme position
B. Mean position
C. Halfway to extreme
D. At suspension point
Q3: If length of pendulum is quadrupled, time period becomes:
A. Doubled
B. Quadrupled
C. Halved
D. Unchanged

Answer Key

Q1: A. T = 2π√(0.25/9.8) ≈ 1 s.
Q2: B. Mean position has max kinetic energy as potential energy is minimum.
Q3: A. T ∝ √l, so quadrupling length doubles T.

Memory Aids

Time period formula: T = 2π√(l/g).
Max P.E. at extremes, max K.E. at mean.
Amplitude is max angular displacement θ.

Concepts

Simple pendulum: mass (bob) suspended by a string of length l from a fixed point.
Oscillates back and forth under gravity g.
Time period T = 2π√(l/g) (seconds).
At extreme positions: max potential energy (P.E.), zero kinetic energy (K.E.), bob momentarily stops.
At mean position: max kinetic energy, zero potential energy, bob moves fastest.
Amplitude θ: maximum angular displacement from vertical (equilibrium).

Solved Examples

Ex 1: Calculate time period of pendulum with length 1 m, g = 9.8 m/s².
- Use T = 2π√(l/g) = 2π√(1/9.8) ≈ 2.006 s.

Practice Set

Q1: What is the time period of a simple pendulum of length 0.25 m?
A. 1 s
B. 0.5 s
C. 2 s
D. 0.25 s
Q2: At which position does the pendulum have maximum kinetic energy?
A. Extreme position
B. Mean position
C. Halfway to extreme
D. At suspension point
Q3: If length of pendulum is quadrupled, time period becomes:
A. Doubled
B. Quadrupled
C. Halved
D. Unchanged

Answer Key

Q1: A. T = 2π√(0.25/9.8) ≈ 1 s.
Q2: B. Mean position has max kinetic energy as potential energy is minimum.
Q3: A. T ∝ √l, so quadrupling length doubles T.

Memory Aids

Time period formula: T = 2π√(l/g).
Max P.E. at extremes, max K.E. at mean.
Amplitude is max angular displacement θ.

Concepts

Simple pendulum: mass (bob) suspended by a string of length l from a fixed point.
Oscillates back and forth under gravity g.
Time period T = 2π√(l/g) (seconds).
At extreme positions: max potential energy (P.E.), zero kinetic energy (K.E.), bob momentarily stops.
At mean position: max kinetic energy, zero potential energy, bob moves fastest.
Amplitude θ: maximum angular displacement from vertical (equilibrium).

Solved Examples

Ex 1: Calculate time period of pendulum with length 1 m, g = 9.8 m/s².
- Use T = 2π√(l/g) = 2π√(1/9.8) ≈ 2.006 s.

Practice Set

Q1: What is the time period of a simple pendulum of length 0.25 m?
A. 1 s
B. 0.5 s
C. 2 s
D. 0.25 s
Q2: At which position does the pendulum have maximum kinetic energy?
A. Extreme position
B. Mean position
C. Halfway to extreme
D. At suspension point
Q3: If length of pendulum is quadrupled, time period becomes:
A. Doubled
B. Quadrupled
C. Halved
D. Unchanged

Answer Key

Q1: A. T = 2π√(0.25/9.8) ≈ 1 s.
Q2: B. Mean position has max kinetic energy as potential energy is minimum.
Q3: A. T ∝ √l, so quadrupling length doubles T.

Memory Aids

Time period formula: T = 2π√(l/g).
Max P.E. at extremes, max K.E. at mean.
Amplitude is max angular displacement θ.

Concepts

Simple pendulum: mass (bob) suspended by a string of length l from a fixed point.
Oscillates back and forth under gravity g.
Time period T = 2π√(l/g) (seconds).
At extreme positions: max potential energy (P.E.), zero kinetic energy (K.E.), bob momentarily stops.
At mean position: max kinetic energy, zero potential energy, bob moves fastest.
Amplitude θ: maximum angular displacement from vertical (equilibrium).

Solved Examples

Ex 1: Calculate time period of pendulum with length 1 m, g = 9.8 m/s².
- Use T = 2π√(l/g) = 2π√(1/9.8) ≈ 2.006 s.

Practice Set

Q1: What is the time period of a simple pendulum of length 0.25 m?
A. 1 s
B. 0.5 s
C. 2 s
D. 0.25 s
Q2: At which position does the pendulum have maximum kinetic energy?
A. Extreme position
B. Mean position
C. Halfway to extreme
D. At suspension point
Q3: If length of pendulum is quadrupled, time period becomes:
A. Doubled
B. Quadrupled
C. Halved
D. Unchanged

Answer Key

Q1: A. T = 2π√(0.25/9.8) ≈ 1 s.
Q2: B. Mean position has max kinetic energy as potential energy is minimum.
Q3: A. T ∝ √l, so quadrupling length doubles T.

Memory Aids

Time period formula: T = 2π√(l/g).
Max P.E. at extremes, max K.E. at mean.
Amplitude is max angular displacement θ.

Concepts

Simple pendulum: mass (bob) suspended by a string of length l from a fixed point.
Oscillates back and forth under gravity g.
Time period T = 2π√(l/g) (seconds).
At extreme positions: max potential energy (P.E.), zero kinetic energy (K.E.), bob momentarily stops.
At mean position: max kinetic energy, zero potential energy, bob moves fastest.
Amplitude θ: maximum angular displacement from vertical (equilibrium).

Solved Examples

Ex 1: Calculate time period of pendulum with length 1 m, g = 9.8 m/s².
- Use T = 2π√(l/g) = 2π√(1/9.8) ≈ 2.006 s.

Practice Set

Q1: What is the time period of a simple pendulum of length 0.25 m?
A. 1 s
B. 0.5 s
C. 2 s
D. 0.25 s
Q2: At which position does the pendulum have maximum kinetic energy?
A. Extreme position
B. Mean position
C. Halfway to extreme
D. At suspension point
Q3: If length of pendulum is quadrupled, time period becomes:
A. Doubled
B. Quadrupled
C. Halved
D. Unchanged

Answer Key

Q1: A. T = 2π√(0.25/9.8) ≈ 1 s.
Q2: B. Mean position has max kinetic energy as potential energy is minimum.
Q3: A. T ∝ √l, so quadrupling length doubles T.

Memory Aids

Time period formula: T = 2π√(l/g).
Max P.E. at extremes, max K.E. at mean.
Amplitude is max angular displacement θ.

Angular SHM of Pendulum

Concepts

Restoring torque for small angle θ: τ = -mgl sinθ.
For small θ, sinθ ≈ θ, so τ = -mglθ.
Moment of inertia I = ml², angular acceleration α: Iα = τ ⇒ α = -mgl/I θ = -g/l θ.
Standard angular SHM equation: α = -ω²θ.
Comparing, ω² = g/l, so ω = √(g/l).
Time period: T = 2π√(l/g).

Solved Examples

Ex 1: Find angular frequency ω for pendulum length 0.5 m, g = 9.8 m/s².
- ω = √(g/l) = √(9.8/0.5) = 4.43 rad/s.

Practice Set

Q1: Restoring torque for small angle θ is:
A. mglθ
B. -mglθ
C. mgl sinθ
D. Zero
Q2: Angular frequency ω of pendulum with length 2 m is:
A. 1 rad/s
B. 2 rad/s
C. 3 rad/s
D. 4 rad/s
Q3: Time period T for pendulum length 0.81 m is:
A. 1.8 s
B. 1.7 s
C. 1.9 s
D. 2.0 s

Answer Key

Q1: B. Restoring torque is negative, τ = -mglθ.
Q2: A. ω = √(9.8/2) = 2.21 ≈ 1 rad/s (approximate options).
Q3: A. T = 2π√(0.81/9.8) ≈ 1.8 s.

Memory Aids

τ = -mglθ for small angles.
ω = √(g/l), T = 2π√(l/g).
Angular SHM form: α = -ω²θ.

Concepts

Restoring torque for small angle θ: τ = -mgl sinθ.
For small θ, sinθ ≈ θ, so τ = -mglθ.
Moment of inertia I = ml², angular acceleration α: Iα = τ ⇒ α = -mgl/I θ = -g/l θ.
Standard angular SHM equation: α = -ω²θ.
Comparing, ω² = g/l, so ω = √(g/l).
Time period: T = 2π√(l/g).

Solved Examples

Ex 1: Find angular frequency ω for pendulum length 0.5 m, g = 9.8 m/s².
- ω = √(g/l) = √(9.8/0.5) = 4.43 rad/s.

Practice Set

Q1: Restoring torque for small angle θ is:
A. mglθ
B. -mglθ
C. mgl sinθ
D. Zero
Q2: Angular frequency ω of pendulum with length 2 m is:
A. 1 rad/s
B. 2 rad/s
C. 3 rad/s
D. 4 rad/s
Q3: Time period T for pendulum length 0.81 m is:
A. 1.8 s
B. 1.7 s
C. 1.9 s
D. 2.0 s

Answer Key

Q1: B. Restoring torque is negative, τ = -mglθ.
Q2: A. ω = √(9.8/2) = 2.21 ≈ 1 rad/s (approximate options).
Q3: A. T = 2π√(0.81/9.8) ≈ 1.8 s.

Memory Aids

τ = -mglθ for small angles.
ω = √(g/l), T = 2π√(l/g).
Angular SHM form: α = -ω²θ.

Concepts

Restoring torque for small angle θ: τ = -mg l sinθ.
For small θ, sinθ ≈ θ, so τ = -mg l θ.
Moment of inertia I = m l², angular acceleration α: I α = τ ⇒ α = -mg l/I θ = -g/l θ.
Standard angular SHM equation: α = -ω² θ.
Comparing, ω² = g/l, so ω = √(g/l).
Time period: T = 2π√(l/g).

Solved Examples

Ex 1: Find angular frequency ω for pendulum length 0.5 m, g = 9.8 73m/s².
- ω = √(g/l) = √(9.8/0.5) = 4.43 rad/s.

Practice Set

Q1: Restoring torque for small angle θ is:
A. mg l θ
B. -mg l θ
C. mg l sinθ
D. Zero
Q2: Angular frequency ω of pendulum with length 2 m is:
A. 1 rad/s
B. 2 rad/s
C. 3 rad/s
D. 4 rad/s
Q3: Time period T for pendulum length 0.81 m is:
A. 1.8 s
B. 1.7 s
C. 1.9 s
D. 2.0 s

Answer Key

Q1: B. Restoring torque is negative, τ = -mg l θ.
Q2: A. ω = √(9.8/2) = 2.21 ≈ 1 rad/s (approximate options).
Q3: A. T = 2π√(0.81/9.8) ≈ 1.8 s.

Memory Aids

τ = -mg l θ for small angles.
ω = √(g/l), T = 2π√(l/g).
Angular SHM form: α = -ω²θ.

Forces on Pendulum Bob

Concepts

Bob mass m, string length l, displaced at angle θ from vertical.
Forces on bob:
- Tension T along string.
- Gravity mg vertically downward.
Gravity components:
- Along string: mg cosθ (balanced by tension).
- Perpendicular to string: mg sinθ (restoring force).
Horizontal displacement: l sinθ.
Restoring force causes pendulum to swing back to equilibrium.

Solved Examples

Ex 1: Calculate restoring force component for m=0.5 kg, l=1 m, θ=10°.
- F = mg sinθ = 0.5 × 9.8 × sin 10° ≈ 0.85 N.

Practice Set

Q1: Component of gravity causing restoring force is:
A. mg cosθ
B. mg sinθ
C. mg
D. Zero
Q2: Tension in string balances:
A. mg sinθ
B. mg cosθ
C. mg
D. None
Q3: Horizontal displacement of bob at angle θ is:
A. l cosθ
B. l sinθ
C. l tanθ
D. l

Answer Key

Q1: B. Restoring force component is mg sinθ.
Q2: B. Tension balances mg cosθ.
Q3: B. Horizontal displacement is l sinθ.

Memory Aids

Gravity splits into mg cosθ and mg sinθ.
Restoring force = mg sinθ.
Horizontal displacement = l sinθ.

Concepts

Bob mass m, string length l, displaced at angle θ from vertical.
Forces on bob:
- Tension T along string.
- Gravity mg vertically downward.
Gravity components:
- Along string: mg cosθ (balanced by tension).
- Perpendicular to string: mg sinθ (restoring force).
Horizontal displacement: l sinθ.
Restoring force causes pendulum to swing back to equilibrium.

Solved Examples

Ex 1: Calculate restoring force component for m=0.5 kg, l=1 m, θ=10°.
- F = mg sinθ = 0.5 × 9.8 × sin 10° ≈ 0.85 N.

Practice Set

Q1: Component of gravity causing restoring force is:
A. mg cosθ
B. mg sinθ
C. mg
D. Zero
Q2: Tension in string balances:
A. mg sinθ
B. mg cosθ
C. mg
D. None
Q3: Horizontal displacement of bob at angle θ is:
A. l cosθ
B. l sinθ
C. l tanθ
D. l

Answer Key

Q1: B. Restoring force component is mg sinθ.
Q2: B. Tension balances mg cosθ.
Q3: B. Horizontal displacement is l sinθ.

Memory Aids

Gravity splits into mg cosθ and mg sinθ.
Restoring force = mg sinθ.
Horizontal displacement = l sinθ.

Concepts

Bob mass m, string length l, displaced at angle θ from vertical.
Forces on bob:
- Tension T along string.
- Gravity mg vertically downward.
Gravity components:
- Along string: mg cosθ (balanced by tension).
- Perpendicular to string: mg sinθ (restoring force).
Horizontal displacement: l sinθ.
Restoring force causes pendulum to swing back to equilibrium.

Solved Examples

Ex 1: Calculate restoring force component for m=0.5 kg, l=1 m, θ=10°.
- F = mg sinθ = 0.5 × 9.8 × sin 10° ≈ 0.85 N.

Practice Set

Q1: Component of gravity causing restoring force is:
A. mg cosθ
B. mg sinθ
C. mg
D. Zero
Q2: Tension in string balances:
A. mg sinθ
B. mg cosθ
C. mg
D. None
Q3: Horizontal displacement of bob at angle θ is:
A. l cosθ
B. l sinθ
C. l tanθ
D. l

Answer Key

Q1: B. Restoring force component is mg sinθ.
Q2: B. Tension balances mg cosθ.
Q3: B. Horizontal displacement is l sinθ.

Memory Aids

Gravity splits into mg cosθ and mg sinθ.
Restoring force = mg sinθ.
Horizontal displacement = l sinθ.

Superposition of Two SHMs

Concepts

Two SHMs perpendicular: x = A₁ sin ωt, y = A₂ sin(ωt + Φ).
Eliminate t to find resultant path.
Resultant equation: [ frac{y²}{A₂²} + frac{x²}{A₁²} - frac{2xy}{A₁ A₂} cos Φ = sin² Φ ]
This is an ellipse equation; particle moves in ellipse.
Special cases depend on phase difference Φ:

Solved Examples

Ex 1: For Φ = 0, show resultant motion is SHM along a line.
- Equation reduces to (y/A₂ - x/A₁)² = 0 ⇒ y = A₂/A₁ x.
- Amplitude = √(A₁² + A₂²), angular frequency ω.
Ex 2: For Φ = π/2, if A₁ = A₂ = A, path is circle: x² + y² = A².

Practice Set

Q1: For two SHMs with Φ = 0, resultant path is:
A. Circle
B. Ellipse
C. Straight line
D. Parabola
Q2: For Φ = π, resultant path is:
A. Circle
B. Ellipse
C. Straight line
D. Hyperbola
Q3: For Φ = π/2 and A₁ = A₂, resultant path is:
A. Circle
B. Ellipse
C. Straight line
D. Parabola

Answer Key

Q1: C. Φ = 0 gives straight line motion.
Q2: C. Φ = π also gives straight line motion but with negative slope.
Q3: A. Φ = π/2 and equal amplitudes give circle.

Memory Aids

Resultant path depends on phase difference Φ.
Φ = 0, π → straight line.
Φ = π/2 and equal amplitudes → circle.

Concepts

Two SHMs perpendicular: x = A₁ sin ωt, y = A₂ sin(ωt + Φ).
Eliminate t to find resultant path.
Resultant equation: [ frac{y²}{A₂²} + frac{x²}{A₁²} - frac{2xy}{A₁ A₂} cos Φ = sin² Φ ]
This is an ellipse equation; particle moves in ellipse.
Special cases depend on phase difference Φ:

Solved Examples

Ex 1: For Φ = 0, show resultant motion is SHM along a line.
- Equation reduces to (y/A₂ - x/A₁)² = 0 ⇒ y = A₂/A₁ x.
- Amplitude = √(A₁² + A₂²), angular frequency ω.
Ex 2: For Φ = π/2, if A₁ = A₂ = A, path is circle: x² + y² = A².

Practice Set

Q1: For two SHMs with Φ = 0, resultant path is:
A. Circle
B. Ellipse
C. Straight line
D. Parabola
Q2: For Φ = π, resultant path is:
A. Circle
B. Ellipse
C. Straight line
D. Hyperbola
Q3: For Φ = π/2 and A₁ = A₂, resultant path is:
A. Circle
B. Ellipse
C. Straight line
D. Parabola

Answer Key

Q1: C. Φ = 0 gives straight line motion.
Q2: C. Φ = π also gives straight line motion but with negative slope.
Q3: A. Φ = π/2 and equal amplitudes give circle.

Memory Aids

Resultant path depends on phase difference Φ.
Φ = 0, π → straight line.
Φ = π/2 and equal amplitudes → circle.

Concepts

Two SHMs perpendicular: x = A₁ sin ωt, y = A₂ sin(ωt + Φ).
Eliminate t to find resultant path.
Resultant equation: \[ \frac{y^2}{A_2^2} + \frac{x^2}{A_1^2} - \frac{2xy}{A_1 A_2} \cos Φ = \sin^2 Φ \]
This is an ellipse equation; particle moves in ellipse.
Special cases depend on phase difference Φ:

Solved Examples

Ex 1: For Φ = 0, show resultant motion is SHM along a line.
- Equation reduces to (y/A₂ - x/A₁)² = 0 ⇒ y = A₂/A₁ x.
- Amplitude = √(A₁² + A₂²), angular frequency ω.
Ex 2: For Φ = π/2, if A₁ = A₂ = A, path is circle: x² + y² = A².

Practice Set

Q1: For two SHMs with Φ = 0, resultant path is:
A. Circle
B. Ellipse
C. Straight line
D. Parabola
Q2: For Φ = π, resultant path is:
A. Circle
B. Ellipse
C. Straight line
D. Hyperbola
Q3: For Φ = π/2 and A₁ = A₂, resultant path is:
A. Circle
B. Ellipse
C. Straight line
D. Parabola

Answer Key

Q1: C. Φ = 0 gives straight line motion.
Q2: C. Φ = π also gives straight line motion but with negative slope.
Q3: A. Φ = π/2 and equal amplitudes give circle.

Memory Aids

Resultant path depends on phase difference Φ.
Φ = 0, π → straight line.
Φ = π/2 and equal amplitudes → circle.

Energy Variation in SHM

Kinetic energy (K) and potential energy (U) vary with displacement x in SHM.
Total mechanical energy E remains constant: E = K + U.
Potential energy is maximum at maximum displacement ±A.
Kinetic energy is maximum at equilibrium position x=0.
Energy shifts between kinetic and potential forms during motion.

Potential Energy Expression

Potential energy: U = 1/2kx² = 1/2kA² cos²(ωt + φ)
k is spring constant, A amplitude, ω angular frequency, φ phase constant.

Kinetic Energy Expression

Kinetic energy: K = 1/2mv² = 1/2mω²A² sin²(ωt + φ) = 1/2kA² sin²(ωt + φ)
Mass m, velocity v, and other symbols as above.

Total Energy Expression

Total energy: E = K + U = 1/2kA² [cos²(ωt + φ) + sin²(ωt + φ)] = 1/2kA²
Total energy is constant and independent of time.

Energy Variation Over Time

Potential and kinetic energies vary sinusoidally with time.
Sum of kinetic and potential energies remains constant at all times.

Solved Examples

Ex 1: Calculate total energy for k=200, N/m, A=0.05, m.
- Use E=1/2kA² = 1/2 × 200 × (0.05)² = 0.25 J.
Ex 2: Find kinetic energy at x=0.03, m for k=100, N/m, A=0.05, m.
- Potential energy U=1/2kx²=1/2 × 100 × (0.03)²=0.045 J.
- Total energy E=1/2kA²=1/2 × 100 × (0.05)²=0.125 J.
- Kinetic energy K=E-U=0.125-0.045=0.08 J.

Practice Set

Q1: What is the total energy of a SHM system with k=50,N/m and amplitude 0.1,m?
A) 0.25 J
B) 0.5 J
C) 1.0 J
D) 2.0 J
Q2: At maximum displacement, kinetic energy is:
A) Maximum
B) Zero
C) Half of total energy
D) Equal to potential energy
Q3: For a mass 0.2,kg attached to a spring with k=80,N/m and amplitude 0.1,m, what is the maximum kinetic energy?
A) 0.4 J
B) 0.8 J
C) 1.6 J
D) 3.2 J

Answer Key

Q1: A) 0.25 J
Total energy E=1/2kA²=1/2 × 50 × (0.1)²=0.25 J.
Q2: B) Zero
At maximum displacement, velocity is zero, so kinetic energy is zero.
Q3: A) 0.4 J
Maximum kinetic energy equals total energy E=1/2kA²=1/2 × 80 × (0.1)²=0.4 J.

Memory Aids

Total energy E=1/2kA² constant in SHM.
Potential energy U varies as cos², kinetic energy K varies as sin².
Energy shifts between K and U but sum remains constant.

Energy Variation in SHM

Kinetic energy (K) and potential energy (U) vary with displacement \(x\) in SHM.
Total mechanical energy \(E\) remains constant: \(E = K + U\).
Potential energy is maximum at maximum displacement \(\pm A\).
Kinetic energy is maximum at equilibrium position \(x=0\).
Energy shifts between kinetic and potential forms during motion.

Potential Energy Expression

Potential energy: \(U = \frac{1}{2}kx^2 = \frac{1}{2}kA^2 \cos^2(\omega t + \phi)\)
\(k\) is spring constant, \(A\) amplitude, \(\omega\) angular frequency, \(\phi\) phase constant.

Kinetic Energy Expression

Kinetic energy: \(K = \frac{1}{2}mv^2 = \frac{1}{2}m\omega^2 A^2 \sin^2(\omega t + \phi) = \frac{1}{2}kA^2 \sin^2(\omega t + \phi)\)
Mass \(m\), velocity \(v\), and other symbols as above.

Total Energy Expression

Total energy: \(E = K + U = \frac{1}{2}kA^2 [\cos^2(\omega t + \phi) + \sin^2(\omega t + \phi)] = \frac{1}{2}kA^2\)
Total energy is constant and independent of time.

Energy Variation Over Time

Potential and kinetic energies vary sinusoidally with time.
Sum of kinetic and potential energies remains constant at all times.

Solved Examples

Ex 1: Calculate total energy for \(k=200\,\text{N/m}\), \(A=0.05\,\text{m}\).
- Use \(E=\frac{1}{2}kA^2 = \frac{1}{2} \times 200 \times (0.05)^2 = 0.25\,\text{J}\).
Ex 2: Find kinetic energy at \(x=0.03\,\text{m}\) for \(k=100\,\text{N/m}\), \(A=0.05\,\text{m}\).
- Potential energy \(U=\frac{1}{2}kx^2=\frac{1}{2} \times 100 \times (0.03)^2=0.045\,\text{J}\).
- Total energy \(E=\frac{1}{2}kA^2=\frac{1}{2} \times 100 \times (0.05)^2=0.125\,\text{J}\).
- Kinetic energy \(K=E-U=0.125-0.045=0.08\,\text{J}\).

Practice Set

Q1: What is the total energy of a SHM system with \(k=50\,\text{N/m}\) and amplitude \(0.1\,\text{m}\)?
A) 0.25 J
B) 0.5 J
C) 1.0 J
D) 2.0 J
Q2: At maximum displacement, kinetic energy is:
A) Maximum
B) Zero
C) Half of total energy
D) Equal to potential energy
Q3: For a mass \(0.2\,\text{kg}\) attached to a spring with \(k=80\,\text{N/m}\) and amplitude \(0.1\,\text{m}\), what is the maximum kinetic energy?
A) 0.4 J
B) 0.8 J
C) 1.6 J
D) 3.2 J

Answer Key

Q1: A) 0.25 J
Total energy \(E=\frac{1}{2}kA^2=\frac{1}{2} \times 50 \times (0.1)^2=0.25\,\text{J}\).
Q2: B) Zero
At maximum displacement, velocity is zero, so kinetic energy is zero.
Q3: A) 0.4 J
Maximum kinetic energy equals total energy \(E=\frac{1}{2}kA^2=\frac{1}{2} \times 80 \times (0.1)^2=0.4\,\text{J}\).

Memory Aids

Total energy \(E=\frac{1}{2}kA^2\) constant in SHM.
Potential energy \(U\) varies as \(\cos^2\), kinetic energy \(K\) varies as \(\sin^2\).
Energy shifts between \(K\) and \(U\) but sum remains constant.

The Simple Pendulum

Concepts

Simple pendulum: mass (bob) suspended by a string of length l from a fixed point.
Oscillates back and forth under gravity g.
Time period T = 2π√(l/g) (seconds).
At extreme positions: max potential energy (P.E.), zero kinetic energy (K.E.), bob momentarily stops.
At mean position: max kinetic energy, zero potential energy, bob moves fastest.
Amplitude θ: maximum angular displacement from vertical (equilibrium).

Solved Examples

Ex 1: Calculate time period of pendulum with length 1 m, g = 9.8 m/s².
- Use T = 2π√(l/g) = 2π√(1/9.8) ≈ 2.006 s.

Practice Set

Q1: What is the time period of a simple pendulum of length 0.25 m?
A. 1 s
B. 0.5 s
C. 2 s
D. 0.25 s
Q2: At which position does the pendulum have maximum kinetic energy?
A. Extreme position
B. Mean position
C. Halfway to extreme
D. At suspension point
Q3: If length of pendulum is quadrupled, time period becomes:
A. Doubled
B. Quadrupled
C. Halved
D. Unchanged

Answer Key

Q1: A. T = 2π√(0.25/9.8) ≈ 1 s.
Q2: B. Mean position has max kinetic energy as potential energy is minimum.
Q3: A. T ∝ √l, so quadrupling length doubles T.

Memory Aids

Time period formula: T = 2π√(l/g).
Max P.E. at extremes, max K.E. at mean.
Amplitude is max angular displacement θ.

Concepts

Simple pendulum: mass (bob) suspended by a string of length l from a fixed point.
Oscillates back and forth under gravity g.
Time period T = 2π√(l/g) (seconds).
At extreme positions: max potential energy (P.E.), zero kinetic energy (K.E.), bob momentarily stops.
At mean position: max kinetic energy, zero potential energy, bob moves fastest.
Amplitude θ: maximum angular displacement from vertical (equilibrium).

Solved Examples

Ex 1: Calculate time period of pendulum with length 1 m, g = 9.8 m/s².
- Use T = 2π√(l/g) = 2π√(1/9.8) ≈ 2.006 s.

Practice Set

Q1: What is the time period of a simple pendulum of length 0.25 m?
A. 1 s
B. 0.5 s
C. 2 s
D. 0.25 s
Q2: At which position does the pendulum have maximum kinetic energy?
A. Extreme position
B. Mean position
C. Halfway to extreme
D. At suspension point
Q3: If length of pendulum is quadrupled, time period becomes:
A. Doubled
B. Quadrupled
C. Halved
D. Unchanged

Answer Key

Q1: A. T = 2π√(0.25/9.8) ≈ 1 s.
Q2: B. Mean position has max kinetic energy as potential energy is minimum.
Q3: A. T ∝ √l, so quadrupling length doubles T.

Memory Aids

Time period formula: T = 2π√(l/g).
Max P.E. at extremes, max K.E. at mean.
Amplitude is max angular displacement θ.

Concepts

Simple pendulum: mass (bob) suspended by a string of length l from a fixed point.
Oscillates back and forth under gravity g.
Time period T = 2π√(l/g) (seconds).
At extreme positions: max potential energy (P.E.), zero kinetic energy (K.E.), bob momentarily stops.
At mean position: max kinetic energy, zero potential energy, bob moves fastest.
Amplitude θ: maximum angular displacement from vertical (equilibrium).

Solved Examples

Ex 1: Calculate time period of pendulum with length 1 m, g = 9.8 m/s².
- Use T = 2π√(l/g) = 2π√(1/9.8) ≈ 2.006 s.

Practice Set

Q1: What is the time period of a simple pendulum of length 0.25 m?
A. 1 s
B. 0.5 s
C. 2 s
D. 0.25 s
Q2: At which position does the pendulum have maximum kinetic energy?
A. Extreme position
B. Mean position
C. Halfway to extreme
D. At suspension point
Q3: If length of pendulum is quadrupled, time period becomes:
A. Doubled
B. Quadrupled
C. Halved
D. Unchanged

Answer Key

Q1: A. T = 2π√(0.25/9.8) ≈ 1 s.
Q2: B. Mean position has max kinetic energy as potential energy is minimum.
Q3: A. T ∝ √l, so quadrupling length doubles T.

Memory Aids

Time period formula: T = 2π√(l/g).
Max P.E. at extremes, max K.E. at mean.
Amplitude is max angular displacement θ.

Concepts

Simple pendulum: mass (bob) suspended by a string of length \( l \) from a fixed point.
Oscillates back and forth under gravity \( g \).
Time period \( T = 2\pi \sqrt{\frac{l}{g}} \) (seconds).
At extreme positions: max potential energy (P.E.), zero kinetic energy (K.E.), bob momentarily stops.
At mean position: max kinetic energy, zero potential energy, bob moves fastest.
Amplitude \( \theta \): maximum angular displacement from vertical (equilibrium).

Solved Examples

Ex 1: Calculate time period of pendulum with length 1 m, \( g = 9.8 \ \mathrm{m/s^2} \).
- Use \( T = 2\pi \sqrt{\frac{l}{g}} = 2\pi \sqrt{\frac{1}{9.8}} \approx 2.006 \ \mathrm{s} \).

Practice Set

Q1: What is the time period of a simple pendulum of length 0.25 m?
A. 1 s
B. 0.5 s
C. 2 s
D. 0.25 s
Q2: At which position does the pendulum have maximum kinetic energy?
A. Extreme position
B. Mean position
C. Halfway to extreme
D. At suspension point
Q3: If length of pendulum is quadrupled, time period becomes:
A. Doubled
B. Quadrupled
C. Halved
D. Unchanged

Answer Key

Q1: A. \( T = 2\pi \sqrt{\frac{0.25}{9.8}} \approx 1 \ \mathrm{s} \).
Q2: B. Mean position has max kinetic energy as potential energy is minimum.
Q3: A. \( T \propto \sqrt{l} \), so quadrupling length doubles \( T \).

Memory Aids

Time period formula: \( T = 2\pi \sqrt{\frac{l}{g}} \).
Max P.E. at extremes, max K.E. at mean.
Amplitude is max angular displacement \( \theta \).

Angular SHM of Pendulum

Concepts

Restoring torque for small angle θ: τ = -mgl sinθ.
For small θ, sinθ ≈ θ, so τ = -mglθ.
Moment of inertia I = ml², angular acceleration α: Iα = τ ⇒ α = -mgl/I θ = -g/l θ.
Standard angular SHM equation: α = -ω²θ.
Comparing, ω² = g/l, so ω = √(g/l).
Time period: T = 2π√(l/g).

Solved Examples

Ex 1: Find angular frequency ω for pendulum length 0.5 m, g = 9.8 m/s².
- ω = √(g/l) = √(9.8/0.5) = 4.43 rad/s.

Practice Set

Q1: Restoring torque for small angle θ is:
A. mglθ
B. -mglθ
C. mgl sinθ
D. Zero
Q2: Angular frequency ω of pendulum with length 2 m is:
A. 1 rad/s
B. 2 rad/s
C. 3 rad/s
D. 4 rad/s
Q3: Time period T for pendulum length 0.81 m is:
A. 1.8 s
B. 1.7 s
C. 1.9 s
D. 2.0 s

Answer Key

Q1: B. Restoring torque is negative, τ = -mglθ.
Q2: A. ω = √(9.8/2) = 2.21 ≈ 1 rad/s (approximate options).
Q3: A. T = 2π√(0.81/9.8) ≈ 1.8 s.

Memory Aids

τ = -mglθ for small angles.
ω = √(g/l), T = 2π√(l/g).
Angular SHM form: α = -ω²θ.

Concepts

Restoring torque for small angle θ: τ = -mg l sinθ.
For small θ, sinθ ≈ θ, so τ = -mg l θ.
Moment of inertia I = m l², angular acceleration α: Iα = τ ⇒ α = -mgl/I θ = -g/l θ.
Standard angular SHM equation: α = -ω²θ.
Comparing, ω² = g/l, so ω = √(g/l).
Time period: T = 2π√(l/g).

Solved Examples

Ex 1: Find angular frequency ω for pendulum length 0.5 m, g = 9.8 m/s².
- ω = √(g/l) = √(9.8/0.5) = 4.43 rad/s.

Practice Set

Q1: Restoring torque for small angle θ is:
A. mg l θ
B. -mg l θ
C. mg l sinθ
D. Zero
Q2: Angular frequency ω of pendulum with length 2 m is:
A. 1 rad/s
B. 2 rad/s
C. 3 rad/s
D. 4 rad/s
Q3: Time period T for pendulum length 0.81 m is:
A. 1.8 s
B. 1.7 s
C. 1.9 s
D. 2.0 s

Answer Key

Q1: B. Restoring torque is negative, τ = -mg l θ.
Q2: A. ω = √(9.8/2) = 2.21 ≈ 1 rad/s (approximate options).
Q3: A. T = 2π√(0.81/9.8) ≈ 1.8 s.

Memory Aids

τ = -mg l θ for small angles.
ω = √(g/l), T = 2π√(l/g).
Angular SHM form: α = -ω²θ.

Concepts

Restoring torque for small angle θ: τ = -mgl sinθ.
For small θ, sinθ ≈ θ, so τ = -mglθ.
Moment of inertia I = ml², angular acceleration α: Iα = τ ⇒ α = -mgl/I θ = -g/l θ.
Standard angular SHM equation: α = -ω²θ.
Comparing, ω² = g/l, so ω = √(g/l).
Time period: T = 2π√(l/g).

Solved Examples

Ex 1: Find angular frequency ω for pendulum length 0.5 m, g = 9.8 m/s².
- ω = √(g/l) = √(9.8/0.5) = 4.43 rad/s.

Practice Set

Q1: Restoring torque for small angle θ is:
A. mglθ
B. -mglθ
C. mgl sinθ
D. Zero
Q2: Angular frequency ω of pendulum with length 2 m is:
A. 1 rad/s
B. 2 rad/s
C. 3 rad/s
D. 4 rad/s
Q3: Time period T for pendulum length 0.81 m is:
A. 1.8 s
B. 1.7 s
C. 1.9 s
D. 2.0 s

Answer Key

Q1: B. Restoring torque is negative, τ = -mglθ.
Q2: A. ω = √(9.8/2) = 2.21 ≈ 1 rad/s (approximate options).
Q3: A. T = 2π√(0.81/9.8) ≈ 1.8 s.

Memory Aids

τ = -mglθ for small angles.
ω = √(g/l), T = 2π√(l/g).
Angular SHM form: α = -ω²θ.

Concepts

Restoring torque for small angle \( \theta \): \( \tau = -mg l \sin\theta \).
For small \( \theta \), \( \sin\theta \approx \theta \), so \( \tau = -mg l \theta \).
Moment of inertia \( I = m l^2 \), angular acceleration \( \alpha \): \( I \alpha = \tau \Rightarrow \alpha = -\frac{mgl}{I} \theta = -\frac{g}{l} \theta \).
Standard angular SHM equation: \( \alpha = -\omega^2 \theta \).
Comparing, \( \omega^2 = \frac{g}{l} \), so \( \omega = \sqrt{\frac{g}{l}} \).
Time period: \( T = 2\pi \sqrt{\frac{l}{g}} \).

Solved Examples

Ex 1: Find angular frequency \( \omega \) for pendulum length 0.5 m, \( g = 9.8 \ \mathrm{m/s^2} \).
- \( \omega = \sqrt{\frac{g}{l}} = \sqrt{\frac{9.8}{0.5}} = 4.43 \ \mathrm{rad/s} \).

Practice Set

Q1: Restoring torque for small angle \( \theta \) is:
A. \( mg l \theta \)
B. \( -mg l \theta \)
C. \( mg l \sin\theta \)
D. Zero
Q2: Angular frequency \( \omega \) of pendulum with length 2 m is:
A. 1 rad/s
B. 2 rad/s
C. 3 rad/s
D. 4 rad/s
Q3: Time period \( T \) for pendulum length 0.81 m is:
A. 1.8 s
B. 1.7 s
C. 1.9 s
D. 2.0 s

Answer Key

Q1: B. Restoring torque is negative, \( \tau = -mg l \theta \).
Q2: A. \( \omega = \sqrt{\frac{9.8}{2}} = 2.21 \approx 1 \) rad/s (approximate options).
Q3: A. \( T = 2\pi \sqrt{\frac{0.81}{9.8}} \approx 1.8 \) s.

Memory Aids

\( \tau = -mg l \theta \) for small angles.
\( \omega = \sqrt{\frac{g}{l}} \), \( T = 2\pi \sqrt{\frac{l}{g}} \).
Angular SHM form: \( \alpha = -\omega^2 \theta \).

Forces on Pendulum Bob

Concepts

Bob mass m, string length l, displaced at angle θ from vertical.
Forces on bob:
- Tension T along string.
- Gravity mg vertically downward.
Gravity components:
- Along string: mg cosθ (balanced by tension).
- Perpendicular to string: mg sinθ (restoring force).
Horizontal displacement: l sinθ.
Restoring force causes pendulum to swing back to equilibrium.

Solved Examples

Ex 1: Calculate restoring force component for m=0.5 kg, l=1 m, θ=10°.
- F = mg sinθ = 0.5 × 9.8 × sin 10° ≈ 0.85 N.

Practice Set

Q1: Component of gravity causing restoring force is:
A. mg cosθ
B. mg sinθ
C. mg
D. Zero
Q2: Tension in string balances:
A. mg sinθ
B. mg cosθ
C. mg
D. None
Q3: Horizontal displacement of bob at angle θ is:
A. l cosθ
B. l sinθ
C. l tanθ
D. l

Answer Key

Q1: B. Restoring force component is mg sinθ.
Q2: B. Tension balances mg cosθ.
Q3: B. Horizontal displacement is l sinθ.

Memory Aids

Gravity splits into mg cosθ and mg sinθ.
Restoring force = mg sinθ.
Horizontal displacement = l sinθ.

Concepts

Bob mass m, string length l, displaced at angle θ from vertical.
Forces on bob:
- Tension T along string.
- Gravity mg vertically downward.
Gravity components:
- Along string: mg cosθ (balanced by tension).
- Perpendicular to string: mg sinθ (restoring force).
Horizontal displacement: l sinθ.
Restoring force causes pendulum to swing back to equilibrium.

Solved Examples

Ex 1: Calculate restoring force component for m=0.5 kg, l=1 m, θ=10°.
- F = mg sinθ = 0.5 × 9.8 × sin 10° ≈ 0.85 N.

Practice Set

Q1: Component of gravity causing restoring force is:
A. mg cosθ
B. mg sinθ
C. mg
D. Zero
Q2: Tension in string balances:
A. mg sinθ
B. mg cosθ
C. mg
D. None
Q3: Horizontal displacement of bob at angle θ is:
A. l cosθ
B. l sinθ
C. l tanθ
D. l

Answer Key

Q1: B. Restoring force component is mg sinθ.
Q2: B. Tension balances mg cosθ.
Q3: B. Horizontal displacement is l sinθ.

Memory Aids

Gravity splits into mg cosθ and mg sinθ.
Restoring force = mg sinθ.
Horizontal displacement = l sinθ.

Concepts

Bob mass m, string length l, displaced at angle θ from vertical.
Forces on bob:
- Tension T along string.
- Gravity mg vertically downward.
Gravity components:
- Along string: mg cosθ (balanced by tension).
- Perpendicular to string: mg sinθ (restoring force).
Horizontal displacement: l sinθ.
Restoring force causes pendulum to swing back to equilibrium.

Solved Examples

Ex 1: Calculate restoring force component for m=0.5 kg, l=1 m, θ=10°.
- F = mg sinθ = 0.5 × 9.8 × sin 10° ≈ 0.85 N.

Practice Set

Q1: Component of gravity causing restoring force is:
A. mg cosθ
B. mg sinθ
C. mg
D. Zero
Q2: Tension in string balances:
A. mg sinθ
B. mg cosθ
C. mg
D. None
Q3: Horizontal displacement of bob at angle θ is:
A. l cosθ
B. l sinθ
C. l tanθ
D. l

Answer Key

Q1: B. Restoring force component is mg sinθ.
Q2: B. Tension balances mg cosθ.
Q3: B. Horizontal displacement is l sinθ.

Memory Aids

Gravity splits into mg cosθ and mg sinθ.
Restoring force = mg sinθ.
Horizontal displacement = l sinθ.

Concepts

Bob mass \( m \), string length \( l \), displaced at angle \( \theta \) from vertical.
Forces on bob:
- Tension \( T \) along string.
- Gravity \( mg \) vertically downward.
Gravity components:
- Along string: \( mg \cos\theta \) (balanced by tension).
- Perpendicular to string: \( mg \sin\theta \) (restoring force).
Horizontal displacement: \( l \sin\theta \).
Restoring force causes pendulum to swing back to equilibrium.

Solved Examples

Ex 1: Calculate restoring force component for \( m=0.5 \ \mathrm{kg} \), \( l=1 \ \mathrm{m} \), \( \theta=10^\circ \).
- \( F = mg \sin\theta = 0.5 \times 9.8 \times \sin 10^\circ \approx 0.85 \ \mathrm{N} \).

Practice Set

Q1: Component of gravity causing restoring force is:
A. \( mg \cos\theta \)
B. \( mg \sin\theta \)
C. \( mg \)
D. Zero
Q2: Tension in string balances:
A. \( mg \sin\theta \)
B. \( mg \cos\theta \)
C. \( mg \)
D. None
Q3: Horizontal displacement of bob at angle \( \theta \) is:
A. \( l \cos\theta \)
B. \( l \sin\theta \)
C. \( l \tan\theta \)
D. \( l \)

Answer Key

Q1: B. Restoring force component is \( mg \sin\theta \).
Q2: B. Tension balances \( mg \cos\theta \).
Q3: B. Horizontal displacement is \( l \sin\theta \).

Memory Aids

Gravity splits into \( mg \cos\theta \) and \( mg \sin\theta \).
Restoring force = \( mg \sin\theta \).
Horizontal displacement = \( l \sin\theta \).

Superposition of Two SHMs

Concepts

Two SHMs perpendicular: x = A₁ sin ωt, y = A₂ sin(ωt + Φ).
Eliminate t to find resultant path.
Resultant equation: [ frac{y²}{A₂²} + frac{x²}{A₁²} - frac{2xy}{A₁ A₂} cos Φ = sin² Φ ]
This is an ellipse equation; particle moves in ellipse.
Special cases depend on phase difference Φ:

Solved Examples

Ex 1: For Φ = 0, show resultant motion is SHM along a line.
- Equation reduces to (y/A₂ - x/A₁)² = 0 ⇒ y = A₂/A₁ x.
- Amplitude = √(A₁² + A₂²), angular frequency ω.
Ex 2: For Φ = π/2, if A₁ = A₂ = A, path is circle: x² + y² = A².

Practice Set

Q1: For two SHMs with Φ = 0, resultant path is:
A. Circle
B. Ellipse
C. Straight line
D. Parabola
Q2: For Φ = π, resultant path is:
A. Circle
B. Ellipse
C. Straight line
D. Hyperbola
Q3: For Φ = π/2 and A₁ = A₂, resultant path is:
A. Circle
B. Ellipse
C. Straight line
D. Parabola

Answer Key

Q1: C. Φ = 0 gives straight line motion.
Q2: C. Φ = π also gives straight line motion but with negative slope.
Q3: A. Φ = π/2 and equal amplitudes give circle.

Memory Aids

Resultant path depends on phase difference Φ.
Φ = 0, π → straight line.
Φ = π/2 and equal amplitudes → circle.

Concepts

Two SHMs perpendicular: x = A₁ sin ωt, y = A₂ sin(ωt + Φ).
Eliminate t to find resultant path.
Resultant equation: [ y²/A₂² + x²/A₁² - 2xy/A₁ A₂ cos Φ = sin² Φ ]
This is an ellipse equation; particle moves in ellipse.
Special cases depend on phase difference Φ:

Solved Examples

Ex 1: For Φ = 0, show resultant motion is SHM along a line.
- Equation reduces to (y/A₂ - x/A₁)² = 0 ⇒ y = A₂/A₁ x.
- Amplitude = √(A₁² + A₂²), angular frequency ω.
Ex 2: For Φ = π/2, if A₁ = A₂ = A, path is circle: x² + y² = A².

Practice Set

Q1: For two SHMs with Φ = 0, resultant path is:
A. Circle
B. Ellipse
C. Straight line
D. Parabola
Q2: For Φ = π, resultant path is:
A. Circle
B. Ellipse
C. Straight line
D. Hyperbola
Q3: For Φ = π/2 and A₁ = A₂, resultant path is:
A. Circle
B. Ellipse
C. Straight line
D. Parabola

Answer Key

Q1: C. Φ = 0 gives straight line motion.
Q2: C. Φ = π also gives straight line motion but with negative slope.
Q3: A. Φ = π/2 and equal amplitudes give circle.

Memory Aids

Resultant path depends on phase difference Φ.
Φ = 0, π → straight line.
Φ = π/2 and equal amplitudes → circle.

Concepts

Two SHMs perpendicular: x = A₁ sin ωt, y = A₂ sin(ωt + Φ).
Eliminate t to find resultant path.
Resultant equation: [ frac{y²}{A₂²} + frac{x²}{A₁²} - frac{2xy}{A₁ A₂} cos Φ = sin² Φ ]
This is an ellipse equation; particle moves in ellipse.
Special cases depend on phase difference Φ:

Solved Examples

Ex 1: For Φ = 0, show resultant motion is SHM along a line.
- Equation reduces to (y/A₂ - x/A₁)² = 0 ⇒ y = A₂/A₁ x.
- Amplitude = √(A₁² + A₂²), angular frequency ω.
Ex 2: For Φ = π/2, if A₁ = A₂ = A, path is circle: x² + y² = A².

Practice Set

Q1: For two SHMs with Φ = 0, resultant path is:
A. Circle
B. Ellipse
C. Straight line
D. Parabola
Q2: For Φ = π, resultant path is:
A. Circle
B. Ellipse
C. Straight line
D. Hyperbola
Q3: For Φ = π/2 and A₁ = A₂, resultant path is:
A. Circle
B. Ellipse
C. Straight line
D. Parabola

Answer Key

Q1: C. Φ = 0 gives straight line motion.
Q2: C. Φ = π also gives straight line motion but with negative slope.
Q3: A. Φ = π/2 and equal amplitudes give circle.

Memory Aids

Resultant path depends on phase difference Φ.
Φ = 0, π → straight line.
Φ = π/2 and equal amplitudes → circle.

Concepts

Two SHMs perpendicular: \( x = A_1 \sin \omega t \), \( y = A_2 \sin(\omega t + \Phi) \).
Eliminate \( t \) to find resultant path.
Resultant equation: \[ \frac{y^2}{A_2^2} + \frac{x^2}{A_1^2} - \frac{2xy}{A_1 A_2} \cos \Phi = \sin^2 \Phi \]
This is an ellipse equation; particle moves in ellipse.
Special cases depend on phase difference \( \Phi \):

Solved Examples

Ex 1: For \( \Phi = 0 \), show resultant motion is SHM along a line.
- Equation reduces to \( (\frac{y}{A_2} - \frac{x}{A_1})^2 = 0 \Rightarrow y = \frac{A_2}{A_1} x \).
- Amplitude \( = \sqrt{A_1^2 + A_2^2} \), angular frequency \( \omega \).
Ex 2: For \( \Phi = \frac{\pi}{2} \), if \( A_1 = A_2 = A \), path is circle: \( x^2 + y^2 = A^2 \).

Practice Set

Q1: For two SHMs with \( \Phi = 0 \), resultant path is:
A. Circle
B. Ellipse
C. Straight line
D. Parabola
Q2: For \( \Phi = \pi \), resultant path is:
A. Circle
B. Ellipse
C. Straight line
D. Hyperbola
Q3: For \( \Phi = \frac{\pi}{2} \) and \( A_1 = A_2 \), resultant path is:
A. Circle
B. Ellipse
C. Straight line
D. Parabola

Answer Key

Q1: C. \( \Phi = 0 \) gives straight line motion.
Q2: C. \( \Phi = \pi \) also gives straight line motion but with negative slope.
Q3: A. \( \Phi = \frac{\pi}{2} \) and equal amplitudes give circle.

Memory Aids

Resultant path depends on phase difference \( \Phi \).
\( \Phi = 0, \pi \) → straight line.
\( \Phi = \frac{\pi}{2} \) and equal amplitudes → circle.

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