Chapter 14: Oscillations

Periodic and Oscillatory Motions

Key Concepts

• Periodic motion: Motion repeated identically after a fixed time interval (period T). Example: Earth's revolution around the Sun (T = 1 year).
• Oscillatory motion: Back-and-forth motion about a fixed mean position within fixed limits (extreme positions). It is periodic and bounded.
• Examples of periodic signals: triangular wave, rectangular pulse, sawtooth wave, each with period T.

Solved Examples

• Ex 1: Identify the period of a triangular wave repeating every 0.5 s.
  • Period T = 0.5 s (given by repetition interval).
• Ex 2: Determine if a motion repeating every 2 s is periodic.
  • Yes, period T = 2 s; motion repeats identically.

Practice Set

• Q1: Which of the following is an example of periodic motion?
  • A) A car accelerating continuously
  • B) Earth revolving around the Sun
  • C) A ball thrown upwards and never returning
  • D) A car stopping at a traffic light
• Q2: Oscillatory motion is characterized by:
  • A) Motion in one direction only
  • B) Motion to and fro about a fixed point
  • C) Random motion
  • D) Motion with increasing amplitude
• Q3: A signal repeats every 0.2 s. What is its frequency?
  • A) 0.2 Hz
  • B) 2 Hz
  • C) 5 Hz
  • D) 10 Hz

Answer Key

• Q1: B - Earth's revolution is periodic motion with fixed period.
• Q2: B - Oscillatory motion is back-and-forth about mean position.
• Q3: C - Frequency ν = 1/T = 1/0.2 = 5 Hz.

Memory Aids

• Periodic motion = repeats after time T.
• Oscillatory = to-and-fro about mean position.
• Frequency ν = 1/Period T.

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) = s⁻¹.
• 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

• Ex 1: Calculate frequency if period T = 0.5 s.
  • ν = 1/T = 1/0.5 = 2 Hz.
• Ex 2: Find angular frequency for ν = 5 Hz.
  • ω = 2π × 5 = 10π rad/s ≈ 31.42 rad/s.

Practice Set

• Q1: If the period of a pendulum is 2 s, its frequency is:
  • A) 0.5 Hz
  • B) 2 Hz
  • C) 1 Hz
  • D) 4 Hz
• Q2: Angular frequency ω for frequency 3 Hz is:
  • A) 6π rad/s
  • B) 3π rad/s
  • C) 2π rad/s
  • D) 9π rad/s
• Q3: A planet completes one orbit in 88 Earth days. Its frequency in Hz is approximately:
  • A) 1.31 × 10⁻⁷ Hz
  • B) 1.31 × 10⁻⁵ Hz
  • C) 1.31 × 10⁻⁶ Hz
  • D) 1.31 × 10⁻⁸ Hz

Answer Key

• Q1: A - ν = 1/2 = 0.5 Hz.
• Q2: A - ω = 2π × 3 = 6π rad/s.
• Q3: A - Frequency = 1/(88 × 24 × 3600) ≈ 1.31 × 10⁻⁷ Hz.

Memory Aids

• Frequency = 1/Period.
• Angular frequency ω = 2π × frequency.
• Period in seconds needed for frequency in Hz.

Displacement and Periodic Function

Key Concepts

• Displacement: Change in physical quantity (position, angle, etc.) with time; can be positive or negative.
• For oscillations, displacement x(t) = A cos(ωt) or A sin(ωt), where A = amplitude.
• Periodic function: Function f(t) satisfying f(t) = f(t + T), T = period.
• Any periodic function can be expressed as sum of sine and cosine functions with different periods.
• Examples of periodic functions: sin(ωt), cos(ωt), sin(ωt) + cos(ωt).
• Non-periodic functions: e^{ωt}, log(ωt).

Solved Examples

• Ex 1: Write displacement for oscillation with amplitude 5 cm and frequency 2 Hz.
  • ω = 2π × 2 = 4π rad/s; x = 5 cos(4πt) cm.
• Ex 2: Check if f(t) = sin(3t) is periodic.
  • Period T = 2π/3; f(t) = f(t + T), so periodic.

Practice Set

• Q1: Displacement of a particle in oscillation is given by x = 3 sin(2πt). Its amplitude is:
  • A) 3 units
  • B) 2π units
  • C) 1 unit
  • D) 6 units
• Q2: The period of function f(t) = cos(5t) is:
  • A) 5 s
  • B) 2π/5 s
  • C) 2π s
  • D) 1/5 s
• Q3: Which of the following is NOT a periodic function?
  • A) sin(ωt)
  • B) cos(ωt)
  • C) e^{ωt}
  • D) sin(ωt) + cos(ωt)

Answer Key

• Q1: A - Amplitude A = 3 units.
• Q2: B - Period T = 2π/5.
• Q3: C - e^{ωt} is non-periodic (exponential growth).

Memory Aids

• Displacement = A sin(ωt) or A cos(ωt).
• Period T = 2π/ω for sin and cos functions.
• Exponential and logarithmic functions are non-periodic.

Simple Harmonic Motion

Definition and Characteristics

• Simple Harmonic Motion (SHM) is a periodic to-and-fro motion about a mean position under a restoring force.
• Acceleration is always directed towards the mean position and proportional to displacement: \( a \propto -x \).
• Mathematically, \( a = -k x \) or \( \frac{d^2x}{dt^2} = -k x \), where \( x \) is displacement.
• Particle oscillates between limits \( -A \) and \( +A \), where \( A \) is amplitude.

Displacement Equation

• Displacement as a function of time: \( x(t) = A \cos(\omega t + \phi) \).
• \( A \): amplitude (maximum displacement).
• \( \omega \): angular frequency (rad/s).
• \( \phi \): phase constant (initial phase).
• Period \( T \) is the time for one complete oscillation; \( \omega = \frac{2\pi}{T} \).

Motion Characteristics

• Speed is maximum at mean position (\( x=0 \)) and zero at extremes (\( x=\pm A \)).
• Motion repeats every period \( T \) regardless of initial phase.

Graphs of SHM

• Displacement \( x \) vs time \( t \) is a cosine wave.
• Amplitude changes shift the wave height but not the period.
• Phase constant \( \phi \) shifts the wave horizontally.
• Different frequencies change the number of oscillations per unit time.

Solved Examples

Ex 1

• Find angular frequency \( \omega \) if period \( T = 0.5 \) s.
• Use formula: \( \omega = \frac{2\pi}{T} = \frac{2\pi}{0.5} = 4\pi \) rad/s.

Ex 2

• Displacement at \( t = 0 \) is \( x = 0.1 \) m, amplitude \( A = 0.2 \) m, \( \phi = 0 \), find \( \omega t \).
• From \( x = A \cos(\omega t + \phi) \), \( 0.1 = 0.2 \cos(\omega t) \) ⇒ \( \cos(\omega t) = 0.5 \) ⇒ \( \omega t = 60^\circ = \pi/3 \) rad.

Practice Set

Q1

Which of the following is true for acceleration in SHM?

• A) Acceleration is constant
• B) Acceleration is proportional to displacement and directed away from mean position
• C) Acceleration is proportional to displacement and directed towards mean position
• D) Acceleration is zero

Q2

A particle executes SHM with amplitude 5 cm and period 2 s. What is its angular frequency?

• A) \( \pi \) rad/s
• B) \( 2\pi \) rad/s
• C) \( \frac{\pi}{2} \) rad/s
• D) \( 4\pi \) rad/s

Q3

Displacement of a particle in SHM is given by \( x = 0.1 \cos(4\pi t + \frac{\pi}{3}) \) m. What is the amplitude and phase constant?

• A) \( A=0.1 \) m, \( \phi=\frac{\pi}{3} \)
• B) \( A=4\pi \) m, \( \phi=0.1 \)
• C) \( A=0.1 \) m, \( \phi=4\pi \)
• D) \( A=\frac{\pi}{3} \) m, \( \phi=0.1 \)

Answer Key

• Q1: C - Acceleration in SHM is always directed towards the mean position and proportional to displacement.
• Q2: A - \( \omega = \frac{2\pi}{T} = \frac{2\pi}{2} = \pi \) rad/s.
• Q3: A - Amplitude is coefficient of cosine (0.1 m), phase constant is \( \frac{\pi}{3} \).

Memory Aids

• Acceleration formula: \( a = -\omega^2 x \) (restoring force direction).
• Angular frequency: \( \omega = \frac{2\pi}{T} \).
• Displacement: \( x = A \cos(\omega t + \phi) \).
• Max speed at mean position, zero at amplitude limits.

Simple Harmonic Motion and Uniform Circular Motion

Concept Overview

• Projection of uniform circular motion (UCM) on any diameter exhibits 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
• Points M and N are perpendicular projections of P on x- and y-axes respectively.
• Both M and N execute SHM with amplitude A, angular frequency \( \omega \), and phase difference \( \pi/2 \).

Solved Examples

• Ex 1: Find x-coordinate of particle at t = 2 s, given A = 5 cm, \( \omega = 3 \text{ rad/s} \).
  • Use x = A \cos \omega t
  • x = 5 \times \cos(3 \times 2) = 5 \cos 6 \text{ cm}
  • Calculate \cos 6 (in radians) for numerical value.
• Ex 2: Determine phase difference between SHMs along x and y axes.
  • Given x = A \cos \omega t, y = A \sin \omega t
  • Phase difference = \( \pi/2 \) radians (90°)

Practice Set

• Q1: The projection of uniform circular motion on a diameter is:
  • A) Uniform circular motion
  • B) Simple harmonic motion
  • C) Random motion
  • D) Oscillatory motion without periodicity
• Q2: For a particle in UCM with radius 10 cm and angular speed 5 rad/s, what is the x-coordinate at t = 0.2 s?
  • A) 10 cm
  • B) 5 cm
  • C) 8.09 cm
  • D) 0 cm
• Q3: The phase difference between SHMs along x and y axes for a particle in UCM is:
  • A) 0
  • B) \( \pi/4 \)
  • C) \( \pi/2 \)
  • D) \( \pi \)

Answer Key

• Q1: B) Simple harmonic motion
  Projection of UCM on diameter is SHM by definition.
• Q2: C) 8.09 cm
  Calculate x = 10 \( \cos(5 \times 0.2) = 10 \cos 1 = 8.09 \) cm.
• Q3: C) \( \pi/2 \)
  SHMs along x and y axes differ by 90° phase.

Memory Aids

• Projection of UCM on any diameter = SHM.
• x = A \cos \omega t, y = A \sin \omega t.
• Phase difference between x and y SHMs = 90° (\( \pi/2 \)).
• Amplitude of SHM = radius of circle A.

Velocity and Acceleration in Simple Harmonic Motion

Displacement in SHM

• Displacement from mean position: \( x(t) = A \cos(\omega t + \phi) \)
• \( A \): amplitude (m), \( \omega \): angular frequency (rad/s), \( \phi \): phase angle (rad)

Velocity in SHM

• Velocity is rate of change of displacement: \( v(t) = \frac{dx}{dt} = -\omega A \sin(\omega t + \phi) \)
• Velocity amplitude: \( v_m = \omega A \)
• At mean position (\( x=0 \)), velocity is maximum: \( v = \pm v_m \)
• At extreme positions (\( x=\pm A \)), velocity is zero
• Alternate form: \( v(t) = -\omega \sqrt{A^2 - x^2} \)

Acceleration in SHM

• Acceleration is rate of change of velocity: \( a(t) = \frac{dv}{dt} = -\omega^2 A \cos(\omega t + \phi) \)
• Relation with displacement: \( a(t) = -\omega^2 x(t) \)
• Acceleration amplitude: \( a_m = \omega^2 A \)
• At mean position (\( x=0 \)), acceleration is zero
• At extreme positions (\( x=\pm A \)), acceleration is maximum and directed opposite to displacement

Solved Examples

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

Practice Set

• Q1: At which position is velocity maximum in SHM?
  • A) At extreme position
  • B) At mean position
  • C) At half amplitude
  • D) Velocity is constant
• Q2: The acceleration of a particle executing SHM is zero at:
  • A) Mean position
  • B) Extreme position
  • C) Half amplitude
  • D) All positions
• Q3: A particle executes SHM with amplitude 0.1 m and angular frequency 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) At mean position — Velocity is maximum where displacement is zero.
• Q2: A) Mean position — Acceleration is zero at mean position where restoring force is zero.
• Q3: B) 10 m/s — Maximum velocity \( v_m = \omega A = 10 \times 0.1 = 1 \) m/s (Corrected: 1 m/s, so option A is correct)

Memory Aids

• Velocity formula: \( v = -\omega A \sin(\omega t + \phi) \)
• Acceleration formula: \( a = -\omega^2 x \)
• Velocity max at mean position, zero at extremes
• Acceleration max at extremes, zero at mean

SHM Graphs

Displacement, Velocity, and Acceleration Graphs

• Displacement \( x(t) \): Cosine wave between +A and -A, period \( T \)
• Velocity \( v(t) \): Sine wave, zero at extremes, max at mean
• Acceleration \( a(t) \): Cosine wave, opposite phase to displacement
• All have same period \( T \) but differ in phase

Solved Examples

• Ex 1: Identify phase difference between displacement and velocity.
  • Velocity leads displacement by \( \frac{\pi}{2} \) (90°) phase.
• Ex 2: At time \( t = 0 \), if displacement is maximum, what is velocity?
  • Velocity is zero at maximum displacement.

Practice Set

• Q1: Which graph represents acceleration in SHM?
  • A) Sine wave in phase with displacement
  • B) Cosine wave opposite phase to displacement
  • C) Constant zero line
  • D) Linear increasing line
• Q2: What is the phase difference between velocity and acceleration in SHM?
  • A) 0°
  • B) 90°
  • C) 180°
  • D) 270°
• Q3: If displacement is \( x = A \cos(\omega t) \), velocity is:
  • A) \( -\omega A \sin(\omega t) \)
  • B) \( \omega A \cos(\omega t) \)
  • C) \( -\omega^2 A \cos(\omega t) \)
  • D) \( \omega^2 A \sin(\omega t) \)

Answer Key

• Q1: B) Cosine wave opposite phase to displacement — Acceleration is opposite in phase to displacement.
• Q2: D) 270° — Velocity leads acceleration by 270° phase difference.
• Q3: A) \( -\omega A \sin(\omega t) \) — Velocity is derivative of displacement.

Memory Aids

• Displacement: cosine wave
• Velocity: sine wave, 90° ahead of displacement
• Acceleration: cosine wave, 180° out of phase with displacement

Force Law for Simple Harmonic Motion

Key Concepts

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

Solved Examples

• Ex 1: Find force on mass \( m = 2\,kg \) displaced by \( x = 0.1\,m \) 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) \( 0 \)
• Q2: If \( \omega = 10\,rad/s \) and \( m = 0.5\,kg \), what is \( k \)?
  • A) 50 N/m
  • B) 5 N/m
  • C) 100 N/m
  • D) 20 N/m
• Q3: For a particle in S.H.M., acceleration is given by?
  • A) \( a = \omega^2 x \)
  • B) \( a = -\omega^2 x \)
  • C) \( a = m\omega^2 x \)
  • D) \( a = 0 \)

Answer Key

• Q1: B) \( -kx \) - Force is restoring, opposite to displacement.
• Q2: A) 50 N/m - \( k = m\omega^2 = 0.5 \times 10^2 = 50 \)
• Q3: B) \( a = -\omega^2 x \) - Acceleration is opposite and proportional to displacement.

Memory Aids

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

Oscillations Due to a Spring

Key Concepts

• Massless spring on smooth surface with mass \( m \) attached
• Restoring force: \( F = -kx \), \( k \) is spring constant
• Negative sign indicates force opposite to displacement
• Oscillation about equilibrium position \( x=0 \)

Solved Examples

• Ex 1: Calculate restoring force if \( k=100\,N/m \) and \( x=0.05\,m \).
  • \( F = -kx = -100 \times 0.05 = -5\,N \)

Practice Set

• Q1: What is the restoring force when spring is compressed by \( x \)?
  • A) \( kx \)
  • B) \( -kx \)
  • C) \( 0 \)
  • D) \( m\omega^2 x \)
• Q2: If \( m=0.2\,kg \), \( k=80\,N/m \), find acceleration at \( x=0.1\,m \).
  • A) \( -40\,m/s^2 \)
  • B) \( -80\,m/s^2 \)
  • C) \( -20\,m/s^2 \)
  • D) \( 0 \)
• Q3: The negative sign in \( F = -kx \) indicates?
  • A) Force is zero
  • B) Force is in direction of displacement
  • C) Force opposes displacement
  • D) Force is constant

Answer Key

• Q1: B) \( -kx \) - Restoring force opposes displacement.
• Q2: A) \( a = F/m = -kx/m = -80 \times 0.1 / 0.2 = -40\,m/s^2 \)
• Q3: C) Force opposes displacement - negative sign shows restoring nature.

Memory Aids

• Restoring force always opposite to displacement
• Force constant \( k \) defines stiffness
• Acceleration \( a = -\frac{k}{m} x \)

Time Period of SHM for Combination of Springs

Key Concepts

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

Solved Examples

• Ex 1: Find \( T \) for series springs with \( k_1=100\,N/m \), \( k_2=200\,N/m \), \( m=0.5\,kg \).
  • \( \frac{1}{k_{eq}} = \frac{1}{100} + \frac{1}{200} = \frac{3}{200} \Rightarrow k_{eq} = \frac{200}{3} \approx 66.67 \)
  • \( T = 2\pi \sqrt{\frac{0.5}{66.67}} \approx 1.72\,s \)
• Ex 2: Find \( T \) for parallel springs with same values.
  • \( k_{eq} = 100 + 200 = 300 \)
  • \( T = 2\pi \sqrt{\frac{0.5}{300}} \approx 0.81\,s \)

Practice Set

• Q1: For two springs in series, equivalent spring constant is?
  • A) \( k_1 + k_2 \)
  • B) \( \frac{1}{k_1} + \frac{1}{k_2} \)
  • C) \( \frac{1}{\frac{1}{k_1} + \frac{1}{k_2}} \)
  • D) \( k_1 k_2 \)
• Q2: Time period for parallel springs with \( k_1=50\,N/m \), \( k_2=150\,N/m \), \( m=1\,kg \) is?
  • A) 0.79 s
  • B) 1.26 s
  • C) 0.63 s
  • D) 1.57 s
• Q3: If \( T = 2\pi \sqrt{\frac{m}{k_{eq}}} \) and \( T=2s \), \( m=0.5kg \), find \( k_{eq} \).
  • A) 2.47 N/m
  • B) 1.23 N/m
  • C) 0.5 N/m
  • D) 3.14 N/m

Answer Key

• Q1: C) \( \frac{1}{\frac{1}{k_1} + \frac{1}{k_2}} \) - Series springs add inversely.
• Q2: A) 0.79 s - \( k_{eq} = 200 \), \( T = 2\pi \sqrt{1/200} \approx 0.79 \)
• Q3: A) \( k_{eq} = \frac{4\pi^2 m}{T^2} = \frac{4 \times 9.87 \times 0.5}{4} = 2.47 \)

Memory Aids

• Series: \( 1/k_{eq} = 1/k_1 + 1/k_2 \)
• 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 \(E = K + U\) remains constant throughout the motion.
• Potential energy: \(U = \frac{1}{2}kx^2 = \frac{1}{2}kA^2 \cos^2(\omega t + \phi)\)
• Kinetic energy: \(K = \frac{1}{2}mv^2 = \frac{1}{2}m\omega^2 A^2 \sin^2(\omega t + \phi) = \frac{1}{2}kA^2 \sin^2(\omega t + \phi)\)
• Total energy: \(E = \frac{1}{2}kA^2 [\cos^2(\omega t + \phi) + \sin^2(\omega t + \phi)] = \frac{1}{2}kA^2\)
• Energy shifts between kinetic and potential forms but total energy is time-independent.

Solved Examples

• Ex 1: Calculate total energy for a spring with \(k=200\,\mathrm{N/m}\) and amplitude \(A=0.05\,\mathrm{m}\).
  • Use \(E=\frac{1}{2}kA^2 = \frac{1}{2} \times 200 \times (0.05)^2 = 0.25\,\mathrm{J}\).
• Ex 2: Find kinetic energy at \(x=0.03\,\mathrm{m}\) for above spring.
  • Potential energy \(U=\frac{1}{2}kx^2=\frac{1}{2} \times 200 \times (0.03)^2=0.09\,\mathrm{J}\).
  • Kinetic energy \(K=E-U=0.25-0.09=0.16\,\mathrm{J}\).

Practice Set

• Q1: What is the total energy of a SHM system with \(k=100\,\mathrm{N/m}\) and amplitude \(0.1\,\mathrm{m}\)?
  • A) 0.5 J
  • B) 1.0 J
  • C) 2.0 J
  • D) 5.0 J
• Q2: At maximum displacement, the kinetic energy is:
  • A) Maximum
  • B) Zero
  • C) Half of total energy
  • D) Equal to potential energy
• Q3: For a mass-spring system, if amplitude doubles, total energy changes by:
  • A) Doubles
  • B) Quadruples
  • C) Halves
  • D) Remains same

Answer Key

• Q1: A) 0.5 J — Using \(E=\frac{1}{2}kA^2=\frac{1}{2} \times 100 \times 0.1^2=0.5\,\mathrm{J}\).
• Q2: B) Zero — At maximum displacement, velocity is zero, so kinetic energy is zero.
• Q3: B) Quadruples — Total energy \(E \propto A^2\), doubling amplitude increases energy by 4 times.

Memory Aids

• Total energy in SHM: \(E=\frac{1}{2}kA^2\) (constant)
• Energy shifts: \(K \leftrightarrow U\) with \(\sin^2 + \cos^2 = 1\)
• At center: max \(K\), zero \(U\); at extremes: max \(U\), zero \(K\)

Energy Variation with Time

Time Dependence of Energies

• Kinetic energy \(K(t) = \frac{1}{2}kA^2 \sin^2(\omega t + \phi)\)
• Potential energy \(U(t) = \frac{1}{2}kA^2 \cos^2(\omega t + \phi)\)
• Total energy \(E = K(t) + U(t) = \frac{1}{2}kA^2\) constant over time.
• Energy oscillates between kinetic and potential forms periodically with angular frequency \(\omega\).

Solved Examples

• Ex 1: Find kinetic energy at \(t=\frac{\pi}{2\omega}\) for \(k=50\,\mathrm{N/m}\), \(A=0.1\,\mathrm{m}\), \(\phi=0\).
  • \(K=\frac{1}{2}kA^2 \sin^2(\omega t) = \frac{1}{2} \times 50 \times 0.01 \times \sin^2(\frac{\pi}{2}) = 0.25\,\mathrm{J}\).
• Ex 2: Calculate potential energy at \(t=0\) for same system.
  • \(U=\frac{1}{2}kA^2 \cos^2(0) = 0.5\,\mathrm{J}\).

Practice Set

• Q1: At time when kinetic energy is maximum, potential energy is:
  • A) Maximum
  • B) Zero
  • C) Half of total energy
  • D) Equal to kinetic energy
• Q2: If \(\omega = 10\,\mathrm{rad/s}\), what is the period of energy oscillation?
  • A) 0.1 s
  • B) 0.2 s
  • C) 0.314 s
  • D) 1 s
• Q3: For \(E=0.5\,\mathrm{J}\), what is \(K(t)\) at \(\omega t + \phi = 45^\circ\)?
  • A) 0.25 J
  • B) 0.35 J
  • C) 0.5 J
  • D) 0.0 J

Answer Key

• Q1: B) Zero — Kinetic energy max when potential energy is zero (energy conservation).
• Q2: B) 0.2 s — Period \(T=\frac{2\pi}{\omega} = \frac{2\pi}{10} = 0.628\,\mathrm{s}\), energy oscillates with same period.
• Q3: A) 0.25 J — \(K=E \sin^2 45^\circ = 0.5 \times 0.5 = 0.25\,\mathrm{J}\).

Memory Aids

• Energy oscillates with period \(T=\frac{2\pi}{\omega}\)
• Use \(\sin^2 + \cos^2 = 1\) to relate \(K(t)\) and \(U(t)\)
• Max kinetic energy at equilibrium position, max potential at amplitude

The Simple Pendulum

Concepts and Formulas

• A simple pendulum consists of a mass (bob) suspended by a light, inextensible string of length \( l \).
• It oscillates under gravity with small angular displacement \( \theta \).
• Time period \( T = 2\pi \sqrt{\frac{l}{g}} \), where \( g \) is acceleration due to gravity.
• At extreme positions: maximum potential energy (P.E.), zero kinetic energy (K.E.), bob momentarily at rest.
• At mean position: maximum kinetic energy, zero potential energy, bob moves fastest.
• Amplitude is the maximum angular displacement \( \theta \).

Solved Examples

Ex 1: Calculate the time period of a pendulum of length 1 m. (Take \( g = 9.8 \ \mathrm{m/s^2} \))

• Use \( T = 2\pi \sqrt{\frac{l}{g}} = 2\pi \sqrt{\frac{1}{9.8}} \approx 2.006 \ \mathrm{s} \).

Practice Set

Q1: The time period of a simple pendulum depends on:

• A) Mass of the bob
• B) Length of the string
• C) Amplitude of oscillation
• D) Both mass and amplitude

Q2: For a pendulum of length 0.5 m, the time period is approximately:

• A) 1.42 s
• B) 0.71 s
• C) 2.0 s
• D) 0.5 s

Q3: At the highest point of the pendulum's swing, the kinetic energy is:

• A) Maximum
• B) Zero
• C) Half of potential energy
• D) Equal to total energy

Answer Key

• Q1: B — Time period depends only on length and gravity, not mass or amplitude (for small angles).
• Q2: A — \( T = 2\pi \sqrt{\frac{0.5}{9.8}} \approx 1.42 \ \mathrm{s} \).
• Q3: B — Kinetic energy is zero at the highest point as the bob momentarily stops.

Memory Aids

• Time period \( T \) ~ \( \sqrt{l} \), independent of mass.
• Maximum P.E. at amplitude, maximum K.E. at equilibrium.
• Small angle approximation: \( \sin \theta \approx \theta \) (in radians).

Angular SHM of Pendulum

Concepts and Formulas

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

Solved Examples

Ex 1: Find the angular frequency of a pendulum of length 0.8 m.

• \( \omega = \sqrt{\frac{g}{l}} = \sqrt{\frac{9.8}{0.8}} = 3.5 \ \mathrm{rad/s} \).

Practice Set

Q1: The restoring torque on a pendulum bob is proportional to:

• A) \( \theta^2 \)
• B) \( \theta \)
• C) \( \sin \theta \)
• D) Constant

Q2: The moment of inertia of pendulum bob about suspension point is:

• A) \( m l \)
• B) \( m l^2 \)
• C) \( m / l^2 \)
• D) \( m g l \)

Q3: For small oscillations, the angular acceleration \( \alpha \) is:

• A) \( -\frac{g}{l} \theta \)
• B) \( \frac{g}{l} \theta \)
• C) \( -\frac{l}{g} \theta \)
• D) Zero

Answer Key

• Q1: B — Torque is proportional to angular displacement \( \theta \) for small angles.
• Q2: B — Moment of inertia for point mass at distance \( l \) is \( m l^2 \).
• Q3: A — Angular acceleration is \( -\frac{g}{l} \theta \) restoring towards equilibrium.

Memory Aids

• Torque \( \tau = -mg l \theta \) for small \( \theta \).
• Angular frequency \( \omega = \sqrt{g/l} \).
• Time period \( T = 2\pi / \omega \).

Forces on Pendulum Bob

Concepts and Formulas

• For pendulum displaced at angle \( \theta \), forces on bob are:
  • Tension \( T \) along string.
  • Weight \( mg \) vertically downward.
• Weight components:
  • Along string: \( mg \cos \theta \) (balanced by tension).
  • Perpendicular to string: \( mg \sin \theta \) (restoring force).
• Horizontal displacement from vertical: \( l \sin \theta \).
• Restoring force causes oscillation back to equilibrium.

Solved Examples

Ex 1: Find the component of weight causing restoring force at \( 30^\circ \).

• \( mg \sin 30^\circ = mg \times 0.5 = 0.5 mg \).

Practice Set

Q1: The tension in the string balances which component of weight?

• A) \( mg \sin \theta \)
• B) \( mg \cos \theta \)
• C) \( mg \)
• D) None

Q2: The restoring force is proportional to:

• A) \( mg \cos \theta \)
• B) \( mg \sin \theta \)
• C) \( T \)
• D) \( mg \)

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

Memory Aids

• Weight components: \( mg \cos \theta \) along string, \( mg \sin \theta \) perpendicular.
• Restoring force causes oscillation.

Superposition of Two SHMs

Concepts and Formulas

• Two SHMs along perpendicular directions:
  • \( x = A_1 \sin \omega t \)
  • \( y = A_2 \sin (\omega t + \Phi) \)
• Eliminate time \( t \) to find trajectory:
• General 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 represents an ellipse centered at origin.
• Special cases based on phase difference \( \Phi \):

Special Cases

• \( \Phi = 0 \): Particle moves in a straight line \( y = \frac{A_2}{A_1} x \).
• \( \Phi = \pi \): Particle moves in a straight line \( y = -\frac{A_2}{A_1} x \).
• \( \Phi = \frac{\pi}{2} \): Particle moves in an ellipse \( \frac{y^2}{A_2^2} + \frac{x^2}{A_1^2} = 1 \).
• If \( A_1 = A_2 \) and \( \Phi = \frac{\pi}{2} \), motion is circular.

Solved Examples

Ex 1: For \( A_1 = 3 \), \( A_2 = 4 \), \( \Phi = 0 \), find amplitude of resultant SHM.

• Amplitude \( = \sqrt{3^2 + 4^2} = 5 \).

Practice Set

Q1: When \( \Phi = 0 \), the resultant motion is:

• A) Ellipse
• B) Circle
• C) Straight line
• D) Parabola

Q2: For \( A_1 = A_2 = A \) and \( \Phi = \frac{\pi}{2} \), the path is:

• A) Ellipse
• B) Circle
• C) Straight line
• D) Hyperbola

Q3: The general equation of the path is:

• A) \( \frac{y^2}{A_2^2} + \frac{x^2}{A_1^2} = 1 \)
• B) \( y = \frac{A_2}{A_1} x \)
• C) \( \frac{y^2}{A_2^2} + \frac{x^2}{A_1^2} - \frac{2xy}{A_1 A_2} \cos \Phi = \sin^2 \Phi \)
• D) \( y^2 + x^2 = A^2 \)

Answer Key

• Q1: C — \( \Phi = 0 \) gives straight line motion.
• Q2: B — Equal amplitudes and phase difference \( \pi/2 \) produce circular motion.
• Q3: C — General ellipse equation with phase difference \( \Phi \).

Memory Aids

• \( \Phi = 0, \pi \) → straight line.
• \( \Phi = \pi/2 \) → ellipse or circle if amplitudes equal.
• Resultant amplitude \( = \sqrt{A_1^2 + A_2^2} \) for \( \Phi = 0, \pi \).