Oscillations — Class 11 Physics

1. Periodic Motion and Oscillation

A motion that repeats itself at regular intervals is periodic motion. An oscillation is a to-and-fro motion about a stable equilibrium position. Examples include a mass attached to a spring, a simple pendulum for small angular displacements, and vibrations of tuning forks.

The time period T is the time taken for one complete oscillation. Frequency is f=1/T, and angular frequency is ω=2πf=2π/T.

2. Simple Harmonic Motion

Simple harmonic motion (SHM) is an oscillatory motion in which the restoring acceleration is directly proportional to the displacement from the mean position and directed towards it:

a=−ω²x.

The corresponding restoring force is F=−kx for a spring system, with ω=√(k/m).

The minus sign indicates that the restoring force or acceleration is opposite to displacement.

3. Displacement, Amplitude and Phase

A standard SHM equation is x=A cos(ωt+φ), where A is amplitude and φ is the initial phase. The amplitude is the maximum magnitude of displacement from the equilibrium position.

Velocity and acceleration are obtained by differentiation:

v=−Aω sin(ωt+φ)

a=−Aω² cos(ωt+φ)=−ω²x.

4. Maximum Speed and Acceleration

At the mean position, displacement is zero and speed is maximum:

v_max=Aω.

At the extreme positions, speed is zero and the magnitude of acceleration is maximum:

a_max=Aω².

5. Velocity–Displacement Relation

Eliminating time from the SHM equations gives:

v²=ω²(A²−x²).

This relation is useful for finding the speed when a particle is at a specified displacement without first finding time.

6. Energy in SHM

For an ideal mass–spring oscillator, total mechanical energy remains constant:

E=½kA²=½mω²A².

Potential energy at displacement x is U=½kx², while kinetic energy is K=½k(A²−x²).

At the mean position, kinetic energy is maximum and potential energy is minimum. At an extreme position, potential energy is maximum and kinetic energy is zero.

7. Graphical Understanding

In SHM, displacement, velocity and acceleration vary sinusoidally. Velocity is 90° out of phase with displacement for a cosine representation, while acceleration is opposite in phase to displacement. Understanding these phase relationships helps determine the direction and magnitude of motion at any instant.

8. Spring–Mass System

For a mass m attached to an ideal spring of force constant k, the equation of motion is m(d²x/dt²)=−kx. Comparing with a=−ω²x gives:

ω=√(k/m) and T=2π√(m/k).

A larger mass increases the time period, while a stiffer spring decreases it.

9. Combination of Springs

For springs in parallel, the effective spring constant is k_eq=k₁+k₂+…. For two springs in series, 1/k_eq=1/k₁+1/k₂.

These results allow the time period of a mass attached to a spring combination to be found using T=2π√(m/k_eq).

10. Simple Pendulum

For a simple pendulum of length L undergoing small angular oscillations, the motion is approximately SHM. Its time period is:

T=2π√(L/g).

The approximation is valid for small angular displacement, where sinθ≈θ when θ is measured in radians. The period is independent of the bob’s mass in the ideal model and, for small amplitudes, is approximately independent of amplitude.

11. Why the Simple Pendulum Performs SHM

The tangential restoring force of a pendulum bob is F_t=−mg sinθ. For small θ, sinθ≈θ and x≈Lθ, so F_t≈−(mg/L)x. This has the SHM form F=−kx with effective constant mg/L, leading to ω=√(g/L).

12. Phase and Initial Conditions

The phase constant depends on the initial position and velocity. For example, if the particle starts at maximum positive displacement with zero velocity, a convenient equation is x=A cosωt. If it starts from the mean position moving in the positive direction, x=A sinωt is convenient.

13. Damped and Forced Oscillations

In real systems, resistive forces cause the amplitude to decrease with time; this is damping. A driven oscillator can be maintained by an external periodic force. When the driving frequency approaches the system’s natural frequency, the amplitude can become large: this phenomenon is called resonance. The magnitude of the response depends on damping and driving conditions.

14. Natural Frequency

The natural frequency is the frequency at which a system oscillates freely after being displaced and released. For an ideal spring–mass system, f=(1/2π)√(k/m). For a small-angle simple pendulum, f=(1/2π)√(g/L).

Worked Numerical 1 — Time Period of a Spring

Question: A 0.5 kg mass is attached to a spring of force constant 200 N/m. Find the time period.

T=2π√(m/k)=2π√(0.5/200)=0.314 s approximately.

Worked Numerical 2 — Maximum Speed

Question: An oscillator has amplitude 0.08 m and angular frequency 5 rad/s. Find its maximum speed.

v_max=Aω=0.08×5=0.40 m/s.

Worked Numerical 3 — Maximum Acceleration

Question: An oscillator has amplitude 0.05 m and frequency 2 Hz. Find its maximum acceleration.

ω=2πf=4π rad/s. Therefore a_max=Aω²=0.05(4π)²≈7.90 m/s².

Worked Numerical 4 — Speed at a Given Displacement

Question: An SHM particle has A=0.10 m and ω=4 rad/s. Find its speed at x=0.06 m.

v²=ω²(A²−x²)=16(0.01−0.0036)=0.1024, so v=0.32 m/s.

Worked Numerical 5 — Simple Pendulum

Question: Find the period of a 1 m long simple pendulum for g=9.8 m/s².

T=2π√(1/9.8)≈2.01 s.

Worked Numerical 6 — Finding Spring Constant

Question: A 0.25 kg mass attached to a spring oscillates with a period of 0.50 s. Find the spring constant.

T=2π√(m/k), so k=4π²m/T²=4π²(0.25)/(0.50)²≈39.5 N/m.

Worked Numerical 7 — Energy of SHM

Question: A spring oscillator has k=100 N/m and amplitude 0.20 m. Find its total mechanical energy.

E=½kA²=½(100)(0.20)²=2.0 J.

Worked Numerical 8 — Change in Pendulum Period

Question: If the length of a simple pendulum is increased by a factor of 4, how does its period change?

Since T∝√L, T₂/T₁=√4=2. The period doubles.

Important Derivation — Equation of SHM for a Spring

For a mass m displaced by x from equilibrium, Hooke’s law gives F=−kx. Newton’s second law gives m(d²x/dt²)=−kx, or d²x/dt²+(k/m)x=0. Comparing with the standard SHM equation d²x/dt²+ω²x=0 gives ω²=k/m, hence T=2π√(m/k).

Important Derivation — Time Period of a Simple Pendulum

For a pendulum displaced through a small angle θ, the tangential restoring force is −mg sinθ≈−mgθ. Since x=Lθ, the force becomes F≈−(mg/L)x. Comparing with F=−kx gives an angular frequency ω=√(g/L). Therefore T=2π√(L/g).

Explained MCQs

1. In SHM, acceleration is proportional to:
   Answer: Negative displacement. The relation is a=−ω²x.
2. Where is the speed maximum in SHM?
   Answer: At the mean position. There x=0 and v_max=Aω.
3. Where is acceleration maximum in magnitude?
   Answer: At the extreme positions. Its magnitude is Aω².
4. If the amplitude of an ideal spring oscillator is doubled, its time period:
   Answer: Remains unchanged. The period depends on m and k, not amplitude, for ideal SHM.
5. If the mass attached to an ideal spring is made four times, the period becomes:
   Answer: Twice. T∝√m.
6. If the length of a simple pendulum is made four times, its frequency becomes:
   Answer: Half. f∝1/√L.

Competency Question

A child on a swing is given a small push and then allowed to oscillate. Explain how the swing’s displacement, speed and acceleration change during one cycle, and identify the positions where kinetic and potential energies are maximum.

HOTS

Two spring–mass oscillators have the same mass but different spring constants. The first spring is stiffer. Predict which oscillator has the higher natural frequency and explain the result using the restoring force and time-period formula.

Common Mistakes

• Forgetting the negative sign in a=−ω²x and interpreting acceleration as being in the direction of displacement.
• Using degrees instead of radians when applying the small-angle approximation sinθ≈θ.
• Confusing angular frequency ω with ordinary frequency f.
• Assuming the simple-pendulum formula is exact for large amplitudes.
• Thinking amplitude changes the ideal spring–mass time period.
• Using T=2π√(L/g) for a pendulum without checking the small-angle condition.

Quick Revision

• f=1/T
• ω=2πf=2π/T
• x=A cos(ωt+φ)
• v=−Aω sin(ωt+φ)
• a=−ω²x
• v²=ω²(A²−x²)
• v_max=Aω
• a_max=Aω²
• E=½kA²=½mω²A²
• T_spring=2π√(m/k)
• T_pendulum=2π√(L/g) for small oscillations
• k_parallel=k₁+k₂+…
• 1/k_series=1/k₁+1/k₂+…