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Physics · Class 11 · Chapter 13

Oscillations

A swinging pendulum, a mass on a spring and the atoms in a solid all move back and forth about a mean position. This chapter builds the mathematics of simple harmonic motion, which you will use again for waves and for AC circuits in Class 12.

In this chapter: periodic and oscillatory motion, period, frequency and displacement, simple harmonic motion, SHM as the projection of uniform circular motion, velocity and acceleration in SHM, the force law, energy in SHM, the spring-block system and the simple pendulum.

Periodic and oscillatory motion

A motion that repeats itself at regular intervals of time is periodic. If a body in periodic motion moves back and forth about a fixed mean (equilibrium) position, the motion is oscillatory. Every oscillatory motion is periodic, but not every periodic motion is oscillatory: the Earth going round the Sun is periodic but not oscillatory.

  • Period T: the smallest interval after which the motion repeats (s).
  • Frequency ν = 1/T: the number of oscillations per second, in hertz (1 Hz = 1 s−1).
  • Angular frequency ω = 2π/T = 2πν: in rad s−1.

At the mean position the net force on the body is zero. When it is displaced, a force acts that tries to bring it back; this is what keeps it oscillating.

Simple harmonic motion

A particle is in simple harmonic motion (SHM) if its displacement from the mean position varies with time as a sine or cosine function:

x(t) = A cos(ωt + φ)A = amplitude (maximum displacement, always positive); (ωt + φ) = phase; φ = phase constant, fixed by the position and velocity at t = 0.

The phase tells you the state of motion (position and direction) at any instant. Two particles with the same A and ω but different φ are at different points of their cycle at a given time.

SHM and uniform circular motion

If a particle P moves uniformly on a circle of radius A with angular speed ω, the foot of the perpendicular from P onto a diameter moves in SHM along that diameter, with amplitude A and angular frequency ω. This picture is the easiest way to remember phase relations. Note that uniform circular motion itself is not SHM; only its projection is.

SHM as the projection of uniform circular motionwww.iitmedicoguide.comxPP′θAx = A cos(ωt + φ)P moves anticlockwiseat constant ωThe foot of theperpendicular P′moves to and fro alongthe diameter in SHMamplitude = radius Aangular frequency = ωperiod T = 2π/ωangle θ = ωt + φwww.iitmedicoguide.com
As P goes round the circle at constant speed, its projection P′ on the x axis oscillates between +A and −A. One revolution of P is one complete oscillation of P′.

Velocity and acceleration in SHM

v(t) = dx/dt = −ωA sin(ωt + φ)a(t) = dv/dt = −ω2A cos(ωt + φ) = −ω2xspeed at displacement x: v = ω√A2 − x2
  • At the mean position (x = 0): speed is maximum, ωA, and acceleration is zero.
  • At the extreme positions (x = ±A): speed is zero and acceleration is maximum, ω2A, directed towards the mean position.
  • Velocity leads displacement in phase by π/2, and acceleration is π out of phase with displacement.
Displacement, velocity and acceleration in SHMwww.iitmedicoguide.comx+A−AA cos ωtv+Aω−Aω−Aω sin ωta+Aω²−Aω²−Aω² cos ωtT/4T/23T/4T5T/4twww.iitmedicoguide.com
For φ = 0, the body starts at x = +A. Velocity is zero at the extremes and largest when x = 0; acceleration always has the opposite sign to x.
Worked example: A body oscillates in SHM according to x = 6.0 cos(3πt + π/3) m. Find its displacement, velocity and acceleration at t = 2.0 s.
Solution: At t = 2 s, the phase is 6π + π/3, which is equivalent to π/3. x = 6.0 cos(π/3) = 3.0 m. v = −(3π)(6.0) sin(π/3) = −18π × 0.866 ≈ −49 m s−1. a = −(3π)2 × 3.0 = −27π2 ≈ −2.7 × 102 m s−2.

Force law for SHM

From a = −ω2x and Newton's second law,

F = −mω2x = −kxA restoring force proportional to displacement and directed towards the mean position is the defining condition for SHM. Here k = mω2 is the force constant.

For a block of mass m on a smooth floor attached to a spring of spring constant k:

ω = √k/m,   T = 2π√m/kA heavier block or a softer spring gives a longer period. T does not depend on the amplitude.

For two springs k1 and k2 joined side by side (parallel), the effective constant is k1 + k2; joined end to end (series), 1/k = 1/k1 + 1/k2. Cutting a spring into two equal halves doubles the spring constant of each half.

Energy in SHM

kinetic energy: K = ½mv2 = ½k(A2 − x2)potential energy: U = ½kx2total energy: E = K + U = ½kA2 = ½mω2A2

The total energy stays constant and is proportional to the square of the amplitude. Kinetic energy is maximum at the mean position and potential energy at the extremes. K and U are equal at x = A/√2. Both K and U vary periodically with time, but with period T/2, that is, at twice the frequency of the oscillation.

Worked example: A 1 kg block attached to a spring of spring constant 50 N m−1 is pulled 10 cm from its equilibrium position on a smooth floor and released. Find its maximum speed, and its kinetic and potential energies when it is 5 cm from the mean position.
Solution: ω = √(50/1) ≈ 7.07 rad s−1, so vmax = ωA = 7.07 × 0.1 ≈ 0.71 m s−1. At x = 0.05 m: U = ½ × 50 × (0.05)2 = 0.0625 J; K = ½ × 50 × (0.01 − 0.0025) = 0.1875 J. Their sum is 0.25 J, which equals ½kA2 = ½ × 50 × 0.01.

The simple pendulum

A simple pendulum is a small heavy bob of mass m hanging from a light, inextensible string of length L. When it is displaced by an angle θ, the restoring torque about the point of suspension is −mgL sin θ. For small angles, sin θ ≈ θ (in radians), so the torque is proportional to θ and the motion is SHM:

T = 2π√L/gValid for small amplitudes (up to about 10° to 15°). T does not depend on the mass of the bob or on the amplitude.
  • A seconds pendulum has a period of 2 s. Its length is L = gT2/4π2 = 9.8 × 4/(4 × 9.87) ≈ 0.99 m, about one metre.
  • On the Moon, where g is about one sixth of its value on Earth, the same pendulum would have a period √6 times longer.
  • In a freely falling lift the effective g is zero, and the pendulum does not oscillate at all.

Real oscillations die out gradually because of friction and air resistance (damping), and a system driven by a periodic force oscillates at the driving frequency; the amplitude becomes very large when the driving frequency is close to the natural frequency (resonance). These ideas are useful background, though the current NCERT text concentrates on undamped SHM.

Common mistakes: (1) Using degrees instead of radians inside sin or cos, or in the small-angle approximation. (2) Saying acceleration is zero at the extremes because velocity is zero there; it is maximum. (3) Thinking the period of a spring-block system depends on g; it does not, even for a vertical spring. (4) Forgetting that K and U oscillate at twice the frequency of x. (5) Calling uniform circular motion SHM; only its projection is.

JEE and NEET focus

  • Finding A, ω, T, φ, v and a from a given equation of SHM.
  • Speed at a given displacement, and the displacement where K = U.
  • Spring-block systems, including springs in series and parallel and a cut spring.
  • Energy in SHM: graphs of K and U against x and t.
  • Simple pendulum: dependence on L and g, pendulum in a lift, seconds pendulum.

Practice questions

In SHM, the acceleration of the particle is:

  1. Constant
  2. Proportional to displacement and in the same direction
  3. Proportional to displacement and opposite to it
  4. Proportional to velocity
Show answer
C. a = −ω2x.

A particle in SHM has amplitude 5 cm and period 0.2π s. Its maximum speed is:

  1. 0.25 m s−1
  2. 0.5 m s−1
  3. 1 m s−1
  4. 5 m s−1
Show answer
B. ω = 2π/0.2π = 10 rad s−1; vmax = 10 × 0.05 = 0.5 m s−1.

In SHM of amplitude A, kinetic and potential energies are equal at a displacement of:

  1. A/2
  2. A/√2
  3. A/4
  4. A
Show answer
B. ½k(A2 − x2) = ½kx2 gives x2 = A2/2.

The length of a simple pendulum is made four times. Its period:

  1. Doubles
  2. Halves
  3. Becomes four times
  4. Stays the same
Show answer
A. T ∝ √L.

A block on a spring of constant k oscillates with period T. Two such springs are joined end to end and the same block is attached. The new period is:

  1. T/2
  2. T/√2
  3. √2 T
  4. 2T
Show answer
C. In series the effective constant is k/2, and T ∝ 1/√k.

The phase difference between displacement and velocity in SHM is:

  1. 0
  2. π/4
  3. π/2
  4. π
Show answer
C. v = −ωA sin(ωt + φ) = ωA cos(ωt + φ + π/2).

If the amplitude of a spring-block oscillator is doubled, its total energy becomes:

  1. Double
  2. Four times
  3. Half
  4. Unchanged
Show answer
B. E = ½kA2.

A particle executes SHM of frequency ν. Its kinetic energy varies periodically with frequency:

  1. ν/2
  2. ν
  3. 2ν
  4. 4ν
Show answer
C. K ∝ sin2(ωt + φ) = ½[1 − cos 2(ωt + φ)].
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