PhysicsCBSE Class 11 · Chapter 13

Oscillations Class 11: Notes, Simple Harmonic Motion and Numericals

Updated 14 October 2026 · Fact-checked

Oscillations is Chapter 13 of CBSE Class 11 Physics. It covers periodic and oscillatory motion, period and frequency, simple harmonic motion and its displacement, velocity and acceleration, phase, the link between SHM and uniform circular motion, oscillations of a loaded spring, kinetic and potential energy in SHM, and the derivation of the period of a simple pendulum.

What Oscillations is about

A swing, a guitar string, the needle of a sewing machine and the atoms in a crystal all move back and forth about a fixed point: they oscillate. Motion that repeats after a fixed time is periodic; the time for one repetition is the period T and the number of repetitions per second is the frequency $ν = \frac{1}{T}$ .

The simplest and most important oscillation is simple harmonic motion (SHM), in which the restoring force is proportional to the displacement and opposite to it: $F = - k x$ . Its displacement is a sine or cosine function of time, $x = A cos (ω t + ϕ)$ , where A is the amplitude, ω the angular frequency and $(ω t + ϕ)$ the phase. SHM is the shadow of uniform circular motion on a diameter, which gives an easy picture of its velocity ( $ω A^{2} - x^{2}$ ) and acceleration ( $- ω^{2} x$ ).

You then study two real oscillators: a mass on a spring, with $T = 2 π \frac{m}{k}$ , and the simple pendulum, with $T = 2 π \frac{L}{g}$ for small swings. In SHM, energy keeps changing between kinetic and potential while the total $\frac{1}{2} k A^{2}$ stays constant. Damped and forced oscillations and resonance have been removed for 2026-27. This chapter leads directly to Waves, and SHM returns in Class 12 in alternating current and electromagnetic waves.

Weightage: This chapter is part of Unit X, Oscillations and Waves, which the syllabus gives 10 of the 70 theory marks in the school annual exam.Source: CBSE Physics (042) syllabus 2026-27, Class XI course structure, Unit X Oscillations and Waves

Key concepts

Periodic and oscillatory motion; period and frequency

NCERT section 13.2

A motion that repeats itself at regular intervals of time is periodic (the Earth round the Sun, the hands of a clock). If a body moves to and fro about a mean position, the motion is oscillatory (or vibratory). Every oscillation is periodic, but not every periodic motion is an oscillation.

Period T: the least time after which the motion repeats (s).
Frequency ν: the number of repetitions per second, $ν = \frac{1}{T}$ ; unit hertz (Hz = s $^{- 1}$ ).
Displacement x(t): the change from the mean (equilibrium) position, which can be positive or negative.

Periodic functions: $sin ω t$ and $cos ω t$ repeat after $T = \frac{2 π}{ω}$ , and any combination such as $a sin ω t + b cos ω t$ is also periodic with the same period. (Any periodic function can be built from sines and cosines of different frequencies, an idea due to Fourier.) Functions such as $e^{- ω t}$ or $lo g ω t$ are not periodic.

Worked example

A hummingbird's wings beat 3000 times a minute. Find the frequency and period of the wing beat. [1 mark]

Try it first, then show the solutionSolution

$ν = \frac{3000}{60} = 50 H z$
$T = \frac{1}{ν} = 0.02 s$

Answer: 50 Hz; 0.02 s

Simple harmonic motion: displacement, amplitude and phase

NCERT section 13.3

A particle executes SHM if its displacement from the mean position varies with time as
$x (t) = A cos (ω t + ϕ)$

A, the amplitude: the largest displacement (always positive).
ω, the angular frequency: $ω = \frac{2 π}{T} = 2 π ν$ (rad s $^{- 1}$ ).
(ωt + φ), the phase: tells you the state of motion (position and direction) at time t.
φ, the phase constant (initial phase): the phase at t = 0, fixed by where the particle starts.

Using sine instead of cosine just changes φ by $\frac{π}{2}$ . Equivalently, SHM is the motion of a particle whose acceleration is proportional to its displacement and always directed towards the mean position.

Worked example

A particle's displacement is $x = 0.05 cos (4 π t + \frac{π}{3})$ m. Find its amplitude, period, frequency, initial phase and displacement at t = 0. [2 marks]

Try it first, then show the solutionSolution

A = 0.05 m; ω = 4π rad s $^{- 1}$ , so $T = \frac{2 π}{4 π} = 0.5 s$ and ν = 2 Hz
φ = π/3; at t = 0, $x = 0.05 cos 60° = 0.025 m$

Answer: A = 5 cm, T = 0.5 s, ν = 2 Hz, φ = π/3, x(0) = 2.5 cm

SHM and uniform circular motion

NCERT section 13.4

Let a particle P move round a circle of radius A at constant angular speed ω. Its projection on a diameter (the foot of the perpendicular from P) moves to and fro along the diameter.

If OP makes angle $(ω t + ϕ)$ with the x-axis, the projection is at $x = A cos (ω t + ϕ)$ : SHM. So:

the radius of the circle is the amplitude;
the angular speed of P is the angular frequency of the SHM;
the angle OP makes is the phase.

The projections of P's velocity ( $ω A$ , along the tangent) and centripetal acceleration ( $ω^{2} A$ , towards O) on the diameter give the velocity and acceleration of the SHM. This 'reference circle' is a powerful way to find times: moving from the mean position to $x = \frac{A}{2}$ needs P to turn through 30°, which takes $\frac{T}{12}$ .

Worked example

A peg on a wheel of radius 10 cm turning at 2 revolutions per second casts a shadow on a wall. Describe the shadow's motion. [1 mark]

Try it first, then show the solutionSolution

The shadow is the projection of uniform circular motion: SHM
Amplitude = 10 cm; ω = 2π × 2 = 4π rad s $^{- 1}$ ; period 0.5 s

Answer: SHM with A = 10 cm, ω = 4π rad s $^{- 1}$ (T = 0.5 s)

Velocity and acceleration in SHM

NCERT section 13.5

Differentiating $x = A cos (ω t + ϕ)$ :

Velocity: $v = - ω A sin (ω t + ϕ)$ ; in terms of position, $∣ v ∣ = ω A^{2} - x^{2}$ . Maximum $ω A$ at the mean position; zero at the extremes.
Acceleration: $a = - ω^{2} A cos (ω t + ϕ) = - ω^{2} x$ . Maximum $ω^{2} A$ at the extremes; zero at the mean position. It always points towards the mean position.

Phase relations: velocity leads displacement by $\frac{π}{2}$ , and acceleration is $π$ out of phase with displacement. The equation $a = - ω^{2} x$ (or $\frac{d ^{2} x}{d t ^{2}} + ω^{2} x = 0$ ) is the test for SHM: if you can show it, the motion is SHM with $T = \frac{2 π}{ω}$ .

Worked example

A body performs SHM with amplitude 5 cm and period 0.2 s. Find its maximum speed and maximum acceleration. [2 marks]

Try it first, then show the solutionSolution

$ω = \frac{2 π}{0.2} = 10 π \approx 31.4 r a d s^{- 1}$
$v_{ma x} = ω A = 31.4 \times 0.05 \approx 1.57 m s^{- 1}$ ; $a_{ma x} = ω^{2} A = 986 \times 0.05 \approx 49.3 m s^{- 2}$

Answer: About 1.57 m s $^{- 1}$ and 49.3 m s $^{- 2}$

Force law for SHM and oscillations of a loaded spring

NCERT section 13.6

From Newton's second law, $F = ma = - m ω^{2} x$ . So SHM needs a restoring force proportional to the displacement:
$F = - k x$ , with $k = m ω^{2}$
k is the force constant (N m $^{- 1}$ ). Then
$ω = \frac{k}{m}$ and $T = 2 π \frac{m}{k}$

Loaded spring: a block of mass m on a smooth surface attached to a light spring of spring constant k. Stretching by x gives a restoring force $- k x$ (Hooke's law), so the block executes SHM with $T = 2 π \frac{m}{k}$ . The period does not depend on the amplitude. A heavier block or a softer spring gives a longer period.

Combinations: two springs side by side (parallel) act as one of constant $k_{1} + k_{2}$ ; end to end (series) as one of constant $\frac{k _{1} k _{2}}{k _{1} + k _{2}}$ . Cutting a spring in half doubles its k.

Worked example

A 0.5 kg block is attached to a spring of constant 200 N m $^{- 1}$ on a smooth table. Find the angular frequency and period of its oscillations. [1 mark]

Try it first, then show the solutionSolution

$ω = \frac{k}{m} = 400 = 20 r a d s^{- 1}$
$T = \frac{2 π}{20} \approx 0.314 s$

Answer: 20 rad s $^{- 1}$ ; about 0.31 s

Energy in SHM

NCERT section 13.7

For a particle in SHM (force constant $k = m ω^{2}$ ):

Kinetic energy: $K = \frac{1}{2} m v^{2} = \frac{1}{2} k (A^{2} - x^{2}) = \frac{1}{2} k A^{2} sin^{2} (ω t + ϕ)$
Potential energy: $U = \frac{1}{2} k x^{2} = \frac{1}{2} k A^{2} cos^{2} (ω t + ϕ)$
Total energy: $E = K + U = \frac{1}{2} k A^{2} = \frac{1}{2} m ω^{2} A^{2}$ , constant.

At the mean position the energy is all kinetic; at the extremes all potential. K and U each vary with period $\frac{T}{2}$ (they go through two cycles in one oscillation). They are equal at $x = \pm \frac{A}{2}$ . Total energy is proportional to the square of the amplitude.

Worked example

A block on a spring (k = 200 N m $^{- 1}$ ) oscillates with amplitude 0.1 m. Find its total energy, and its kinetic and potential energies at x = 0.05 m. [2 marks]

Try it first, then show the solutionSolution

$E = \frac{1}{2} \times 200 \times 0.01 = 1 J$
$U = \frac{1}{2} \times 200 \times 0.0025 = 0.25 J$ ; $K = 1 - 0.25 = 0.75 J$

Answer: E = 1 J; K = 0.75 J, U = 0.25 J

The simple pendulum

NCERT section 13.8

A simple pendulum is a small heavy bob of mass m on a light, inextensible string of length L, swinging in a vertical plane.

Derivation: when the string makes angle θ with the vertical, the tension passes through the pivot and has no torque; gravity gives a restoring torque $τ = - m g L sin θ$ . With moment of inertia $I = m L^{2}$ about the pivot, $τ = I α$ gives
$m L^{2} α = - m g L sin θ$
For small angles, $sin θ \approx θ$ (in radians; good to about 1% up to 15°), so
$α = - \frac{g}{L} θ$
This is the SHM equation with $ω^{2} = \frac{g}{L}$ :
$T = 2 π \frac{L}{g}$

The period does not depend on the mass of the bob or (for small swings) on the amplitude. A seconds pendulum has a period of 2 s; its length is about 0.99 m. A pendulum is used to measure g precisely.

Worked example

Find the period of a simple pendulum 0.64 m long, and the length needed for a period of 3.0 s. (g = 9.8 m s $^{- 2}$ ) [2 marks]

Try it first, then show the solutionSolution

$T = 2 π \frac{0.64}{9.8} \approx 1.61 s$
For T = 3.0 s: $L = \frac{g T ^{2}}{4 π ^{2}} = \frac{9.8 \times 9}{4 π ^{2}} \approx 2.23 m$

Answer: About 1.61 s; about 2.23 m

Physics in real life: Oscillations

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Simple pendulum on the Moon

Suppose a rover on Mars carries a pendulum clock that keeps correct time on the Earth. On Mars, g is about 0.38 of its value on Earth.

How does the period of the pendulum change, and does the clock run fast or slow?

Try it first, then show the solutionSolution

$T \propto \frac{1}{g}$ , so $\frac{T _{M a r s}}{T _{E a r t h}} = \frac{1}{0.38} \approx 1.62$
Each swing takes about 1.6 times longer, so the clock runs slow

Answer: Period about 1.6 times longer; the clock loses time

Memory hook: Weaker gravity, lazier pendulum.

Spring-mass oscillations

Imagine a car of mass 1000 kg resting on four identical springs, each with k = $2.5 \times 1 0^{4}$ N m $^{- 1}$ , after it goes over a bump.

Suppose the weight is shared equally and the shock absorbers are ignored. Find the period of the car's up-and-down oscillation.

Try it first, then show the solutionSolution

Each spring carries 250 kg: $T = 2 π \frac{250}{2.5 \times 1 0 ^{4}} = 2 π \times 0.1$
T ≈ 0.63 s (equivalently, four springs in parallel, 4k, holding 1000 kg)

Answer: About 0.63 s

Memory hook: Heavier load or softer spring: slower bounce.

Period independent of mass

Suppose two children of different masses sit on two identical swings with ropes 2.5 m long and are given small pushes.

Whose swing has the longer period? Find it. (g = 9.8 m s $^{- 2}$ )

Try it first, then show the solutionSolution

$T = 2 π \frac{L}{g}$ contains no mass, so the periods are equal
$T = 2 π \frac{2.5}{9.8} \approx 3.2 s$

Answer: Equal, about 3.2 s each

Memory hook: A pendulum cares about length and g, not the rider.

Topics in this chapter

Oscillations formula sheet

Period, frequency, angular frequency: $ν = \frac{1}{T}, ω = \frac{2 π}{T} = 2 π ν$
Displacement in SHM: $x = A cos (ω t + ϕ)$
Velocity in SHM: $v = - ω A sin (ω t + ϕ), ∣ v ∣ = ω A^{2} - x^{2}, v_{ma x} = ω A$
Acceleration in SHM: $a = - ω^{2} x, a_{ma x} = ω^{2} A$
Force law: $F = - k x, k = m ω^{2}$
Spring-mass system: $T = 2 π \frac{m}{k}$
Springs in parallel and series: $k_{p} = k_{1} + k_{2}, \frac{1}{k _{s}} = \frac{1}{k _{1}} + \frac{1}{k _{2}}$
Energy in SHM: $K = \frac{1}{2} k (A^{2} - x^{2}), U = \frac{1}{2} k x^{2}, E = \frac{1}{2} k A^{2}$
Simple pendulum: $T = 2 π \frac{L}{g}$

How the school exam asks Oscillations

The question patterns to expect, and the method that works for each.

Question type	How it is asked	Method
Periodic and SHM basics	Which functions are periodic or SHM; amplitude, period and phase from an equation; combining sin and cos terms.	SHM needs a single sine or cosine of ωt (or a sum of the same frequency); $a sin ω t + b cos ω t$ has amplitude $a^{2} + b^{2}$ .
Velocity, acceleration and phase	Speed at a given displacement; maximum speed and acceleration; v and a at a given time.	$v = ω A^{2} - x^{2}$ , $a = - ω^{2} x$ ; differentiate x(t) for time-based questions.
Spring systems	Period of a block on a spring; effect of cutting or combining springs; showing motion is SHM.	Find the effective k; $T = 2 π m / k$ .
Energy in SHM	K, U and E at a point; where K = U; graphs of energy against x or t.	E = ½kA² constant; K = E − U; K = U at $x = A / 2$ .
Simple pendulum	Derive the period; seconds pendulum; effect of length, g, mass, lift acceleration, the Moon.	Torque about the pivot with small-angle approximation; $T \propto L / g_{e f f}$ .

Tricks to remember Oscillations

Short memory aids. Say each one aloud once, then use it on a question the same day.

Mean: fast and free; extreme: stopped and pulled
At the mean position v is maximum and a = 0. At the extremes v = 0 and a is maximum.
Use it when: Comparing quantities at positions.
a = −ω²x is the fingerprint of SHM
If you can write acceleration as minus a constant times displacement, it is SHM with ω = √constant.
Use it when: Proving motion is SHM.
m over k, L over g
Spring: $T = 2 π m / k$ . Pendulum: $T = 2 π L / g$ . 'What resists' over 'what restores'.
Use it when: Period formulas.
Energy is half k A squared
E = ½kA² always; K = U at A/√2.
Use it when: Energy questions.
Phase ladder x, v, a
v is π/2 ahead of x, a is π ahead of x (opposite).
Use it when: Phase-difference MCQs.

Oscillations: exam-level solved questions

Ordered from exam level to HOTS. Cover the solution, write your own, then compare step by step.

Example 1: Where is acceleration largest? · 1 mark · MCQExam level

In simple harmonic motion, the acceleration of the particle is maximum:
(a) at the mean position
(b) at the extreme positions
(c) halfway between
(d) it is constant

Try it first, then show the solutionSolution

$a = - ω^{2} x$ is largest where |x| = A.

Answer: (b) at the extreme positions

Example 2: Periodic but not SHM · 1 mark · Assertion-reasonExam level

Assertion (A): The motion of the Earth round the Sun is periodic but not simple harmonic.
Reason (R): SHM is a to-and-fro motion about a mean position with a restoring force proportional to displacement.
(a) Both true, R explains A. (b) Both true, R does not explain A. (c) A true, R false. (d) A false, R true.

Try it first, then show the solutionSolution

The Earth repeats its orbit (periodic) but does not move to and fro about a mean position, so R explains why it is not SHM.

Answer: (a)

Example 3: Amplitude of a combination · 2 marks · Very short answerHarder exam level

The displacement of a particle is $x = 3 sin 2 t + 4 cos 2 t$ (x in cm, t in s). Find the amplitude and period, and the maximum speed. [2 marks]

Try it first, then show the solutionSolution

$A = 3^{2} + 4^{2} = 5 c m$ ; ω = 2 rad s $^{- 1}$ , so T = π s ≈ 3.14 s
$v_{ma x} = ω A = 10 c m s^{- 1}$

Answer: 5 cm; π s; 10 cm s $^{- 1}$

Example 4: Block on a spring · 3 marks · Short answerHarder exam level

A 2 kg block attached to a spring (k = 200 N m $^{- 1}$ ) oscillates on a smooth floor with amplitude 10 cm. Find (a) the period, (b) the maximum speed, (c) the speed when the block is 6 cm from the mean position. [3 marks]

Try it first, then show the solutionSolution

(a) $ω = \frac{200}{2} = 10 r a d s^{- 1}$ ; $T = \frac{2 π}{10} \approx 0.63 s$
(b) $v_{ma x} = ω A = 10 \times 0.10 = 1.0 m s^{- 1}$
(c) $v = ω A^{2} - x^{2} = 10 0.01 - 0.0036 = 10 \times 0.08 = 0.8 m s^{- 1}$

Answer: (a) 0.63 s (b) 1.0 m s $^{- 1}$ (c) 0.8 m s $^{- 1}$

Example 5: Pendulum on the Moon · 3 marks · Short answerHarder exam level

A pendulum has a period of 2 s on the Earth. What is its period on the Moon, where g is one-sixth of the Earth's value? What length would give a 2 s period there? (Earth g = 9.8 m s $^{- 2}$ ) [3 marks]

Try it first, then show the solutionSolution

$T \propto \frac{1}{g}$ : $T_{M oo n} = 2 \times 6 \approx 4.9 s$
For T = 2 s on the Moon: $L = \frac{g _{M} T ^{2}}{4 π ^{2}} = \frac{( 9.8/6 ) \times 4}{4 π ^{2}}$
$L \approx 0.165 m$ (one-sixth of the Earth's 0.993 m)

Answer: About 4.9 s; about 16.5 cm

Example 6: SHM, energy and the pendulum · 5 marks · Long answerHarder exam level

(a) Show that a block attached to a horizontal spring on a smooth surface executes SHM, and find its period. (b) Write expressions for its KE and PE at displacement x and show the total is constant. (c) For k = 400 N m $^{- 1}$ , m = 1 kg and A = 5 cm, find the total energy and the maximum speed. (d) At what displacement are KE and PE equal? [5 marks]

Try it first, then show the solutionSolution

(a) Restoring force $F = - k x$ , so $ma = - k x$ , $a = - \frac{k}{m} x$ : SHM with $ω = \frac{k}{m}$ , $T = 2 π \frac{m}{k}$
(b) $K = \frac{1}{2} k (A^{2} - x^{2})$ , $U = \frac{1}{2} k x^{2}$ , so $E = \frac{1}{2} k A^{2}$ , the same at every x
(c) $E = \frac{1}{2} \times 400 \times 0.0025 = 0.5 J$
$v_{ma x} = \frac{2 E}{m} = 1 m s^{- 1}$ (also ωA = 20 × 0.05)
(d) $\frac{1}{2} k x^{2} = \frac{1}{2} \cdot \frac{1}{2} k A^{2} \Rightarrow x = \frac{A}{2} \approx 3.5 c m$

Answer: $T = 2 π m / k$ ; E = 0.5 J; 1 m s $^{- 1}$ ; x ≈ 3.5 cm

Example 7: Springs combined · 3 marks · Short answerHOTS / competency

A block on a spring of constant k has period T. Find the period if (a) two such springs are used side by side, (b) two are joined end to end, (c) the original spring is cut in half and one half is used. [3 marks]

Try it first, then show the solutionSolution

(a) Parallel: 2k, so $T^{'} = \frac{T}{2} \approx 0.71 T$
(b) Series: $\frac{k}{2}$ , so $T^{'} = 2 T \approx 1.41 T$
(c) Half the length: 2k, so $T^{'} = \frac{T}{2}$

Answer: (a) T/√2 (b) √2 T (c) T/√2

Example 8: Tunnel through the Earth · 3 marks · Short answerHOTS / competency

Imagine a straight tunnel through the centre of the Earth (uniform density). Using $g (d) = g (1 - d / R)$ , show that a ball dropped into it executes SHM and find its period. (R = $6.4 \times 1 0^{6}$ m, g = 9.8 m s $^{- 2}$ ) [3 marks]

Try it first, then show the solutionSolution

At distance r from the centre, the acceleration is $g \frac{r}{R}$ towards the centre: $a = - \frac{g}{R} r$
This is SHM with $ω^{2} = \frac{g}{R}$ , so $T = 2 π \frac{R}{g}$
$T = 2 π \frac{6.4 \times 1 0 ^{6}}{9.8} \approx 5.1 \times 1 0^{3} s \approx 85 min$ (same as a satellite skimming the surface)

Answer: SHM; about 85 minutes

Case-study practice (Section E style)

Read the situation, then answer every part. In a school exam set on the board pattern this is a 4-mark question split into 1 + 1 + 1 + 1 marks.

A grandfather clock

Suppose a grandfather clock uses a simple pendulum that should tick once every second (a period of 2 s). Take g = 9.8 m s $^{- 2}$ .

(i) · 1 mark
What length should the pendulum have?
Answer
Answer: About 0.993 m
$L = \frac{g T ^{2}}{4 π ^{2}}$
(ii) · 1 mark
If the pendulum is made 1% too long, does the clock run fast or slow?
Answer
Answer: Slow
Longer L gives longer T.
(iii) · 1 mark
By roughly what percentage does the period change?
Answer
Answer: About 0.5%
$T \propto L$ : half the percentage change in L.
(iv) · 1 mark
Would a heavier bob change the timekeeping?
Answer
Answer: No
T does not depend on the mass of the bob.

Bouncing on a spring

Suppose a 0.5 kg toy hangs from a spring of constant 50 N m $^{- 1}$ and is pulled down 4 cm from equilibrium and released.

(i) · 1 mark
Find its angular frequency.
Answer
Answer: 10 rad s $^{- 1}$
$50/0.5$
(ii) · 1 mark
Find its period.
Answer
Answer: About 0.63 s
2π/10
(iii) · 1 mark
Find its maximum speed.
Answer
Answer: 0.4 m s $^{- 1}$
ωA = 10 × 0.04
(iv) · 1 mark
Find the total energy of the oscillation.
Answer
Answer: 0.04 J
$\frac{1}{2} \times 50 \times 0.0 4^{2}$

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Show that in SHM the speed of the particle at displacement x is $ω A^{2} - x^{2}$ . [2 marks] (2 marks)

Step	Marks
$x = A cos (ω t + ϕ)$ , so $v = - ω A sin (ω t + ϕ)$	1
$sin (ω t + ϕ) = 1 - \frac{x ^{2}}{A ^{2}}$ , giving $∣ v ∣ = ω A^{2} - x^{2}$	1

Show that the total mechanical energy of a particle in SHM is constant and equal to $\frac{1}{2} k A^{2}$ . [3 marks] (3 marks)

Step	Marks
$K = \frac{1}{2} m v^{2} = \frac{1}{2} m ω^{2} A^{2} sin^{2} (ω t + ϕ) = \frac{1}{2} k A^{2} sin^{2} (ω t + ϕ)$	1
$U = \frac{1}{2} k x^{2} = \frac{1}{2} k A^{2} cos^{2} (ω t + ϕ)$	1
$E = K + U = \frac{1}{2} k A^{2} (sin^{2} + cos^{2}) = \frac{1}{2} k A^{2}$ , independent of time	1

(a) Derive the expression for the period of a simple pendulum. (b) State two factors it does not depend on. (c) Find the period of a 1.0 m pendulum (g = 9.8 m s $^{- 2}$ ) and its period if it hangs in a lift accelerating upwards at g/3. [5 marks] (5 marks)

Step	Marks
(a) Restoring torque about the pivot: $τ = - m g L sin θ$ ; tension has zero torque	1
$I α = τ$ with $I = m L^{2}$ : $α = - \frac{g}{L} sin θ \approx - \frac{g}{L} θ$ for small θ: SHM with $ω^{2} = \frac{g}{L}$ , so $T = 2 π \frac{L}{g}$	1
(b) The mass of the bob and (for small swings) the amplitude	1
(c) $T = 2 π \frac{1.0}{9.8} \approx 2.01 s$	1
In the lift $g_{e f f} = g + \frac{g}{3} = \frac{4 g}{3}$ , so $T^{'} = T \frac{3}{4} \approx 1.74 s$	1

Where students lose marks in Oscillations

The slips examiners see most often, and what to write instead.

Mistake: Using degrees in ωt
Calculator in degree mode.
Do this: ωt and φ are in radians; switch the calculator to radian mode.
Usually costs: Whole numerical.
Mistake: Forgetting the small-angle condition in the pendulum derivation
Skipping the approximation step.
Do this: Write 'for small θ, sin θ ≈ θ' explicitly; it is a marked step.
Usually costs: 1 mark.
Mistake: Saying the period of a pendulum depends on the mass of the bob
Intuition from heavier swings.
Do this: T depends only on L and g (for small amplitude).
Usually costs: 1 mark.
Mistake: Using the total length of a combined spring wrongly
Not working out the effective k.
Wrong working
Spring cut in half: same k
Right working
Half the length: k doubles, T becomes $T / 2$
Do this: k is inversely proportional to the length of a uniform spring.
Usually costs: 1-2 marks.
Mistake: Writing about damped or forced oscillations and resonance
Old notes.
Do this: These were removed for 2026-27.
Usually costs: Time.

NCERT exercise guide: Oscillations

What each exercise in the NCERT chapter tests, how to approach it, and a question of the same kind solved step by step. Questions here are written in our own words.

Exercises 13.1-13.6 (periodic motion and SHM)

What it tests: Identifying periodic and SHM motions, x-t plots and functions, and acceleration-displacement relations.

How to approach it: SHM needs $a \propto - x$ ; a function is periodic if it repeats after a fixed time; combine terms of the same frequency.

Same kind as Exercise 13.4

Is $x = 3 sin ω t + 4 cos ω t$ simple harmonic? If so, find its amplitude and period.

Show the solutionSolution

Write it as $5 (\frac{3}{5} sin ω t + \frac{4}{5} cos ω t) = 5 sin (ω t + ϕ)$ with $tan ϕ = \frac{4}{3}$
A single sine of ωt: SHM with A = 5 and $T = \frac{2 π}{ω}$

Answer: Yes; amplitude 5, period 2π/ω

Exercises 13.7-13.14 (equations, springs and reference circles)

What it tests: Writing x(t) for given conditions, spring balances and spring-mass systems, reference circles and a locomotive piston.

How to approach it: Find ω from $k / m$ , A from the starting displacement, φ from where the motion starts.

Same kind as Exercise 13.9

A 4 kg block on a smooth floor is attached to a spring of constant 100 N m $^{- 1}$ . It is pulled 20 cm and released. Find its period, maximum speed and maximum acceleration.

Show the solutionSolution

$ω = \frac{100}{4} = 5 r a d s^{- 1}$ , $T = \frac{2 π}{5} \approx 1.26 s$
$v_{ma x} = ω A = 5 \times 0.2 = 1 m s^{- 1}$
$a_{ma x} = ω^{2} A = 25 \times 0.2 = 5 m s^{- 2}$

Answer: 1.26 s; 1 m s $^{- 1}$ ; 5 m s $^{- 2}$

Exercises 13.15-13.18 (pendulums and other oscillators)

What it tests: Pendulums on the Moon and in accelerating vehicles, a floating cork and a liquid column in a U-tube.

How to approach it: E = ½kA²; for a pendulum in an accelerating frame use the effective g.

Same kind as Exercise 13.16

A simple pendulum has a period of 2 s on the ground. What is its period in a lift that is (a) accelerating upwards at 2.2 m s $^{- 2}$ , (b) falling freely? (g = 9.8 m s $^{- 2}$ )

Show the solutionSolution

(a) $g_{e f f} = 9.8 + 2.2 = 12$ m s $^{- 2}$ ; $T^{'} = 2 \frac{9.8}{12} \approx 1.81 s$
(b) $g_{e f f} = 0$ : no restoring force, the pendulum does not oscillate (T infinite)

Answer: (a) about 1.81 s (b) it does not oscillate

Practice questions on Oscillations

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQExam level
The period of a simple pendulum does not depend on:
(a) its length
(b) the value of g
(c) the mass of the bob
(d) the place on Earth
Hint
$T = 2 π L / g$ .
Answer
(c) the mass of the bob
Q2 · 1 mark · MCQExam level
In SHM, the phase difference between displacement and velocity is:
(a) 0
(b) π/4
(c) π/2
(d) π
Hint
cos and −sin.
Answer
(c) π/2
Q3 · 1 mark · Assertion-reasonExam level
Assertion (A): The total energy of a particle in SHM stays constant.
Reason (R): Its kinetic and potential energies change, but their sum is always $\frac{1}{2} k A^{2}$ when there is no friction.
(a) Both true, R explains A. (b) Both true, R does not explain A. (c) A true, R false. (d) A false, R true.
Hint
sin² + cos² = 1.
Answer
(a)
Q4 · 2 marks · Very short answerExam level
Define simple harmonic motion and write its differential equation.
Hint
a ∝ −x.
Answer
SHM is periodic motion in which the acceleration is proportional to the displacement from the mean position and directed towards it: $\frac{d ^{2} x}{d t ^{2}} = - ω^{2} x$ .
Q5 · 1 mark · Very short answerHarder exam level
A particle in SHM has amplitude 10 cm and angular frequency 4 rad s $^{- 1}$ . Find its speed when it is 6 cm from the mean position.
Hint
$v = ω A^{2} - x^{2}$ .
Answer
$v = 4 100 - 36 = 32$ cm s $^{- 1}$ .
Q6 · 2 marks · Very short answerHarder exam level
The displacement of a particle is $x = 0.2 cos (5 t + \frac{π}{6})$ m. Find its velocity and acceleration at t = 0.
Hint
Differentiate.
Answer
$v = - 0.2 \times 5 sin \frac{π}{6} = - 0.5$ m s $^{- 1}$ ; $a = - 25 \times 0.2 cos \frac{π}{6} \approx - 4.33$ m s $^{- 2}$ .
Q7 · 1 mark · Very short answerHarder exam level
Find the length of a simple pendulum whose period is 1.5 s at a place where g = 9.8 m s $^{- 2}$ .
Hint
$L = g T^{2} /4 π^{2}$ .
Answer
$L = \frac{9.8 \times 2.25}{4 π ^{2}} \approx 0.56$ m.
Q8 · 2 marks · Very short answerHarder exam level
A 0.75 kg block on a spring of constant 300 N m $^{- 1}$ oscillates with amplitude 4 cm. Find its total energy and its kinetic energy when it is 2 cm from the mean position.
Hint
K = ½k(A² − x²).
Answer
$E = \frac{1}{2} \times 300 \times 0.0016 = 0.24$ J; $K = \frac{1}{2} \times 300 \times (0.0016 - 0.0004) = 0.18$ J.
Q9 · 2 marks · Very short answerHarder exam level
Using the reference circle, find the time a particle in SHM of period T takes to go from the mean position to half the amplitude.
Hint
Angle turned on the circle.
Answer
$x = A sin ω t = \frac{A}{2} \Rightarrow ω t = \frac{π}{6}$ , so $t = \frac{π /6}{2 π / T} = \frac{T}{12}$ .
Q10 · 2 marks · Very short answerHOTS / competency
A pendulum clock keeps correct time on the ground floor. It is taken to the top of a very tall tower. Will it gain or lose time? Explain.
Hint
g decreases with height.
Answer
g is slightly smaller at the top, so $T = 2 π L / g$ is slightly longer: each swing takes longer and the clock loses time.
Q11 · 3 marks · Short answerHOTS / competency
Show that the displacement in SHM can equally be written as $x = A sin (ω t + ϕ^{'})$ . How are φ and φ′ related? Why do both forms describe the same motion?
Hint
$cos θ = sin (θ + \frac{π}{2})$ .
Answer
$A cos (ω t + ϕ) = A sin (ω t + ϕ + \frac{π}{2})$ , so $ϕ^{'} = ϕ + \frac{π}{2}$ . Both have the same amplitude and ω, and the phase constant only reflects where the particle is at t = 0, so they describe the same SHM with a different choice of starting reference.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Define simple harmonic motion.Show answer
Periodic motion in which the acceleration is proportional to the displacement from the mean position and directed towards it ( $a = - ω^{2} x$ ).
2.What is the phase of SHM?Show answer
The quantity (ωt + φ) that fixes the position and direction of motion at time t.
3.Write the speed of a particle in SHM at displacement x.Show answer
$v = ω A^{2} - x^{2}$ .
4.Write the period of a mass on a spring.Show answer
$T = 2 π m / k$ .
5.What is the total energy of a particle in SHM?Show answer
$\frac{1}{2} k A^{2} = \frac{1}{2} m ω^{2} A^{2}$ .
6.Write the period of a simple pendulum.Show answer
$T = 2 π L / g$ (small oscillations).
7.What is a seconds pendulum?Show answer
A pendulum with period 2 s; its length is about 0.99 m on Earth.
8.What is the phase difference between displacement and acceleration in SHM?Show answer
π (they are opposite).

Best order to study it

Learn periodic and oscillatory motion, period, frequency and periodic functions.
Learn the SHM equation and the meaning of amplitude, angular frequency and phase.
Connect SHM to uniform circular motion (reference circle).
Derive velocity and acceleration and learn where each is maximum.
Learn the force law and the spring-mass system, including spring combinations.
Learn kinetic, potential and total energy in SHM.
Derive the period of the simple pendulum.

Quick revision checklist

The night before the exam: can you do each of these without looking?

Can you tell whether a given function of time is periodic or SHM?
Can you read A, ω, T and φ from an SHM equation?
Can you explain SHM as the projection of circular motion?
Can you find speed and acceleration at a given displacement?
Can you show that a block on a spring executes SHM?
Can you find the period after cutting or combining springs?
Can you find K and U at any displacement?
Can you sketch K, U and E against x?
Can you derive the period of a simple pendulum?
Can you predict the effect of a lift's acceleration on a pendulum's period?

Oscillations: frequently asked questions

What was removed from this chapter for 2026-27?

Damped simple harmonic motion and forced oscillations with resonance were removed. Everything up to the simple pendulum (NCERT 13.1 to 13.8) is in scope.

Do I need to derive the period of a simple pendulum?

Yes. The syllabus says 'simple pendulum derivation of expression for its time period'. Use torque about the pivot, I = mL^2 and the small-angle approximation.

Is every oscillation simple harmonic?

No. SHM needs a restoring force exactly proportional to displacement. A pendulum is SHM only for small angles; a ball bouncing on the floor oscillates but is not SHM.

Why do K and U oscillate with half the period of the motion?

They depend on \sin^2 and \cos^2 of the phase, which repeat every half cycle. K is maximum at the mean position, which the particle passes twice per period.

How does this chapter connect to later topics?

SHM of each particle of a medium makes a wave (next chapter), and in Class 12 alternating current and the fields of an electromagnetic wave vary sinusoidally, described exactly like SHM with amplitude, ω and phase.

Based on: CBSE Physics (042) syllabus 2026-27 (Classes XI-XII); NCERT Physics Part II, Textbook for Class XI, Chapter 13 (Reprint 2026-27)