Oscillations — JEE Main Physics Notes & Formulas

1. Chapter at a Glance

An oscillation is a to-and-fro motion about a mean position. The simplest and most important kind is simple harmonic motion (SHM), where the restoring force is proportional to displacement. Springs, pendulums, vibrating strings and AC circuits all follow the same mathematics. The chapter covers the SHM equation, velocity and acceleration, energy, spring and pendulum systems, and an introduction to damped and forced oscillations. JEE Main asks one question almost every year.

What JEE asks most

Displacement, velocity, acceleration and phase in SHM.
Kinetic and potential energy at a given displacement.
Time period of spring systems (series/parallel) and pendulums (in lifts, on other planets).
Time taken between two positions.

2. Simple Harmonic Motion

SHM happens when F = −kx, i.e. a = −ω²x. The solution is:

x = A sin(ωt + φ) v = ω√(A² − x²) a = −ω²x

A = amplitude, ω = angular frequency, T = 2π/ω, f = 1/T, (ωt + φ) = phase, φ = initial phase.
Maximum speed v_max = Aω at the mean position; maximum acceleration a_max = Aω² at the extreme positions.
Velocity leads displacement by π/2; acceleration is π out of phase with displacement.
SHM is the projection of uniform circular motion on a diameter.

In SHM, velocity is a quarter cycle ahead of displacement, and acceleration is always opposite to displacement.

3. Energy in SHM

KE = ½mω²(A² − x²) PE = ½mω²x² E = ½mω²A² = ½kA²

Total energy is constant; KE and PE interchange. They are equal at x = A/√2.
KE and PE each oscillate with frequency 2f (period T/2).
Average KE = average PE = E/2 over a cycle.

4. Spring and Pendulum Systems

System	Time period
Mass on a spring	T = 2π√(m/k)
Springs in series	1/k = 1/k₁ + 1/k₂
Springs in parallel (or on both sides of a block)	k = k₁ + k₂
Spring cut into n equal parts	each part has constant nk
Simple pendulum (small angle)	T = 2π√(l/g)
Pendulum in lift (acceleration a up/down)	T = 2π√(l/(g ± a))
Liquid in U-tube (column length L)	T = 2π√(L/2g)
Floating body (submerged depth h)	T = 2π√(h/g)
Physical pendulum	T = 2π√(I/mgd)

A pendulum with period exactly 2 s is a "seconds pendulum" (length ≈ 1 m on Earth). The period of a spring–mass system does not depend on g, so it is the same on the Moon or in a lift.

5. Damped and Forced Oscillations

Damped oscillation: amplitude decreases with time because of friction or drag: A(t) = A₀e^−bt/2m.
Forced oscillation: a periodic driving force makes the system oscillate at the driver's frequency.
Resonance: when the driving frequency equals the natural frequency, the amplitude becomes very large (limited only by damping). Examples: soldiers break step on bridges; tuning a radio.

6. Formula Master Table

Quantity	Formula
Condition for SHM	a = −ω²x
Displacement	x = A sin(ωt + φ)
Velocity	v = ω√(A² − x²); v_max = Aω
Acceleration	a_max = Aω²
Total energy	E = ½kA²
Spring	T = 2π√(m/k)
Pendulum	T = 2π√(l/g_eff)
Time from mean to A/2	T/12

7. Mind Map

Oscillations

SHM basics

a = −ω²x
x, v, a and phase

Energy

KE + PE = ½kA²
Frequency 2f

Systems

Springs (series/parallel)
Pendulum, g_eff
U-tube, floating body

Damping

Decaying amplitude
Forced, resonance

8. Tips and Tricks

Shortcuts

Use the phasor (reference circle) picture for time questions: mean → A/2 takes T/12, A/2 → A takes T/6.
If two speeds v₁, v₂ at displacements x₁, x₂ are given: ω² = (v₁² − v₂²)/(x₂² − x₁²).
In any accelerated frame, replace g by the effective g for a pendulum.
A block between two springs attached to walls on either side: k_eff = k₁ + k₂.

Mistakes students make

Using the pendulum formula for large angles (it is valid only for small angles).
Thinking a spring–mass period changes on the Moon.
Confusing angular frequency ω with frequency f (ω = 2πf).

9. Important One-liners

In SHM, speed is maximum at the mean position and zero at the extremes.
Acceleration is maximum at the extremes and zero at the mean position.
The period of SHM does not depend on amplitude.
A pendulum does not oscillate in a freely falling lift.
Kinetic energy oscillates at twice the frequency of the motion.
At resonance, the amplitude of a forced oscillation is maximum.

10. Solved Numericals: Average to Advanced

Five average-level and five advanced-level problems. Click to reveal each answer and its full working. Take g = 10 m s⁻² and π² ≈ 9.87 where needed.

Average level

N1Average

A 0.5 kg mass is attached to a spring of constant 50 N m⁻¹. Find ω and the time period.

Show answer & solution

Concept: ω = √(k/m); T = 2π/ω.

ω = √(50/0.5) = √100 = 10 rad s⁻¹.
T = 2π/10 ≈ 0.628 s.

Answer: 10 rad s⁻¹; 0.628 s

N2Average

A particle moves as x = 5 sin(2t + π/6) cm (t in s). Find its amplitude, period, maximum speed and maximum acceleration.

Show answer & solution

Concept: compare with x = A sin(ωt + φ).

A = 5 cm; ω = 2 rad s⁻¹ ⇒ T = 2π/2 = π ≈ 3.14 s.
v_max = Aω = 10 cm s⁻¹.
a_max = Aω² = 20 cm s⁻².

Answer: 5 cm; π s; 10 cm s⁻¹; 20 cm s⁻²

N3Average

Find the length of a simple pendulum whose period is 2 s.

Show answer & solution

Concept: T = 2π√(l/g) ⇒ l = gT²/4π².

l = 10 × 4/(4 × 9.87).
≈ 1.01 m.

Answer: about 1.0 m

N4Average

For a particle in SHM of amplitude A, find (a) the ratio KE : PE and (b) the speed, when the displacement is A/2.

Show answer & solution

Concept: KE ∝ (A² − x²), PE ∝ x².

PE ∝ A²/4; KE ∝ A² − A²/4 = 3A²/4.
KE : PE = 3 : 1.
v = ω√(A² − A²/4) = (√3/2)Aω ≈ 0.87 v_max.

Answer: 3 : 1; (√3/2)Aω

N5Average

A 0.5 kg block on a spring (k = 50 N m⁻¹) oscillates with amplitude 0.1 m. Find its total energy and maximum speed.

Show answer & solution

Concept: E = ½kA²; at the mean position E is all kinetic.

E = ½ × 50 × 0.01 = 0.25 J.
½ × 0.5 × v_max² = 0.25 ⇒ v_max = 1 m s⁻¹ (also Aω = 0.1 × 10).

Answer: 0.25 J; 1 m s⁻¹

Advanced level

N6Advanced

Two identical springs, each 100 N m⁻¹, support the same mass, first in series and then in parallel. Find the effective spring constants and the ratio of the periods T_series : T_parallel.

Show answer & solution

Concept: series 1/k = 1/k₁ + 1/k₂; parallel k = k₁ + k₂; T ∝ 1/√k.

k_s = 100 × 100/200 = 50 N m⁻¹.
k_p = 200 N m⁻¹.
T_s/T_p = √(k_p/k_s) = √4 = 2.

Answer: 50 and 200 N m⁻¹; 2 : 1

N7Advanced

A 1 m simple pendulum hangs in a lift. Find its period when the lift (a) is at rest, (b) accelerates upward at 2.5 m s⁻², (c) falls freely.

Show answer & solution

Concept: T = 2π√(l/g_eff), with g_eff = g + a (upward acceleration).

(a) T = 2π√(1/10) ≈ 1.99 s.
(b) g_eff = 12.5 m s⁻²: T = 2π√(1/12.5) ≈ 1.78 s.
(c) g_eff = 0: the pendulum does not oscillate (T → ∞).

Answer: 1.99 s; 1.78 s; no oscillation

N8Advanced

A particle executes SHM with period 12 s, starting from the mean position. Find the time it takes to reach (a) half the amplitude, (b) the extreme position from half the amplitude.

Show answer & solution

Concept: x = A sin(ωt), ω = 2π/T.

(a) A/2 = A sin(ωt) ⇒ ωt = π/6 ⇒ t = (π/6)/(2π/12) = 1 s (= T/12).
Time from mean to extreme = T/4 = 3 s.
(b) From A/2 to A = 3 − 1 = 2 s (= T/6).

Answer: 1 s; 2 s

N9Advanced

(a) A liquid column of total length 0.4 m oscillates in a U-tube. Find its period. (b) A block of mass 1 kg is fixed between two walls by springs of 100 N m⁻¹ and 300 N m⁻¹ on either side. Find its period.

Show answer & solution

Concept: (a) U-tube: restoring force = 2ρAgx for mass ρAL ⇒ T = 2π√(L/2g). (b) Both springs push back ⇒ k = k₁ + k₂.

(a) T = 2π√(0.4/20) = 2π√0.02 ≈ 0.89 s.
(b) k = 100 + 300 = 400 N m⁻¹.
T = 2π√(1/400) = 2π/20 ≈ 0.314 s.

Answer: (a) about 0.89 s; (b) about 0.31 s

N10Advanced

A particle in SHM has speed 3 cm s⁻¹ at 4 cm from the mean position and 4 cm s⁻¹ at 3 cm from it. Find the amplitude and period.

Show answer & solution

Concept: v² = ω²(A² − x²) at both points.

9 = ω²(A² − 16) and 16 = ω²(A² − 9).
Subtract: 7 = ω²(16 − 9) = 7ω² ⇒ ω = 1 rad s⁻¹.
Then 9 = A² − 16 ⇒ A² = 25 ⇒ A = 5 cm.
T = 2π/ω = 2π ≈ 6.28 s.

Answer: 5 cm; 2π s

11. PYQ-pattern Questions with Solutions

These questions are original and modelled on concepts frequently tested in JEE Main.

Q1Easy

In SHM, the acceleration is maximum:

At the mean position
At the extreme positions
Midway
Everywhere the same

Show solution

a = −ω²x is largest in magnitude when |x| = A.

Answer: (B)

Q2Easy

A particle performs SHM with frequency f. Its kinetic energy varies with frequency:

Show solution

KE ∝ cos²(ωt) = (1 + cos 2ωt)/2, so it repeats at 2f.

Answer: (C)

Q3Medium

A pendulum clock is taken to the Moon (g_moon = g/6). Its period becomes:

T/6
6T
√6 T
T/√6

Show solution

T ∝ 1/√g, so T′ = T√6.

Answer: (C)

Q4Hard

The phase difference between displacement and acceleration of a particle in SHM is:

0
π/2
π
3π/2

Show solution

a = −ω²x = ω²x sin(… + π): acceleration is always opposite to displacement.

Answer: (C)

Q5Numerical

A spring of constant 100 N m⁻¹ is cut into two equal halves. Find the spring constant (in N m⁻¹) of each half.

Show solution

k ∝ 1/length, so each half has 2 × 100 = 200 N m⁻¹.

Answer: 200

Q6Numerical

A particle moves as x = 4 cos(πt) m. Find its period in seconds.

Show solution

ω = π ⇒ T = 2π/π = 2 s.

Answer: 2

12. Quick Revision Summary

Remember before the exam

SHM: a = −ω²x; x = A sin(ωt + φ); v = ω√(A² − x²).
v_max = Aω; a_max = Aω²; E = ½kA².
Spring T = 2π√(m/k); series/parallel rules; cut spring stiffer.
Pendulum T = 2π√(l/g_eff).
Mean → A/2: T/12; A/2 → A: T/6.

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