Oscillations

Class 11

12 previous-year questions from this chapter every option and the correct answer, free · where the marks are

1. Simple harmonic motion: equations, energy

Exam focus: Total energy in SHM is constant and proportional to amplitude squared, but keeps trading between KE (maximum at the mean position) and PE (maximum at the extremes) — and acceleration itself is never constant, since a = −ω²x.

SHM: Equations and Energy

Physics Wallah - Alakh Pandey

Directly matches: equations of SHM and energy in SHM.

PhysicsNEET 2024show ▾

x = 5 sin (π t + \frac{π}{3})

m represents the motion of a particle executing simple harmonic motion, the amplitude and time period of the motion, respectively, are:

A.5 m, 2 s✓
B.5 cm, 1 s
C.5 m, 1 s
D.5 cm, 2 s

Solution

Comparing with $x = A sin (ω t + ϕ)$ : amplitude $A = 5$ m and $ω = π$ rad/s, so $T = \frac{2 π}{ω} = 2$ s.

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PhysicsNEET 2023show ▾

The

x

–

t

graph of a particle performing simple harmonic motion is shown in the figure. The acceleration of the particle at

t = 2

s is:

Question 38 as printed in the NEET 2023 booklet

A. $\frac{π ^{2}}{8} m s^{- 2}$
B. $- \frac{π ^{2}}{8} m s^{- 2}$
C. $\frac{π ^{2}}{16} m s^{- 2}$
D. $- \frac{π ^{2}}{16} m s^{- 2}$ ✓

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PhysicsNEET 2021show ▾

A body is executing simple harmonic motion with frequency

n

. The frequency of its potential energy is:

A. $n$
B. $2 n$ ✓
C. $3 n$
D. $4 n$

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PhysicsNEET 2020show ▾

The phase difference between the displacement and the acceleration of a particle in simple harmonic motion is:

A. $π$ rad✓
B. $\frac{3 π}{2}$ rad
C. $\frac{π}{2}$ rad
D.Zero

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PhysicsNEET 2019show ▾

The average velocity of a particle executing SHM in one complete vibration is:

A. $\frac{A ω}{2}$
B. $A ω$
C. $\frac{A ω ^{2}}{2}$
D.Zero✓

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PhysicsNEET 2019show ▾

The displacement of a particle executing simple harmonic motion is given by

y = A_{0} + A sin ω t + B cos ω t

. Then the amplitude of its oscillation is given by:

A. $A_{0} + A^{2} + B^{2}$
B. $A^{2} + B^{2}$ ✓
C. $A_{0}^{2} + (A + B)^{2}$
D. $A + B$

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PhysicsNEET 2019show ▾

A particle P moves in a circle of radius 3 m, centred at the origin in the

x y

-plane, with period

T = 4

s. At

t = 0

, P is at the topmost point of the circle (maximum

y

-displacement). The

y

-projection of the position vector of P, as a function of time, is:

A. $y (t) = - 3 cos 2 π t$ m
B. $y (t) = 4 sin (\frac{π t}{2})$ m
C. $y (t) = 3 cos (\frac{3 π t}{2})$ m
D. $y (t) = 3 cos (\frac{π t}{2})$ m✓

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2. Spring–mass system and simple pendulum

Exam focus: A spring-mass system's period, T = 2π√(m/k), doesn't involve g and works the same on the Moon, while a pendulum's period, T = 2π√(L/g), depends on effective g and simply stops swinging in a freely falling lift.

Spring-Block Oscillations and Time Period of SHM

Physics Wallah - Alakh Pandey

Thoroughly covers spring-mass/spring-block oscillations and time-period derivation; the simple pendulum is covered by the same channel in a separate lecture, not included here.

PhysicsNEET 2025show ▾

Two identical point masses P and Q, suspended from two separate massless springs of spring constants

k_{1}

and

k_{2}

respectively, oscillate vertically. If their maximum speeds are the same, the ratio

A_{Q} / A_{P}

of the amplitude

A_{Q}

of mass Q to the amplitude

A_{P}

of mass P is:

A. $\frac{k _{2}}{k _{1}}$
B. $\frac{k _{1}}{k _{2}}$
C. $\frac{k _{2}}{k _{1}}$
D. $\frac{k _{1}}{k _{2}}$ ✓

Solution

Maximum speed $= A ω$ , equal for both: $A_{Q} ω_{Q} = A_{P} ω_{P}$ . With $ω \propto k$ for equal masses, $\frac{A _{Q}}{A _{P}} = \frac{ω _{P}}{ω _{Q}} = \frac{k _{1}}{k _{2}}$ .

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PhysicsNEET 2025show ▾

In an oscillating spring–mass system, a spring is connected to a box filled with sand. As the box oscillates, sand leaks slowly out of the box vertically so that the average frequency

ω (t)

and average amplitude

A (t)

of the system change with time

t

. Which one of the following options (graphs shown) schematically depicts these changes correctly? Question 23 as printed in the NEET 2025 booklet

A.Graph (1)
B.Graph (2)✓
C.Graph (3)
D.Graph (4)

Solution

$ω = k / m$ , so as sand leaves and $m$ falls, $ω$ rises and levels off. Each grain that leaves carries away its share of the kinetic energy, so the oscillator's energy $\frac{1}{2} k A^{2}$ decreases and the amplitude falls — option (2).

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PhysicsNEET 2024show ▾

If the mass of the bob in a simple pendulum is increased to thrice its original mass and its length is made half its original length, then the new time period of oscillation is

\frac{x}{2}

times its original time period. Then the value of

x

is:

A.2✓
B. $23$
C.4
D.3

Solution

$T = 2 π \frac{l}{g}$ is independent of mass. Halving $l$ gives $T^{'} = \frac{T}{2} = \frac{2}{2} T$ , so $x = 2$ .

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PhysicsNEET 2022show ▾

Two pendulums of length 121 cm and 100 cm start vibrating in phase. At some instant, the two are at their mean position in the same phase. The minimum number of vibrations of the shorter pendulum after which the two are again in phase at the mean position is:

A.8
B.11✓
C.9
D.10

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PhysicsNEET 2021show ▾

A spring is stretched by 5 cm by a force of 10 N. The time period of the oscillations when a mass of 2 kg is suspended by it is:

A.0.0628 s
B.6.28 s
C.3.14 s
D.0.628 s✓

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3. Damped, forced oscillations and resonance

Exam focus: Resonance occurs when the driving frequency matches the system's own natural frequency, producing a sharp buildup in amplitude — the textbook reason soldiers break step crossing a bridge rather than march across it in rhythm.

Damped Oscillations, Angular SHM and Resonance

Physics Wallah - Alakh Pandey

Covers damped oscillations and resonance directly, plus angular SHM as a bonus topic.

Oscillations

1. Simple harmonic motion: equations, energy

Simple harmonic motion: equations and energy

2. Spring–mass system and simple pendulum

Spring–mass system and simple pendulum

3. Damped, forced oscillations and resonance

Damped, forced oscillations and resonance