Oscillations – JEE Main Practice Set MCQ 25 Questions

Q1.

In SHM, the kinetic energy becomes three times the potential energy at displacement:

A) $\frac{A}{2}$ 2A
B) $\frac{A}{\sqrt{2}}$ 2A
C) $\frac{A}{2\sqrt{2}}$ 22A
D) $\frac{A}{2} \sqrt{3}$ 2A3

Q2.

In an SHM system, 36% of total energy is kinetic at some displacement. The particle is at:

A) $0.8A$ 0.8A
B) $0.6A$ 0.6A
C) $0.5A$ 0.5A
D) $0.4A$ 0.4A

Q3.

For motion given by $a = -25x$ a=−25x, the time period is:

A) $\frac{2\pi}{5}$ 52π
B) $\frac{\pi}{5}$ 5π
C) $\frac{2\pi}{25}$ 252π
D) $\frac{\pi}{25}$ 25π

Q4.

Speed becomes half of maximum speed at displacement:

A) $\frac{A}{2}$ 2A
B) $\frac{\sqrt{3}A}{2}$ 23A
C) $\frac{A}{\sqrt{2}}$ 2A
D) $\frac{A}{\sqrt{3}}$ 3A

Q5.

Superposition of
$x_1 = A \sin \omega t$ x1=Asinωt
$x_2 = A \sin(\omega t + \pi/2)$ x2=Asin(ωt+π/2)
gives resultant amplitude:

A) $A$ A
B) $\sqrt{2}A$ 2A
C) $2A$ 2A
D) $A/2$ A/2

Q6.

If mass in spring system increases by 44%, time period increases by:

A) 20%
B) 22%
C) 44%
D) 10%

Q7.

At $x = \frac{A}{3}$ x=3A, acceleration is what fraction of maximum acceleration?

A) 1/3
B) 1/9
C) 1/2
D) 2/3

Q8.

If gravity becomes $4g$ 4g, time period of pendulum becomes:

A) $T/2$ T/2
B) $T/4$ T/4
C) $2T$ 2T
D) $T$ T

Q9.

At displacement $A/2$ A/2, potential energy is:

A) 1/2 of total
B) 1/4 of total
C) 3/4 of total
D) 1/3 of total

Q10.

Time taken from extreme to mean position is:

A) $T/4$ T/4
B) $T/2$ T/2
C) $T/8$ T/8
D) $T/6$ T/6

Q11.

If amplitude = 0.2 m and time period = 2 s, maximum speed is:

A) 0.2π
B) 0.4π
C) 0.1π
D) 0.8π

Q12.

Springs $k$ k and $3k$ 3k in series have equivalent constant:

A) $\frac{3k}{4}$ 43k
B) $4k$ 4k
C) $\frac{k}{4}$ 4k
D) $2k$ 2k

Q13.

Ratio of time from 0 to $A/2$ A/2 and $A/2$ A/2 to $A$ A is:

A) 1:1
B) 1:2
C) 2:1
D) 1:3

Q14.

If spring breaks at mean position, mass rises to height:

A) $\frac{A^2 \omega^2}{2g}$ 2gA2ω2
B) $\frac{A\omega}{g}$ gAω
C) $A$ A
D) $\frac{A^2}{g}$ gA2

Q15.

Using velocity values at two positions, angular frequency depends on:

A) Difference in displacement
B) Difference in squares of velocities
C) Difference in squares of displacement
D) Both B and C

Q16.

Pendulum taken inside Earth will:

A) Gain time
B) Lose time
C) No change
D) Stop oscillating

Q17.

Time spent near mean position compared to near extreme is:

A) Less
B) More
C) Equal
D) Zero

Q18.

Mass between two identical springs (k each) has effective constant:

A) k
B) 2k
C) k/2
D) 4k

Q19.

Amplitude when projected from mean with speed $v_0$ v0 is:

A) $v_0/\omega$ v0/ω
B) $v_0 \omega$ v0ω
C) $v_0^2/\omega$ v02/ω
D) $\omega/v_0$ ω/v0

Q20.

Cutting spring into two equal halves makes new constant:

A) k/2
B) 2k
C) k
D) 4k

Q21.

Superposition with phase difference $\pi$ π results in:

A) Amplitude 2A
B) Zero motion
C) A
D) √2A

Q22.

Average speed over one complete oscillation equals:

A) $\frac{2A}{T}$ T2A
B) $\frac{4A}{T}$ T4A
C) $\frac{\pi A}{T}$ TπA
D) $\frac{A}{T}$ TA

Q23.

Time from $A/2$ A/2 to $\sqrt{3}A/2$ 3A/2 equals:

A) $T/12$ T/12
B) $T/6$ T/6
C) $T/8$ T/8
D) $T/4$ T/4

Q24.

If total energy is 8 J and amplitude 0.1 m, spring constant is:

A) 800
B) 1600
C) 400
D) 200

Q25.

If pendulum length increases by 21%, time period increases by approximately:

A) 10%
B) 21%
C) 5%
D) 42%

Answer Only

Question No.	Answer
1	D
2	A
3	A
4	B
5	B
6	A
7	A
8	A
9	B
10	A
11	B
12	A
13	C
14	A
15	D
16	B
17	B
18	B
1

Solutions

Oscillations – Detailed Solutions

Q1

KE = 3 × PE

Total Energy: $E = \frac{1}{2}kA^2$ E=21kA2

At displacement $x$ x: $PE = \frac{1}{2}kx^2$ PE=21kx2 $KE = \frac{1}{2}k(A^2 – x^2)$ KE=21k(A2−x2)

Given: $\frac{1}{2}k(A^2 – x^2) = 3 \times \frac{1}{2}kx^2$ 21k(A2−x2)=3×21kx2 $A^2 – x^2 = 3x^2$ A2−x2=3×2 $A^2 = 4x^2$ A2=4×2 $x = \frac{A}{2}$ x=2A

Correct option: D

Q2

36% energy kinetic → 64% potential $\frac{1}{2}kx^2 = 0.64 \times \frac{1}{2}kA^2$ 21kx2=0.64×21kA2 $x^2 = 0.64A^2$ x2=0.64A2 $x = 0.8A$ x=0.8A

Correct option: A

Q3

Given: $a = -25x$ a=−25x

Compare with: $a = -\omega^2 x$ a=−ω2x $\omega^2 = 25$ ω2=25 $\omega = 5$ ω=5 $T = \frac{2\pi}{\omega} = \frac{2\pi}{5}$ T=ω2π=52π

Correct option: A

Q4

$v = \omega \sqrt{A^2 – x^2}$ v=ωA2−x2

Given: $v = \frac{v_{max}}{2} = \frac{\omega A}{2}$ v=2vmax=2ωA $\omega^2 (A^2 – x^2) = \frac{\omega^2 A^2}{4}$ ω2(A2−x2)=4ω2A2 $A^2 – x^2 = \frac{A^2}{4}$ A2−x2=4A2 $x^2 = \frac{3A^2}{4}$ x2=43A2 $x = \frac{\sqrt{3}A}{2}$ x=23A

Correct option: B

Q5

Phase difference = π/2

Resultant amplitude: $R = \sqrt{A^2 + A^2} = \sqrt{2}A$ R=A2+A2=2A

Correct option: B

Q6

$T = 2\pi \sqrt{\frac{m}{k}}$ T=2πkm

If mass increases 44%: $m’ = 1.44m$ m′=1.44m $T’ = T\sqrt{1.44} = 1.2T$ T′=T1.44=1.2T

20% increase

Correct option: A

Q7

$a = -\omega^2 x$ a=−ω2x $a_{max} = \omega^2 A$ amax=ω2A $\frac{a}{a_{max}} = \frac{x}{A}$ amaxa=Ax $= \frac{1}{3}$ =31

Correct option: A

Q8

$T = 2\pi \sqrt{\frac{L}{g}}$ T=2πgL

If $g \to 4g$ g→4g: $T’ = 2\pi \sqrt{\frac{L}{4g}} = \frac{T}{2}$ T′=2π4gL=2T

Correct option: A

Q9

$\frac{PE}{E} = \frac{x^2}{A^2}$ EPE=A2x2 $= \frac{(A/2)^2}{A^2}$ =A2(A/2)2 $= \frac{1}{4}$ =41

Correct option: B

Q10

Extreme to mean = quarter period $T/4$ T/4

Correct option: A

Q11

$v_{max} = \omega A$ vmax=ωA $\omega = \frac{2\pi}{T} = \pi$ ω=T2π=π $v_{max} = \pi \times 0.2 = 0.2\pi$ vmax=π×0.2=0.2π

Correct option: B

Q12

Series: $\frac{1}{k_{eq}} = \frac{1}{k} + \frac{1}{3k}$ keq1=k1+3k1 $= \frac{4}{3k}$ =3k4 $k_{eq} = \frac{3k}{4}$ keq=43k

Correct option: A

Q13

Time from 0 to A/2: $\sin \theta = 1/2$ sinθ=1/2 $\theta = \pi/6$ θ=π/6

Time = T/12

From A/2 to A: $\pi/2 – \pi/6 = \pi/3$ π/2−π/6=π/3

Time = T/6

Ratio: $1:2$ 1:2

Correct option: C

Q14

At mean: $v_{max} = A\omega$ vmax=Aω

Height: $\frac{1}{2}mv^2 = mgh$ 21mv2=mgh $h = \frac{A^2\omega^2}{2g}$ h=2gA2ω2

Correct option: A

Q15

From: $v^2 = \omega^2(A^2 – x^2)$ v2=ω2(A2−x2)

Subtract two equations → depends on both

Correct option: D

Q16

Inside Earth: $g’ = g(1 – d/R)$ g′=g(1−d/R)

T increases → clock loses time

Correct option: B

Q17

Speed highest near mean → spends less time

Spends more time near extreme

Correct option: B

Q18

Effective constant = 2k

Correct option: B

Q19

$v_{max} = A\omega$ vmax=Aω $A = \frac{v_0}{\omega}$ A=ωv0

Correct option: A

Q20

New length = L/2 $k’ = 2k$ k′=2k

Correct option: B

Q21

Phase difference π

Complete cancellation

Correct option: B

Q22

Total distance in one oscillation: $4A$ 4A

Average speed: $\frac{4A}{T}$ T4A

Correct option: B

Q23

$\sin^{-1}(1/2) = \pi/6$ sin−1(1/2)=π/6 $\sin^{-1}(\sqrt{3}/2) = \pi/3$ sin−1(3/2)=π/3

Difference = π/6

Time: $\frac{T}{12}$ 12T

Correct option: A

Q24

$E = \frac{1}{2}kA^2$ E=21kA2 $8 = \frac{1}{2}k(0.1)^2$ 8=21k(0.1)2 $8 = 0.005k$ 8=0.005k $k = 1600$ k=1600

Correct option: B

Q25

$T \propto \sqrt{L}$ T∝L $L’ = 1.21L$ L′=1.21L $T’ = T\sqrt{1.21} = 1.1T$ T′=T1.21=1.1T

10% increase

Correct option: A

Oscillations – JEE Main Practice Set MCQ 25 Questions

Q1.

Q2.

Q3.

Q4.

Q5.

Q6.

Q7.

Q8.

Q9.

Q10.

Q11.

Q12.

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Q14.

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Q17.

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Q19.

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Q21.

Q22.

Q23.

Q24.

Q25.

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Q20

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Q25

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