Simple Harmonic Motion
MHT CET / Physics / Mechanics / 188 questions
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Simple Harmonic Motion Questions
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1Simple Harmonic Motion
Two particles ' $A$ ' and ' $B$ ' execute SHMs of periods ' $T$ ' and $\frac{3 T}{2}$. If they start from the mean position then the phase difference between them, when the particle ' $A$ ' completes two oscillations will be
MCQ+1 / -02025
2Simple Harmonic Motion
The displacement of particle in S.H.M. is $\mathrm{x}=\mathrm{A} \cos (\omega \mathrm{t}+\pi / 6)$. Its speed will be maximum at time $\left(\sin 90^{\circ}=1\right)$
MCQ+1 / -02025
3Simple Harmonic Motion
As shown in the figure, $S_1$ and $S_2$ are identical springs with spring constant K each. The oscillation frequency of the mass ' $m$ ' is ' $f$ '. If the spring $\mathrm{S}_2$ is removed, the oscillation frequency will become
MCQ+1 / -02025
4Simple Harmonic Motion
Two simple pendulums have first (A) bob of mass ' $M_1$ ' and length ' $L_1$ ', second (B) of mass ' $\mathrm{M}_2$ ' and length ' $\mathrm{L}_2$ '. $\mathrm{M}_1=\mathrm{M}_2$ and $\mathrm{L}_1=2 \mathrm{~L}_2$. If their total energies are...
MCQ+1 / -02025
5Simple Harmonic Motion
A particle starts oscillating simple harmonically from its mean position with time period ' $T$ '. At time $\mathrm{t}=\frac{\mathrm{T}}{6}$, the ratio of the potential energy to kinetic energy of the particle is
$$ \left[\sin 30^{\circ}=\c...
$$ \left[\sin 30^{\circ}=\c...
MCQ+1 / -02025
6Simple Harmonic Motion
The amplitude of a damped oscillator becomes $\left(\frac{1}{3}\right)^{\mathrm{rd}}$ of original amplitude in 2 seconds. If its amplitude after 6 second become $\left(\frac{1}{n}\right)$ times the original amplitude, the value of $n$ is ( ...
MCQ+1 / -02025
7Simple Harmonic Motion
A particle is executing linear S.H.M. starting from mean position. The ratio of the kinetic energy to the potential energy of the particle at a point of half the amplitude is
MCQ+1 / -02025
8Simple Harmonic Motion
A particle is executing S.H.M. of amplitude ' $A$ '. When the potential energy of the particle is half of its maximum value during the oscillation, its displacement from the equilibrium position is
MCQ+1 / -02025
9Simple Harmonic Motion
An object of mass 0.2 kg executes simple harmonic oscillations along the x -axis with frequency of $\left(\frac{25}{\pi}\right) \mathrm{Hz}$. At the position $x=0.04 \mathrm{~m}$, the object has kinetic energy 1 J and potential energy 0.6 J...
MCQ+1 / -02025
10Simple Harmonic Motion
The motion of the particle is given by the equation $\mathrm{x}=\mathrm{A} \sin \omega \mathrm{t}+\mathrm{B} \cos \omega \mathrm{t}$.
The motion of the particle is
The motion of the particle is
MCQ+1 / -02025
11Simple Harmonic Motion
For a particle performing S.H.M.; the total energy is ' $n$ ' times the kinetic energy, when the displacement of a particle from mean position is $\frac{\sqrt{3}}{2} \mathrm{~A}$, where A is the amplitude of S.H.M. The value of ' $n$ ' is
MCQ+1 / -02025
12Simple Harmonic Motion
A particle performing linear S.H.M. has period 8 seconds. At time $\mathrm{t}=0$, it is in the mean position. The ratio of the distances travelled by the particle in the $1^{\text {st }}$ and $2^{\text {nd }}$ second is $\left(\cos 45^{\cir...
MCQ+1 / -02025
13Simple Harmonic Motion
A small spherical ball of radius ' $r$ ' is rolling on a curved surface which is frictionless and has a radius of curvature ' R '. Its motion is simple harmonic. Then its tine period of oscillation is proportional to ( $\mathrm{g}=$ acceler...
MCQ+1 / -02025
14Simple Harmonic Motion
A particle executes S.H.M. starting from the mean position. Its amplitude is ' a ' and its periodic time is ' $T$ '. At a certain instant, its speed ' $u$ ' is half that of maximum speed $\mathrm{V}_{\text {max }}$. The displacement of the ...
MCQ+1 / -02025
15Simple Harmonic Motion
A simple pendulum has time period ' $\mathrm{T}_1$ '. The point of suspension is now moved upward according to equation $\mathrm{y}=\mathrm{kt}^2$ where $\mathrm{k}=1 \mathrm{~m} / \mathrm{s}^2$. If new time period is ' $\mathrm{T}_2$ ' the...
MCQ+1 / -02025
16Simple Harmonic Motion
Two simple harmonic motions of angular frequency $300 \mathrm{rad} / \mathrm{s}$ and $3000 \mathrm{rad} / \mathrm{s}$ have same amplitude. The ratio of their maximum accelerations is
MCQ+1 / -02025
17Simple Harmonic Motion
A mass $m$ is suspended from a spring of negligible mass. The spring is pulled a little and then released, so that mass executes S.H.M. of time period $T$. If the mass is increased by $m_0$, the periodic time becomes $\frac{5 \mathrm{~T}}{4...
MCQ+1 / -02025
18Simple Harmonic Motion
The length of the simple pendulum is made 3 times the original length. If ' T ' is its original time period, then the new time period will be
MCQ+1 / -02025
19Simple Harmonic Motion
The period of S. H.M. of a particle is 16 second. The phase difference between the positions at $\mathrm{t}=2 \mathrm{~s}$ and $\mathrm{t}=4 \mathrm{~s}$ will be
MCQ+1 / -02025
20Simple Harmonic Motion
A particle oscillates in straight line simple harmonically with period 8 second and amplitude $4 \sqrt{2} \mathrm{~m}$. Particle starts from mean position. The ratio of the distance travelled by it in $1^{\text {st }}$ second of its motion ...
MCQ+1 / -02025
21Simple Harmonic Motion
If the period of a oscillation of mass ' m ' suspended from a spring is 2 s , then the period of suspended mass ' 4 m ' with the same spring will be
MCQ+1 / -02025
22Simple Harmonic Motion
A mass $x$ gram is suspended from a light spring. It is pulled in downward direction and released so that mass performs S.H.M. of period T. If mass is increased by Y gram, the period becomes $\frac{4 \mathrm{~T}}{3}$. The ratio of $\mathrm{...
MCQ+1 / -02025
23Simple Harmonic Motion
A particle executes a linear S.H.M. In two of its positions the velocities are $\mathrm{V}_1, \mathrm{~V}_2$ and accelerations are $\mathrm{a}_1$ and $\mathrm{a}_2$ respectively $\left(0
MCQ+1 / -02025
24Simple Harmonic Motion
A mass attached to a spring performs S.H.M. whose displacement is $\mathrm{x}=3 \times 10^{-3} \cos 2 \pi \mathrm{t}$ metre. The time taken to obtain maximum speed for the first time is
MCQ+1 / -02025
25Simple Harmonic Motion
The time period of a simple pendulum inside a stationary lift is $\sqrt{3}$ second. When the lift moves upwards with an acceleration $g / 3$, the time period will be ( $\mathrm{g}=$ acceleration due to gravity)
MCQ+1 / -02025
26Simple Harmonic Motion
' $P$ ' and ' $Q$ ' are fixed points in same plane and mass ' $m$ ' is tied by string as shown in figure. If the mass is displaced slightly out of this plane and released, it will oscillate with time period $(\mathrm{PQ}=2 \mathrm{~d}, \mat...
MCQ+1 / -02025
27Simple Harmonic Motion
A mass suspended from a vertical spring performs S.H.M. of period 0.1 second. The spring is unstretched at the highest point of suspension. Maximum speed of the mass is (Gravitational acceleration $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2$ ...
MCQ+1 / -02025
28Simple Harmonic Motion
For a particle performing; S.H.M. the displacement - time graph is shown.
For that particle the force - time graph is correctly shown in graph
For that particle the force - time graph is correctly shown in graph
MCQ+1 / -02025
29Simple Harmonic Motion
If the length of the oscillating simple pendulum is made $\frac{1}{3}$ times the original keeping amplitude same then increase in its total energy at a place will be
MCQ+1 / -02025
30Simple Harmonic Motion
A vertical spring oscillates with period 6 second with mass $m$ is suspended from it. When the mass is at rest, the spring is stretched through a distance of (Take, acceleration due to gravity, $\mathrm{g}=\pi^2=10 \mathrm{~m} / \mathrm{s}^...
MCQ+1 / -02025
31Simple Harmonic Motion
A particle is performing S.H.M. with an amplitude 4 cm . At the mean position the velocity of the particle is $12 \mathrm{~cm} / \mathrm{s}$. When the speed of the particle becomes $6 \mathrm{~cm} / \mathrm{s}$, the distance of the particle...
MCQ+1 / -02024
32Simple Harmonic Motion
In S.H.M. the displacement of a particle at an instant is $Y=A \cos 30^{\circ}$, where $A=40 \mathrm{~cm}$ and kinetic energy is 200 J . If force constant is $1 \times 10^{\times} \mathrm{N} / \mathrm{m}$, then x will be $\left(\cos 30^{\ci...
MCQ+1 / -02024
33Simple Harmonic Motion
A particle is executing a linear simple harmonic motion. Let ' $\mathrm{V}_1$ ' and ' $\mathrm{V}_2$ ' are its speed at distance ' $x_1$ ' and ' $x_2$ ' from the equilibrium position. The amplitude of oscillation is
MCQ+1 / -02024
34Simple Harmonic Motion
Two bodies A and B of equal mass are suspended from two separate massless springs of spring constants $\mathrm{K}_1$ and $\mathrm{K}_2$ respectively. The two bodies oscillate vertically such that their maximum velocities are equal. The rati...
MCQ+1 / -02024
35Simple Harmonic Motion
The velocity of particle executing S.H.M. varies with displacement $(\mathrm{x})$ as $4 \mathrm{~V}^2=50-\mathrm{x}^2$. The time period of oscillation is $\frac{x}{7}$ second. The value of ' $x$ ' is (Take $\pi=\frac{22}{7}$)
MCQ+1 / -02024
36Simple Harmonic Motion
A simple pendulum of length $l_1$ has time period $\mathrm{T}_1$. Another simple pendulum of length $l_2\left(l_1>l_2\right)$ has time period $T_2$. Then the time period of the pendulum of length $\left(l_1-l_2\right)$ will be
MCQ+1 / -02024
37Simple Harmonic Motion
The maximum velocity and maximum acceleration of a particle performing a linear S.H.M. is ' $\alpha$ ' and ' $\beta$ ' respectively. Then the path length of the particle is
MCQ+1 / -02024
38Simple Harmonic Motion
A mass ' $m$ ' attached to a spring oscillates with a period of 3 second. If the mass is increased by 0.6 kg , the period increases by 3 second. The initial mass ' $m$ ' is equal to
MCQ+1 / -02024
39Simple Harmonic Motion
A tube of uniform bore of cross-sectional area ' $A$ ' has been set up vertically with open end facing up. Now ' $M$ ' gram of a liquid of density ' $d$ ' is poured into it. The column of liquid in this tube will oscillate with a period ' T...
MCQ+1 / -02024
40Simple Harmonic Motion
A small sphere oscillates simple harmonically in a watch glass whose radius of curvature is 1.6 m . The period of oscillation of the sphere in second is (acceleration due to gravity, $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2$ )
MCQ+1 / -02024
41Simple Harmonic Motion
When a particle in linear S.H.M. completes two oscillations, its phase increases by
MCQ+1 / -02024
42Simple Harmonic Motion
Frequency of a particle performing S.H.M. is 10 Hz . The particle is suspended from a vertical spring. At the highest point of its oscillation the spring is unstretched. Maximum speed of the particle is $\left(\mathrm{g}=10 \mathrm{~m} / \m...
MCQ+1 / -02024
43Simple Harmonic Motion
The period of a simple pendulum gets doubled when
MCQ+1 / -02024
44Simple Harmonic Motion
Three masses $500 \mathrm{~g}, 300 \mathrm{~g}$ and 100 g are suspended at the end of spring as shown in figure and are in equilibrium. When the 500 g mass is removed, the system oscillates with a period of 3 second. When the 300 g mass is ...
MCQ+1 / -02024
45Simple Harmonic Motion
A particle is performing simple harmonic motion and if the oscillations are Camped oscillations then the angular frequency is given by
MCQ+1 / -02024
46Simple Harmonic Motion
Choose the correct answer.
When a point of suspension of pendulum is moved vertically upward with acceleration ' $a$ ', its period of oscillation
When a point of suspension of pendulum is moved vertically upward with acceleration ' $a$ ', its period of oscillation
MCQ+1 / -02024
47Simple Harmonic Motion
For a body performing simple harmonic motion, its potential energy is $\mathrm{E}_{\mathrm{x}}$ at displacement x and $\mathrm{E}_{\mathrm{y}}$ at displacement y from mean position. The potential energy $E_0$ at displacement $(x+y)$ is
MCQ+1 / -02024
48Simple Harmonic Motion
A spring has a certain mass suspended from it and its period of vertical oscillations is $T_1$. The spring is now cut into two equal halves and the same mass is suspended from one of the halves. The period of vertical oscillations is now $\...
MCQ+1 / -02024
49Simple Harmonic Motion
The displacement of a particle performing S.H.M. is given by $Y=A \cos [\pi(t+\phi)]$. If at $\mathrm{t}=0$, the displacement is $\mathrm{y}=2 \mathrm{~cm}$ and velocity is $2 \pi \mathrm{~cm} / \mathrm{s}$, the value of amplitude $A$ in cm...
MCQ+1 / -02024
50Simple Harmonic Motion
A particle executing S.H.M. has velocities ' $\mathrm{V}_1$ ' and ' $\mathrm{V}_2$ ' at distances ' $x_1$ ' and ' $x_2$ ' respectively, from the mean position. Its frequency is
MCQ+1 / -02024
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