Simple Harmonic Motion PYQs - Last 5 Years
TS EAMCET / Physics / Mechanics / 23 recent questions
PhysicsMechanics2021-2025
Practice 23 TS EAMCET Physics questions from Simple Harmonic Motion. Use the year-wise and type-wise breakdown to prioritize recent PYQs, then continue into the question list below.
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Last 5 Years Simple Harmonic Motion Questions
Showing 23 of 23 filtered questions.
1Simple Harmonic Motion
If the amplitudes of a damped harmonic oscillator at times $t=0, t_1$ and $t_2$ are $A_0, A_1$ and $A_2$ respectively, then the amplitude of the oscillator at a time of $\left(t_1+t_2\right)$ is
MCQ+1 / -02025
2Simple Harmonic Motion
The force ( $F$ in newton) acting on a particle of mass 90 g executing simple harmonic motion is given by $F+0.04 \pi^2 y=0$, where $y$ is displacement of the particle in metre. If the amplitude of the particle is $\frac{6}{\pi} \mathrm{~m}...
MCQ+1 / -02025
3Simple Harmonic Motion
If the amplitude of a damped harmonic oscillator becomes half of its initial amplitude in a time of 10 s , then the time taken for the mechanical energy of the oscillator to become half of its initial mechanical energy is
MCQ+1 / -02025
4Simple Harmonic Motion
At a given place, to increase the number of oscillations made by a simple pendulum in one minute from 72 to 90 , the length of the pendulum is to be decreased by
MCQ+1 / -02025
5Simple Harmonic Motion
The amplitude of a particle executing simple harmonic motion is 6 cm . The distance of the point from the mean position at which the ratio of the potential and kinetic energies of the particle becomes $4: 5$ is
MCQ+1 / -02025
6Simple Harmonic Motion
A particle is executing simple harmonic motion. If the force acting on the particle at a position is $86.6 \%$ of the maximum force on it, then the ratio of its velocity at that point and its maximum velocity is
MCQ+1 / -02025
7Simple Harmonic Motion
In a time $t$ amplitude of vibrations of a damped oscillator becomes half of its initial value, then the mechanical energy of the oscillator decreases by
MCQ+1 / -02024
8Simple Harmonic Motion
A particle of mass 4 mg is executing simple harmonic motion along $X$-axis with an angular frequency of $40 \mathrm{rad} \mathrm{s}^{-1}$. If the potential energy of the particle is $V(x)=a+b x^2$, where $V(x)$ is in joule and $x$ is in met...
MCQ+1 / -02024
9Simple Harmonic Motion
A massless spring of length $l$ and spring constant $k$ oscillates with a time period $T$ when loaded with a mass $m$. The spring is now cut into three equal parts and are connected in parallel. The frequency of oscillation of the combinati...
MCQ+1 / -02024
10Simple Harmonic Motion
In a simple pendulum experiment for the determination of acceleration due to gravity, the error in the measurement of the length of the pendulum is $1 \%$ and the error in the measurement of the time period is $2 \%$. The error in the estim...
MCQ+1 / -02024
11Simple Harmonic Motion
If a body dropped freely from a height of 20 m reaches the surface of a planet with a velocity of $31.4 \mathrm{~ms}^{-1}$. then the length of a simple pendulum that ticks seconds on the planet is
MCQ+1 / -02024
12Simple Harmonic Motion
The displacement of a particle executing simple harmonic motion is given by $x=2 \cos (t)$ where $t$ is the time in seconds then the time period of the particle is
MCQ+1 / -02023
13Simple Harmonic Motion
The displacement of a particle is given by the relation $x=4(\cos \pi t+\sin \pi t)$. The amplitude of the particle is
MCQ+1 / -02023
14Simple Harmonic Motion
A pendulum has a time period $T$ in air. Whạt it is made to oscillate in water its time period is $\sqrt{2} T$. Then the relative density of the material of the bob of the pendulum is (neglect damping)
MCQ+1 / -02023
15Simple Harmonic Motion
A force of 6.4 N stretches a vertical spring by 0.1 m . If it were to oscillate with a period of $\pi / 4$, then the mass that is to be suspended from the spring is
MCQ+1 / -02023
16Simple Harmonic Motion
For a particle executing simple harmonic motion, the kinetic energy of the particle at a distance of 4 cm from the mean position is $1 / 3$ rd of the maximum kinetic energy. The amplitude of the motion is
MCQ+1 / -02023
17Simple Harmonic Motion
A clock is designed based on the oscillation of a spring-block system suspended vertically in the absence of air-resistance. Assume it shows the correct time when a spring of stiffness $k$ and block is mass $m$ are used. If the block is rep...
MCQ+1 / -02023
18Simple Harmonic Motion
A particle performs simple harmonic motion with a time period of 16 s . At a time $t=2 \mathrm{~s}$, the particle passes through the origin and at $t=4 \mathrm{~s}$ its velocity is $4 \mathrm{~m} / \mathrm{s}$. The amplitude of the motion i...
MCQ+1 / -02022
19Simple Harmonic Motion
A block is in simple harmonic motion (SHM) on the end of the spring with position given by $x=5 \cos \left(\omega t+\frac{\pi}{4}\right) \mathrm{cm}$. If the total mechanical energy used is 100 J to achieve maximum displacement, then the po...
MCQ+1 / -02022
20Simple Harmonic Motion
A simple pendulum of length 1 m and having a bob of mass 100 g is suspended in a car, moving on a circular track of radius 100 m with uniform speed $10 \mathrm{~m} / \mathrm{s}$. If the pendulum makes small oscillation in a radial direction...
MCQ+1 / -02022
21Simple Harmonic Motion
The amplitude of a damped oscillator varies with time as $A(t)=A_0 \exp (-b t / 2 \mathrm{~m})$, where $b=70 \mathrm{~g} / \mathrm{s}$ and $m=200$ g. How long does it take for the mechanical energy to drop to one-fourth of its initial value...
MCQ+1 / -02022
22Simple Harmonic Motion
A body starting at $t=0$ from origin oscillates simple harmonically with a period of 4 s . After what time will its kinetic energy by $75 \%$ of its total energy?
MCQ+1 / -02022
23Simple Harmonic Motion
A simple pendulum consists of a small sphere of mass $m$ suspended by a thread of length $l$. The sphere carries a positive charge $q$. The pendulum is allowed to do small oscillations in uniform electric field $E$ with direction vertically...
MCQ+1 / -02022
