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Oscillations — MH-CET Physics MCQs with Solutions
Free MH-CET Physics Oscillations MCQs with step-by-step solutions covering Simple Harmonic Motion Basics, Energy in SHM, Pendulum Motion, Damped Oscillations, Forced Oscillations & Resonance, Superposition of Oscillations. Practise online on Prepizo — no login needed.
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Sample questions with solutions
Q1 — Simple Harmonic Motion Basics · easy · theory
A motion is simple harmonic if the acceleration of the particle is:
A. Proportional to the velocity of the particle
B. Proportional to the displacement and directed towards the mean position ✓ Correct
C. Proportional to the displacement and directed away from the mean position
D. Constant in magnitude and direction
Solution: The defining relation is $a = -\omega^2 x$. The minus sign shows the acceleration is always a restoring one, pointing back towards the equilibrium position.
Q2 — Simple Harmonic Motion Basics · easy · numerical
A particle executes S.H.M. described by $x = 0.05\sin(100t + \pi/6)\text{ m}$. Its maximum velocity is:
A. $0.5\text{ m/s}$
B. $2.5\text{ m/s}$
C. $50\text{ m/s}$
D. $5\text{ m/s}$ ✓ Correct
Solution: Here $A = 0.05\text{ m}$ and $\omega = 100\text{ rad/s}$, so $v_{\max} = A\omega = 0.05 \times 100 = 5\text{ m/s}$.
Q3 — Simple Harmonic Motion Basics · easy · theory
The maximum acceleration of a particle performing S.H.M. of amplitude $A$ and angular frequency $\omega$ is:
A. $A\omega^2$ ✓ Correct
B. $A^2\omega$
C. $A\omega$
D. $\dfrac{A}{\omega^2}$
Solution: Since $a = -\omega^2 x$, the magnitude is greatest at the extreme positions where $|x| = A$, giving $a_{\max} = A\omega^2$.
Q4 — Simple Harmonic Motion Basics · easy · theory
At the mean position of a particle performing S.H.M.:
A. Both velocity and acceleration are maximum
B. The velocity is zero and the acceleration is maximum
C. The velocity is maximum and the acceleration is zero ✓ Correct
D. Both velocity and acceleration are zero
Solution: At $x = 0$ the restoring force vanishes so $a = 0$, while all the energy is kinetic so the speed is at its greatest.
Q5 — Simple Harmonic Motion Basics · easy · theory
At the extreme positions of a particle performing S.H.M.:
A. The velocity is maximum and the acceleration is zero
B. The velocity is zero and the acceleration is maximum ✓ Correct
C. Both are maximum
D. Both are zero
Solution: At $|x| = A$ the particle momentarily stops before reversing, and the restoring force — and hence the acceleration — is at its largest.
Q6 — Simple Harmonic Motion Basics · easy · theory
The period of oscillation of a mass $m$ suspended from a spring of force constant $k$ is:
A. $2\pi\dfrac{m}{k}$
B. $2\pi\sqrt{\dfrac{k}{m}}$
C. $2\pi\sqrt{\dfrac{m}{k}}$ ✓ Correct
D. $\dfrac{1}{2\pi}\sqrt{\dfrac{m}{k}}$
Solution: With $\omega = \sqrt{\dfrac{k}{m}}$, the period is $T = \dfrac{2\pi}{\omega} = 2\pi\sqrt{\dfrac{m}{k}}$.
Q7 — Simple Harmonic Motion Basics · easy · theory
The quantity $(\omega t + \phi)$ in the S.H.M. equation $x = A\sin(\omega t + \phi)$ is called the:
A. Frequency of the motion
B. Amplitude of the motion
C. Time period
D. Phase of the motion ✓ Correct
Solution: The phase specifies the state of the oscillator at a given instant, and $\phi$, its value at $t = 0$, is the initial phase or epoch.
Q8 — Simple Harmonic Motion Basics · easy · theory
A particle performs S.H.M. with angular frequency $\omega$. The period of its oscillation is:
A. $\dfrac{\omega}{2\pi}$
B. $\dfrac{\pi}{\omega}$
C. $\dfrac{2\pi}{\omega}$ ✓ Correct
D. $2\pi\omega$
Solution: One complete oscillation corresponds to a phase change of $2\pi$, so $T = \dfrac{2\pi}{\omega}$.
Q9 — Simple Harmonic Motion Basics · easy · numerical
If the mass attached to a spring is increased four times, the period of oscillation becomes:
A. Twice as large ✓ Correct
B. Unchanged
C. Four times as large
D. Half as large
Solution: $T = 2\pi\sqrt{\dfrac{m}{k}} \propto \sqrt{m}$, so quadrupling the mass doubles the period.
Q10 — Energy in SHM · easy · theory
The total energy of a particle of mass $m$ performing S.H.M. of amplitude $A$ and angular frequency $\omega$ is:
A. $\dfrac{1}{2}m\omega A^2$
B. $\dfrac{1}{2}m\omega^2A$
C. $m\omega^2A^2$
D. $\dfrac{1}{2}m\omega^2A^2$ ✓ Correct
Solution: At an extreme position all the energy is potential: $E = \dfrac{1}{2}kA^2 = \dfrac{1}{2}m\omega^2A^2$, since $k = m\omega^2$.
Q11 — Energy in SHM · easy · theory
The potential energy of a particle performing S.H.M. at displacement $x$ from the mean position is:
A. $\dfrac{1}{2}k(A^2 - x^2)$
B. $kx^2$
C. $\dfrac{1}{2}kA^2$
D. $\dfrac{1}{2}kx^2$ ✓ Correct
Solution: The restoring force is $F = -kx$, so the work stored in displacing the particle to $x$ is $U = \dfrac{1}{2}kx^2$.
Q12 — Energy in SHM · easy · theory
The total mechanical energy of a particle performing S.H.M. (in the absence of damping) is:
A. Constant throughout the motion ✓ Correct
B. Zero at the mean position
C. Maximum at the extreme positions only
D. Maximum at the mean position only
Solution: Kinetic and potential energies interchange continuously, but their sum $\dfrac{1}{2}kA^2$ stays fixed because the restoring force is conservative.
Q13 — Energy in SHM · easy · theory
The total energy of a particle performing S.H.M. is proportional to:
A. The inverse of the amplitude
B. The amplitude
C. The cube of the amplitude
D. The square of the amplitude ✓ Correct
Solution: $E = \dfrac{1}{2}kA^2$, so $E \propto A^2$.
Q14 — Energy in SHM · easy · numerical
If the amplitude of a simple harmonic oscillator is doubled without changing its frequency, its total energy becomes:
A. Half
B. Unchanged
C. Four times ✓ Correct
D. Twice
Solution: Since $E \propto A^2$, doubling the amplitude multiplies the energy by $4$.
Q15 — Energy in SHM · easy · theory
The kinetic energy of a particle performing S.H.M. is maximum at the:
A. Mean position ✓ Correct
B. Position $x = A/2$
C. Extreme positions
D. Position $x = A/\sqrt{2}$
Solution: The speed is greatest at $x = 0$, where the entire energy of the oscillator is kinetic.
Q16 — Energy in SHM · easy · theory
The graph of the potential energy of a simple harmonic oscillator against displacement is:
A. A sine curve
B. A circle
C. A parabola ✓ Correct
D. A straight line
Solution: Since $U = \dfrac{1}{2}kx^2$ is quadratic in $x$, the graph is a parabola with its minimum at the mean position.
Q17 — Energy in SHM · easy · theory
When a simple harmonic oscillator passes through its mean position, its potential energy is:
A. Equal to the total energy
B. Zero, and the kinetic energy equals the total energy ✓ Correct
C. Half the total energy
D. Maximum
Solution: At $x = 0$, $U = \dfrac{1}{2}kx^2 = 0$, so the whole of the constant total energy appears as kinetic energy.
Q18 — Pendulum Motion · easy · theory
The period of a simple pendulum of length $L$ at a place where the acceleration due to gravity is $g$ is:
A. $2\pi\sqrt{\dfrac{L}{g}}$ ✓ Correct
B. $\dfrac{1}{2\pi}\sqrt{\dfrac{L}{g}}$
C. $2\pi\dfrac{L}{g}$
D. $2\pi\sqrt{\dfrac{g}{L}}$
Solution: For small oscillations the restoring force gives $\omega = \sqrt{\dfrac{g}{L}}$, so $T = 2\pi\sqrt{\dfrac{L}{g}}$.
Q19 — Pendulum Motion · easy · theory
The period of a simple pendulum is independent of:
A. All of these
B. The length of the pendulum
C. The mass of the bob ✓ Correct
D. The acceleration due to gravity
Solution: The mass cancels from the equation of motion, so $T = 2\pi\sqrt{\dfrac{L}{g}}$ contains only the length and $g$.
Q20 — Pendulum Motion · easy · numerical
If the length of a simple pendulum is made four times its original value, its period becomes:
A. Unchanged
B. Twice as long ✓ Correct
C. Four times as long
D. Half as long
Solution: $T \propto \sqrt{L}$, so quadrupling the length doubles the period.
Q21 — Pendulum Motion · easy · theory
A seconds pendulum is one whose period is:
A. $1\text{ s}$
B. $0.5\text{ s}$
C. $4\text{ s}$
D. $2\text{ s}$ ✓ Correct
Solution: A seconds pendulum takes one second for each swing from one extreme to the other, so a complete oscillation takes $2\text{ s}$. Its length on Earth is about $0.993\text{ m}$.
Q22 — Pendulum Motion · easy · numerical
If the acceleration due to gravity at a place is doubled, the period of a given simple pendulum becomes:
A. $\dfrac{1}{\sqrt{2}}$ times its original value ✓ Correct
B. Twice its original value
C. $\sqrt{2}$ times its original value
D. Half its original value
Solution: $T \propto \dfrac{1}{\sqrt{g}}$, so doubling $g$ divides the period by $\sqrt{2}$.
Q23 — Pendulum Motion · easy · theory
The effective length of a simple pendulum is measured from the point of suspension to the:
A. Centre of gravity of the bob ✓ Correct
B. Top of the bob
C. Midpoint of the string only
D. Bottom of the bob
Solution: The bob behaves as a point mass located at its centre of gravity, so the effective length is the string length plus the radius of the bob.
Q24 — Damped Oscillations · easy · theory
In a damped oscillation, the damping force is usually taken to be proportional to the:
A. Square of the displacement
B. Displacement of the oscillator
C. Velocity of the oscillator ✓ Correct
D. Acceleration of the oscillator
Solution: For slow motion through a fluid the resistive force is $F = -bv$, opposing the velocity. This linear form makes the damped equation solvable.
Q25 — Damped Oscillations · easy · theory
The energy lost by a damped oscillator is:
A. Completely destroyed
B. Converted into potential energy of the spring
C. Dissipated as heat in the surrounding medium ✓ Correct
D. Stored as kinetic energy of the oscillator
Solution: Work done against the resistive force is converted into internal energy of the oscillator and the medium, appearing ultimately as heat.
Q26 — Damped Oscillations · easy · theory
The shock absorbers fitted in a motor car are designed to provide:
A. Resonance with the road bumps
B. Zero damping, so that the car oscillates freely
C. Damping close to critical, so that oscillations die out quickly ✓ Correct
D. An increase in the amplitude of oscillation
Solution: Near-critical damping lets the suspension absorb a bump and return to normal ride height at once, without the car bouncing repeatedly.
Q27 — Damped Oscillations · easy · theory
The amplitude of a real pendulum swinging in air gradually decreases mainly because of:
A. An increase in the length of the string
B. A decrease in the acceleration due to gravity
C. A change in the mass of the bob
D. Air resistance and friction at the support ✓ Correct
Solution: Work done against air drag and friction removes mechanical energy from the pendulum, so each successive swing is smaller.
Q28 — Damped Oscillations · easy · theory
A free oscillation differs from a damped oscillation in that a free oscillation has:
A. No definite period
B. A frequency that changes with time
C. Steadily increasing amplitude
D. Constant amplitude, since no energy is dissipated ✓ Correct
Solution: An ideal free oscillation has no resistive force, so its mechanical energy and hence its amplitude remain constant indefinitely.
Q29 — Forced Oscillations & Resonance · easy · theory
Resonance in a forced oscillation occurs when the frequency of the driving force is:
A. Much greater than the natural frequency
B. Much smaller than the natural frequency
C. Equal to the natural frequency of the system ✓ Correct
D. Exactly zero
Solution: At this match, energy is fed into the system in step with its own motion each cycle, producing the largest possible amplitude.
Q30 — Forced Oscillations & Resonance · easy · theory
At resonance, the amplitude of a forced oscillation is:
A. Zero
B. Independent of damping
C. Maximum ✓ Correct
D. Minimum
Solution: The driving force works with the motion throughout each cycle, so the energy transfer per cycle is maximum and the amplitude peaks.