Damped Oscillations — MH-CET Physics MCQs with Solutions
Free MH-CET Physics Damped Oscillations MCQs with step-by-step solutions (32 questions). Part of Oscillations. Practise online on Prepizo — no login needed.
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Questions with solutions
Q1 — 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.
Q2 — Damped Oscillations · medium · theory
The amplitude of a lightly damped oscillator varies with time as:
A. $A_0 e^{-bt/m}$
B. $A_0 e^{+bt/2m}$
C. $A_0 (1 - bt)$
D. $A_0 e^{-bt/2m}$ ✓ Correct
Solution: Solving $m\ddot{x} + b\dot{x} + kx = 0$ for light damping gives an oscillation whose amplitude decays exponentially as $A_0 e^{-bt/2m}$.
Q3 — Damped Oscillations · hard · theory
The total mechanical energy of a lightly damped oscillator decays with time as:
A. $E_0 e^{-bt/m}$ ✓ Correct
B. $E_0 (1 - bt/m)$
C. $E_0 e^{-bt/2m}$
D. $E_0 e^{-2bt/m}$
Solution: Energy is proportional to the square of the amplitude, so $E \propto \left(e^{-bt/2m}\right)^2 = e^{-bt/m}$ — energy decays twice as fast as amplitude.
Q4 — Damped Oscillations · medium · theory
The angular frequency of a damped oscillator compared with its natural (undamped) angular frequency is:
A. Slightly smaller ✓ Correct
B. Slightly larger
C. Always zero
D. Exactly equal
Solution: $\omega' = \sqrt{\dfrac{k}{m} - \dfrac{b^2}{4m^2}}$, which is less than $\omega_0 = \sqrt{\dfrac{k}{m}}$ because the damping term is subtracted.
Q5 — 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.
Q6 — Damped Oscillations · medium · theory
The differential equation governing a damped harmonic oscillator is:
A. $m\dfrac{d^2x}{dt^2} + kx = 0$
B. $m\dfrac{dx}{dt} + kx = 0$
C. $m\dfrac{d^2x}{dt^2} + b\dfrac{dx}{dt} + kx = 0$ ✓ Correct
D. $m\dfrac{d^2x}{dt^2} - b\dfrac{dx}{dt} + kx = 0$
Solution: Newton's second law with both a restoring force $-kx$ and a damping force $-b\dfrac{dx}{dt}$ yields this equation.
Q7 — Damped Oscillations · medium · theory
A critically damped system is one that:
A. Takes the longest possible time to reach equilibrium
B. Oscillates forever with constant amplitude
C. Returns to equilibrium in the shortest time without oscillating ✓ Correct
D. Oscillates with steadily growing amplitude
Solution: At critical damping the system just fails to oscillate and settles fastest. Heavier (overdamped) systems creep back more slowly.
Q8 — 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.
Q9 — Damped Oscillations · medium · theory
A galvanometer whose coil comes to rest quickly without oscillating about the final reading is described as:
A. Overdamped
B. Resonant
C. Undamped
D. Dead-beat (critically damped) ✓ Correct
Solution: Electromagnetic damping from eddy currents in the metal former is adjusted to be nearly critical, so the pointer settles directly on the reading.
Q10 — Damped Oscillations · medium · theory
An overdamped system, when displaced and released, will:
A. Oscillate with constant amplitude
B. Return to equilibrium slowly without oscillating ✓ Correct
C. Never return to equilibrium
D. Oscillate with slowly decreasing amplitude
Solution: With damping greater than critical the resistive force dominates, so the system creeps back to equilibrium taking longer than the critically damped case.
Q11 — 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.
Q12 — Damped Oscillations · medium · theory
In a lightly damped oscillator, the period of oscillation compared with the undamped case is:
A. Slightly shorter
B. Slightly longer ✓ Correct
C. Infinitely long
D. Exactly the same
Solution: Damping lowers the angular frequency slightly, and since $T = \dfrac{2\pi}{\omega'}$, a smaller $\omega'$ means a marginally longer period.
Q13 — Damped Oscillations · medium · theory
The SI unit of the damping constant $b$ in the expression $F = -bv$ is:
A. $\text{kg}/\text{s}$ ✓ Correct
B. $\text{kg}\cdot\text{s}$
C. $\text{N}/\text{m}$
D. $\text{N}\cdot\text{s}^2/\text{m}$
Solution: $b = \dfrac{F}{v}$ has units $\dfrac{\text{N}}{\text{m/s}} = \dfrac{\text{kg}\cdot\text{m/s}^2}{\text{m/s}} = \text{kg}/\text{s}$.
Q14 — Damped Oscillations · hard · theory
For a damped oscillation to remain oscillatory (underdamped), the damping constant must satisfy:
A. $\dfrac{b^2}{4m^2} > \dfrac{k}{m}$
B. $b = 0$ only
C. $\dfrac{b^2}{4m^2} = \dfrac{k}{m}$
D. $\dfrac{b^2}{4m^2} < \dfrac{k}{m}$ ✓ Correct
Solution: The damped angular frequency $\omega' = \sqrt{\dfrac{k}{m} - \dfrac{b^2}{4m^2}}$ is real only when the term under the root is positive.
Q15 — 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.
Q16 — Damped Oscillations · hard · numerical
In a damped oscillation, the amplitude falls to $\dfrac{1}{e}$ of its initial value after a time:
A. $\dfrac{m}{2b}$
B. $\dfrac{m}{b}$
C. $\dfrac{2m}{b}$ ✓ Correct
D. $\dfrac{b}{2m}$
Solution: Setting $e^{-bt/2m} = e^{-1}$ gives $\dfrac{bt}{2m} = 1$, so $t = \dfrac{2m}{b}$.
Q17 — Damped Oscillations · hard · numerical
A damped oscillator has mass $0.2\text{ kg}$ and damping constant $0.04\text{ kg/s}$. The time in which its amplitude falls to $\dfrac{1}{e}$ of the initial value is:
A. $0.2\text{ s}$
B. $5\text{ s}$
C. $20\text{ s}$
D. $10\text{ s}$ ✓ Correct
Solution: The amplitude varies as $e^{-bt/2m}$, so the required time is $t = \dfrac{2m}{b} = \dfrac{2 \times 0.2}{0.04} = 10\text{ s}$.
Q18 — Damped Oscillations · hard · numerical
For a damped oscillator of mass $0.2\text{ kg}$ and damping constant $0.04\text{ kg/s}$, the time in which the energy falls to $\dfrac{1}{e}$ of its initial value is:
A. $2.5\text{ s}$
B. $5\text{ s}$ ✓ Correct
C. $20\text{ s}$
D. $10\text{ s}$
Solution: Energy decays as $e^{-bt/m}$, so the time constant is $\dfrac{m}{b} = \dfrac{0.2}{0.04} = 5\text{ s}$ — half the amplitude time constant.
Q19 — Damped Oscillations · hard · numerical
A damped oscillator has $m = 1\text{ kg}$, $k = 100\text{ N/m}$ and $b = 6\text{ kg/s}$. Its damped angular frequency is approximately:
A. $8.00\text{ rad/s}$
B. $3.00\text{ rad/s}$
C. $10.0\text{ rad/s}$
D. $9.54\text{ rad/s}$ ✓ Correct
Solution: $\omega' = \sqrt{\dfrac{k}{m} - \dfrac{b^2}{4m^2}} = \sqrt{100 - \dfrac{36}{4}} = \sqrt{91} \approx 9.54\text{ rad/s}$.
Q20 — Damped Oscillations · hard · numerical
A damped oscillator has $m = 0.5\text{ kg}$ and $b = 0.1\text{ kg/s}$. After $10\text{ s}$, its amplitude is approximately:
A. $0.5A_0$
B. $0.607A_0$
C. $0.368A_0$ ✓ Correct
D. $0.135A_0$
Solution: $A = A_0e^{-bt/2m} = A_0e^{-(0.1 \times 10)/(2 \times 0.5)} = A_0e^{-1} \approx 0.368A_0$.
Q21 — Damped Oscillations · hard · numerical
A damped oscillator has $m = 0.5\text{ kg}$ and $b = 0.05\text{ kg/s}$. The time for its amplitude to fall to $\dfrac{1}{e}$ of the initial value is:
A. $20\text{ s}$ ✓ Correct
B. $40\text{ s}$
C. $5\text{ s}$
D. $10\text{ s}$
Solution: Amplitude decays as $e^{-bt/2m}$, so the time constant is $\dfrac{2m}{b} = \dfrac{1}{0.05} = 20\text{ s}$.
Q22 — Damped Oscillations · hard · numerical
For a damped oscillator with $m = 0.5\text{ kg}$ and $b = 0.05\text{ kg/s}$, the time for the energy to fall to $\dfrac{1}{e}$ of its initial value is:
A. $40\text{ s}$
B. $10\text{ s}$ ✓ Correct
C. $5\text{ s}$
D. $20\text{ s}$
Solution: Energy decays as $e^{-bt/m}$, so its time constant is $\dfrac{m}{b} = \dfrac{0.5}{0.05} = 10\text{ s}$ — half the amplitude time constant.
Q23 — Damped Oscillations · hard · numerical
A damped oscillator has $m = 2\text{ kg}$, $k = 200\text{ N/m}$ and $b = 8\text{ kg/s}$. Its damped angular frequency is approximately:
A. $9.80\text{ rad/s}$ ✓ Correct
B. $9.95\text{ rad/s}$
C. $8.00\text{ rad/s}$
D. $10.0\text{ rad/s}$
Solution: $\omega' = \sqrt{\dfrac{k}{m} - \dfrac{b^2}{4m^2}} = \sqrt{100 - \dfrac{64}{16}} = \sqrt{96} \approx 9.80\text{ rad/s}$.
Q24 — Damped Oscillations · hard · numerical
A damped oscillator has $m = 1\text{ kg}$ and $b = 0.2\text{ kg/s}$. After $5\text{ s}$, its amplitude is approximately:
A. $0.368A_0$
B. $0.5A_0$
C. $0.135A_0$
D. $0.607A_0$ ✓ Correct
Solution: $A = A_0e^{-bt/2m} = A_0e^{-(0.2 \times 5)/2} = A_0e^{-0.5} \approx 0.607A_0$.
Q25 — Damped Oscillations · hard · numerical
A damped oscillator has $m = 0.4\text{ kg}$ and $b = 0.08\text{ kg/s}$. After $20\text{ s}$, its amplitude is approximately:
A. $0.607A_0$
B. $0.135A_0$ ✓ Correct
C. $0.05A_0$
D. $0.368A_0$
Solution: $\dfrac{bt}{2m} = \dfrac{0.08 \times 20}{0.8} = 2$, so $A = A_0e^{-2} \approx 0.135A_0$.
Q26 — Damped Oscillations · hard · numerical
A damped oscillator has $m = 0.5\text{ kg}$ and $b = 0.1\text{ kg/s}$. After $10\text{ s}$, its energy is approximately:
A. $0.5E_0$
B. $0.607E_0$
C. $0.135E_0$ ✓ Correct
D. $0.368E_0$
Solution: Energy decays as $e^{-bt/m}$, and $\dfrac{bt}{m} = \dfrac{0.1 \times 10}{0.5} = 2$, so $E = E_0e^{-2} \approx 0.135E_0$.
Q27 — Damped Oscillations · easy · numerical
The natural angular frequency of a system with $k = 64\text{ N/m}$ and $m = 4\text{ kg}$ is:
A. $16\text{ rad/s}$
B. $2\text{ rad/s}$
C. $8\text{ rad/s}$
D. $4\text{ rad/s}$ ✓ Correct
Solution: $\omega_0 = \sqrt{\dfrac{k}{m}} = \sqrt{\dfrac{64}{4}} = \sqrt{16} = 4\text{ rad/s}$.
Q28 — Damped Oscillations · hard · numerical
A damped oscillator has $m = 1\text{ kg}$, $k = 100\text{ N/m}$ and $b = 2\text{ kg/s}$. Its damped angular frequency is approximately:
A. $9.80\text{ rad/s}$
B. $8.00\text{ rad/s}$
C. $9.95\text{ rad/s}$ ✓ Correct
D. $10.0\text{ rad/s}$
Solution: $\omega' = \sqrt{100 - \dfrac{4}{4}} = \sqrt{99} \approx 9.95\text{ rad/s}$.
Q29 — Damped Oscillations · medium · numerical
A damped oscillator has $m = 1.5\text{ kg}$ and $b = 0.3\text{ kg/s}$. Its amplitude time constant $\dfrac{2m}{b}$ is:
A. $0.2\text{ s}$
B. $10\text{ s}$ ✓ Correct
C. $20\text{ s}$
D. $5\text{ s}$
Solution: $\dfrac{2m}{b} = \dfrac{2 \times 1.5}{0.3} = 10\text{ s}$.
Q30 — Damped Oscillations · hard · numerical
A damped oscillator has $m = 0.5\text{ kg}$ and $b = 0.1\text{ kg/s}$. The time for its amplitude to halve is approximately ($\ln 2 = 0.693$):
A. $10.0\text{ s}$
B. $3.47\text{ s}$
C. $13.9\text{ s}$
D. $6.93\text{ s}$ ✓ Correct
Solution: Setting $e^{-bt/2m} = \dfrac{1}{2}$ gives $t = \dfrac{2m\ln 2}{b} = \dfrac{1 \times 0.693}{0.1} \approx 6.93\text{ s}$.