Forced Oscillations & Resonance — MH-CET Physics MCQs with Solutions
Free MH-CET Physics Forced Oscillations & Resonance 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 — Forced Oscillations & Resonance · medium · theory
In a forced oscillation, after the steady state is reached, the system oscillates with the:
A. Difference of the two frequencies
B. Average of the two frequencies
C. Frequency of the applied periodic force ✓ Correct
D. Natural frequency of the system
Solution: Transients at the natural frequency die out because of damping, leaving the system oscillating at the driving frequency.
Q2 — 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.
Q3 — 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.
Q4 — Forced Oscillations & Resonance · medium · theory
Increasing the damping of a resonant system has the effect of:
A. Reducing the amplitude at resonance and broadening the resonance curve ✓ Correct
B. Leaving the resonance curve unaltered
C. Shifting the resonant frequency to a much higher value
D. Increasing the amplitude at resonance
Solution: Damping limits the maximum amplitude and makes the response less sharply peaked, so the system responds over a wider band of frequencies.
Q5 — Forced Oscillations & Resonance · medium · theory
In the ideal case of zero damping, the amplitude of a system driven exactly at resonance would:
A. Oscillate between two fixed values
B. Fall to zero
C. Remain equal to the driving amplitude
D. Grow without limit ✓ Correct
Solution: With nothing to dissipate the energy fed in each cycle, the amplitude would in principle increase indefinitely. Real systems always have some damping.
Q6 — Forced Oscillations & Resonance · easy · theory
Soldiers are ordered to break step while marching across a suspension bridge because:
A. Marching in step produces a magnetic effect
B. Breaking step reduces friction on the roadway
C. Their rhythmic steps might resonate with a natural frequency of the bridge ✓ Correct
D. Marching in step increases the total weight on the bridge
Solution: A periodic driving force matching a natural frequency of the structure could build up dangerously large oscillations through resonance.
Q7 — Forced Oscillations & Resonance · medium · theory
The sharpness of resonance of an oscillating system is greatest when the damping is:
A. Very large
B. Very small ✓ Correct
C. Critical
D. Equal to the driving force
Solution: Low damping gives a tall, narrow resonance peak, meaning the system responds strongly only very close to its natural frequency.
Q8 — Forced Oscillations & Resonance · easy · theory
Tuning a radio receiver to a particular station is an everyday example of:
A. Total internal reflection
B. Electrical resonance ✓ Correct
C. The Doppler effect
D. Mechanical damping
Solution: The tuning knob adjusts a capacitance so that the circuit's natural frequency matches the carrier frequency of the desired station, which is then amplified far more than the rest.
Q9 — Forced Oscillations & Resonance · easy · theory
The natural frequency of an oscillating system depends on:
A. The physical properties of the system, such as its mass and stiffness ✓ Correct
B. The frequency of the applied driving force
C. The amplitude of the driving force
D. The duration for which the force is applied
Solution: For a spring-mass system $f_0 = \dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}$ — a property of the system alone, independent of how it is driven.
Q10 — Forced Oscillations & Resonance · easy · theory
A child on a swing is pushed gently once every cycle at just the right moment, making the swing rise higher and higher. This illustrates:
A. Resonance ✓ Correct
B. Interference
C. Total internal reflection
D. Damping
Solution: The pushes are applied at the swing's natural frequency and in phase with its motion, so each push adds energy and the amplitude grows.
Q11 — Forced Oscillations & Resonance · hard · theory
A freely vibrating system loses energy to its surroundings, but a system in a steady forced oscillation maintains a constant amplitude because:
A. The driving force supplies exactly the energy lost to damping in each cycle ✓ Correct
B. The natural frequency of the system becomes zero
C. No energy is dissipated in forced oscillations
D. The damping force becomes zero in the steady state
Solution: Steady state is a balance: energy input per cycle from the driver equals energy dissipated per cycle by damping.
Q12 — Forced Oscillations & Resonance · easy · theory
The collapse of the Tacoma Narrows bridge in 1940 is commonly cited as an example of the destructive effect of:
A. Resonant build-up of oscillations driven by the wind ✓ Correct
B. Simple thermal expansion
C. The photoelectric effect
D. Total internal reflection
Solution: Periodic forcing from the wind matched a natural torsional mode of the deck, and the oscillations grew until the structure failed.
Q13 — Forced Oscillations & Resonance · medium · theory
In Melde's experiment, standing waves of large amplitude appear on a string when:
A. The string is made as heavy as possible
B. The tuning fork is held stationary
C. The frequency of the tuning fork matches a natural frequency of the string ✓ Correct
D. The tension in the string is reduced to zero
Solution: The experiment demonstrates resonance: adjusting the length or tension until a natural frequency of the string equals the driving frequency produces large, stable loops.
Q14 — Forced Oscillations & Resonance · medium · theory
A tuning fork held over a tube of adjustable air column produces a loud sound at certain lengths. This is because:
A. The tuning fork changes its own frequency
B. The air column resonates when its natural frequency equals the fork frequency ✓ Correct
C. The air in the tube is compressed permanently
D. Sound travels faster in the shorter column
Solution: At the resonant lengths a standing wave is set up in the air column, and the amplitude of vibration — and hence the loudness — becomes large.
Q15 — Forced Oscillations & Resonance · easy · theory
A forced oscillation differs from a free oscillation chiefly because a forced oscillation:
A. Occurs only in electrical systems
B. Cannot be damped
C. Is maintained by an external periodic force ✓ Correct
D. Has no definite amplitude
Solution: A free oscillation happens at the system's own frequency after an initial disturbance; a forced oscillation is continuously driven from outside.
Q16 — Forced Oscillations & Resonance · hard · theory
When the driving frequency is very much greater than the natural frequency of a system, the amplitude of the forced oscillation is:
A. Independent of the driving frequency
B. Very small ✓ Correct
C. Maximum
D. Equal to the resonant amplitude
Solution: The system's inertia prevents it from following such rapid forcing, so it barely responds and the amplitude falls well below the resonant value.
Q17 — Forced Oscillations & Resonance · medium · numerical
A mass of $1\text{ kg}$ on a spring of force constant $400\text{ N/m}$ has a natural frequency of approximately:
A. $6.37\text{ Hz}$
B. $3.18\text{ Hz}$ ✓ Correct
C. $0.314\text{ Hz}$
D. $20.0\text{ Hz}$
Solution: $\omega_0 = \sqrt{\dfrac{k}{m}} = \sqrt{400} = 20\text{ rad/s}$, so $f_0 = \dfrac{20}{2\pi} \approx 3.18\text{ Hz}$.
Q18 — Forced Oscillations & Resonance · medium · numerical
A mass of $0.25\text{ kg}$ hangs from a spring of force constant $100\text{ N/m}$. Resonance occurs when the driving angular frequency is:
A. $20\text{ rad/s}$ ✓ Correct
B. $400\text{ rad/s}$
C. $25\text{ rad/s}$
D. $10\text{ rad/s}$
Solution: Resonance requires the driving frequency to match the natural one: $\omega_0 = \sqrt{\dfrac{k}{m}} = \sqrt{\dfrac{100}{0.25}} = \sqrt{400} = 20\text{ rad/s}$.
Q19 — Forced Oscillations & Resonance · hard · numerical
A simple pendulum of length $1\text{ m}$ has a natural frequency of approximately ($g = 10\text{ m/s}^2$):
A. $3.16\text{ Hz}$
B. $0.16\text{ Hz}$
C. $0.50\text{ Hz}$ ✓ Correct
D. $1.99\text{ Hz}$
Solution: $\omega_0 = \sqrt{\dfrac{g}{L}} = \sqrt{10} = 3.16\text{ rad/s}$, so $f_0 = \dfrac{3.16}{6.283} \approx 0.50\text{ Hz}$.
Q20 — Forced Oscillations & Resonance · easy · numerical
A mass-spring system has a natural period of $0.5\text{ s}$. The driving frequency that produces resonance is:
A. $2\text{ Hz}$ ✓ Correct
B. $4\text{ Hz}$
C. $1\text{ Hz}$
D. $0.5\text{ Hz}$
Solution: Resonance occurs when the driving frequency equals the natural frequency $f_0 = \dfrac{1}{T} = \dfrac{1}{0.5} = 2\text{ Hz}$.
Q21 — Forced Oscillations & Resonance · medium · numerical
A mass of $0.25\text{ kg}$ on a spring of force constant $100\text{ N/m}$ has a natural frequency of approximately:
A. $0.32\text{ Hz}$
B. $3.18\text{ Hz}$ ✓ Correct
C. $6.37\text{ Hz}$
D. $20.0\text{ Hz}$
Solution: $\omega_0 = \sqrt{\dfrac{100}{0.25}} = 20\text{ rad/s}$, so $f_0 = \dfrac{20}{2\pi} \approx 3.18\text{ Hz}$.
Q22 — Forced Oscillations & Resonance · easy · numerical
A mass of $1\text{ kg}$ on a spring of force constant $900\text{ N/m}$ resonates when driven at an angular frequency of:
A. $900\text{ rad/s}$
B. $4.77\text{ rad/s}$
C. $30\text{ rad/s}$ ✓ Correct
D. $15\text{ rad/s}$
Solution: Resonance occurs at the natural frequency $\omega_0 = \sqrt{\dfrac{900}{1}} = 30\text{ rad/s}$.
Q23 — Forced Oscillations & Resonance · medium · numerical
A simple pendulum of length $0.4\text{ m}$ has a natural angular frequency of ($g = 10\text{ m/s}^2$):
A. $5\text{ rad/s}$ ✓ Correct
B. $2.5\text{ rad/s}$
C. $0.2\text{ rad/s}$
D. $25\text{ rad/s}$
Solution: $\omega_0 = \sqrt{\dfrac{g}{L}} = \sqrt{\dfrac{10}{0.4}} = \sqrt{25} = 5\text{ rad/s}$.
Q24 — Forced Oscillations & Resonance · easy · numerical
A mass-spring system has a natural period of $0.2\text{ s}$. Resonance occurs at a driving frequency of:
A. $0.2\text{ Hz}$
B. $10\text{ Hz}$
C. $2\text{ Hz}$
D. $5\text{ Hz}$ ✓ Correct
Solution: Resonance requires the driving frequency to equal $f_0 = \dfrac{1}{T} = \dfrac{1}{0.2} = 5\text{ Hz}$.
Q25 — Forced Oscillations & Resonance · easy · numerical
A mass of $4\text{ kg}$ on a spring of force constant $1600\text{ N/m}$ has a natural angular frequency of:
A. $400\text{ rad/s}$
B. $10\text{ rad/s}$
C. $20\text{ rad/s}$ ✓ Correct
D. $40\text{ rad/s}$
Solution: $\omega_0 = \sqrt{\dfrac{1600}{4}} = \sqrt{400} = 20\text{ rad/s}$.
Q26 — Forced Oscillations & Resonance · easy · numerical
A mass of $0.1\text{ kg}$ on a spring of force constant $40\text{ N/m}$ resonates at an angular frequency of:
A. $4\text{ rad/s}$
B. $2\text{ rad/s}$
C. $400\text{ rad/s}$
D. $20\text{ rad/s}$ ✓ Correct
Solution: $\omega_0 = \sqrt{\dfrac{40}{0.1}} = \sqrt{400} = 20\text{ rad/s}$.
Q27 — Forced Oscillations & Resonance · medium · numerical
A simple pendulum of length $2.5\text{ m}$ has a natural angular frequency of ($g = 10\text{ m/s}^2$):
A. $4\text{ rad/s}$
B. $2\text{ rad/s}$ ✓ Correct
C. $25\text{ rad/s}$
D. $0.5\text{ rad/s}$
Solution: $\omega_0 = \sqrt{\dfrac{10}{2.5}} = \sqrt{4} = 2\text{ rad/s}$.
Q28 — Forced Oscillations & Resonance · easy · numerical
A system has a natural period of $4\text{ s}$. Its resonant frequency is:
A. $4\text{ Hz}$
B. $1\text{ Hz}$
C. $0.5\text{ Hz}$
D. $0.25\text{ Hz}$ ✓ Correct
Solution: $f_0 = \dfrac{1}{T} = \dfrac{1}{4} = 0.25\text{ Hz}$.
Q29 — Forced Oscillations & Resonance · easy · numerical
A mass of $1\text{ kg}$ on a spring of force constant $256\text{ N/m}$ has a natural angular frequency of:
A. $8\text{ rad/s}$
B. $256\text{ rad/s}$
C. $32\text{ rad/s}$
D. $16\text{ rad/s}$ ✓ Correct
Solution: $\omega_0 = \sqrt{256} = 16\text{ rad/s}$.
Q30 — Forced Oscillations & Resonance · hard · numerical
A load stretches a spring by $0.04\text{ m}$ at equilibrium. The natural angular frequency of the resulting oscillation is ($g = 10\text{ m/s}^2$):
A. $250\text{ rad/s}$
B. $2.5\text{ rad/s}$
C. $5.0\text{ rad/s}$
D. $15.8\text{ rad/s}$ ✓ Correct
Solution: At equilibrium $\dfrac{k}{m} = \dfrac{g}{x}$, so $\omega_0 = \sqrt{\dfrac{10}{0.04}} = \sqrt{250} \approx 15.8\text{ rad/s}$.