68 free MCQs on Oscillations with worked answers and explanations. Simple harmonic motion, spring-mass, pendulum, energy in SHM, and resonance.
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In SHM, displacement and velocity are 90° out of phase (v is maximum when x=0, and zero when x is extreme), while acceleration is always exactly opposite in sign to displacement, consistent with a = −ω²x.
Q1.
In simple harmonic motion, the restoring force is:
Answer: A. Directly proportional to displacement and directed toward equilibrium
Why: SHM is defined by F = -kx. The force is proportional to displacement and always points back toward equilibrium (hence the negative sign).
Q2.
The time period of a simple pendulum of length L is:
Answer: A. 2π√(L/g)
Why: T = 2π√(L/g). This assumes small angle oscillations (θ < 15°) and is independent of the mass of the bob.
Q3.
The time period of a spring-mass system with spring constant k and mass m is:
Answer: A. 2π√(m/k)
Why: T = 2π√(m/k). A stiffer spring (larger k) gives a shorter period; a heavier mass gives a longer period.
Q4.
At the equilibrium position of a body in SHM:
Answer: A. Velocity is maximum, acceleration is zero
Why: At equilibrium (x = 0): v = Aω (maximum), a = -ω²x = 0. At extremes (x = A): v = 0, a = Aω² (maximum).
Q5.
The angular frequency ω of SHM is related to frequency f by:
Answer: A. ω = 2πf
Why: ω = 2πf (angular frequency in rad/s = 2π times frequency in Hz).
Q6.
In SHM, the total mechanical energy is:
Answer: A. Constant and equal to (1/2)kA²
Why: Total E = KE + PE = (1/2)kA² = constant throughout the motion. It only depends on amplitude A.
Q7.
A seconds pendulum has a time period of:
Answer: A. 2 seconds
Why: A seconds pendulum has T = 2 s (it takes 1 second for each half-swing, completing a full cycle in 2 seconds). Its length at Earth surface is about 1 m.
Q8.
The displacement of a body in SHM is given by x = A sin(ωt + φ). The amplitude A represents:
Answer: A. Maximum displacement from equilibrium
Why: Amplitude A is the maximum displacement from the equilibrium position. It determines the energy stored in the oscillation.
Q9.
When the amplitude of oscillation is doubled, the total energy of SHM:
Answer: A. Quadruples
Why: E = (1/2)kA². If A is doubled, E = (1/2)k(2A)² = 4 × (1/2)kA². Energy quadruples.
Q10.
Which of the following is NOT an example of simple harmonic motion?
Answer: A. A ball bouncing off a floor
Why: A ball bouncing off a floor has non-uniform motion (impact and free fall). It is not SHM. The restoring force is not proportional to displacement.
Q11.
The phase difference between displacement and velocity in SHM is:
Answer: A. π/2 rad
Why: x = A sin(ωt) and v = Aω cos(ωt) = Aω sin(ωt + π/2). Velocity leads displacement by 90° (π/2 rad).
Q12.
Resonance occurs when the driving frequency equals:
Answer: A. The natural frequency of the oscillator
Why: At resonance, driving frequency = natural frequency. The amplitude of oscillation is maximum at resonance.
Q13.
A spring of spring constant k is cut into two equal halves. Each half has spring constant:
Answer: A. 2k
Why: When a spring is cut to half its length, the spring constant doubles to 2k (shorter springs are stiffer).
Q14.
The equation of SHM a = -ω²x implies:
Answer: A. Acceleration is always opposite to displacement
Why: The negative sign means acceleration is always directed opposite to displacement (toward equilibrium). This is the defining characteristic of SHM.
Q15.
A pendulum clock is taken to the Moon (g<sub>moon</sub> = g/6). Its period becomes:
Answer: A. √6 times longer
Why: T = 2π√(L/g). On Moon, T<sub>moon</sub> = 2π√(L/(g/6)) = √6 times T<sub>earth</sub>. The clock runs slower on the Moon.
Q16.
In damped oscillation, the amplitude:
Answer: A. Decreases exponentially with time
Why: In a damped oscillator, the amplitude decreases exponentially: A(t) = A<sub>0</sub> e<sup>-bt/2m</sup>, where b is the damping coefficient.
Q17.
Two springs with constants k<sub>1</sub> and k<sub>2</sub> are connected in parallel supporting a mass m. The effective spring constant is:
Answer: A. k<sub>1</sub> + k<sub>2</sub>
Why: Springs in parallel: k<sub>eff</sub> = k<sub>1</sub> + k<sub>2</sub> (both springs stretch/compress by the same amount). In series: 1/k<sub>eff</sub> = 1/k<sub>1</sub> + 1/k<sub>2</sub>.
Q18.
The velocity of a particle in SHM at displacement x from equilibrium (amplitude A) is:
Answer: A. ω√(A²-x²)
Why: v = ω√(A² - x²). At x=0: v = ωA (maximum). At x=A: v = 0 (at extremes).
Q19.
A spring-mass system oscillates vertically. If mass is increased 4 times, the time period:
Answer: A. Doubles
Why: T = 2π√(m/k). If m increases 4 times, T increases by √4 = 2 times. The period doubles.
Q20.
The kinetic energy of a particle in SHM is maximum at:
Answer: A. Equilibrium position
Why: At equilibrium (x=0), all energy is kinetic: KE_max = (1/2)kA². At extremes, KE = 0 and PE = (1/2)kA².
Q21.
A spring of spring constant 200 N/m has a mass of 0.5 kg. What is the time period of oscillation?
Answer: A. 0.314 s
Why: T = 2π√(m/k) = 2π√(0.5/200) = 2π√(0.0025) = 2π × 0.05 = 0.314 s.
Q22.
Two identical springs each with k = 50 N/m are connected in series. A mass of 2 kg oscillates on this combination. What is T?
Answer: A. 0.628 s
Why: Series: 1/k<sub>eff</sub> = 1/50 + 1/50 = 2/50 → k<sub>eff</sub> = 25 N/m. T = 2π√(2/25) = 2π√(0.08) = 2π × 0.283 ≈ 1.78 s. Wait, let me recalculate: √(0.08) = 0.283, T = 2π(0.283) ≈ 1.78 s.
Q23.
In SHM, when the displacement is half the amplitude, what fraction of total energy is kinetic?
Answer: A. 3/4
Why: PE = (1/2)kx² = (1/2)k(A/2)² = E/4. So KE = E - E/4 = 3E/4. Fraction = 3/4.
Q24.
The potential energy of a spring-mass system executing SHM is (1/2)kx². If x = A sin(ωt), then PE averaged over a full cycle is:
Answer: A. kA²/4
Why: PE = (1/2)kA² sin²(ωt). Average of sin² over a full cycle = 1/2. So avg PE = (1/2)kA² × (1/2) = kA²/4. Same as average KE.
Q25.
A physical pendulum (rigid body) has MI I about pivot and its CM is at distance d. Its time period is:
Answer: A. 2π√(I/Mgd)
Why: T = 2π√(I/Mgd). For a simple pendulum, I = mL² and d = L, giving T = 2π√(L/g).
Q26.
A mass is attached to two springs of constants k<sub>1</sub> and k<sub>2</sub> in series. If k<sub>1</sub> = k<sub>2</sub> = k, the effective constant is:
Answer: A. k/2
Why: Series: 1/k<sub>eff</sub> = 1/k + 1/k = 2/k → k<sub>eff</sub> = k/2. Series connection makes the effective spring softer.
Q27.
A particle performs SHM with amplitude 0.1 m and frequency 5 Hz. What is its maximum velocity?
Answer: A. π m/s
Why: v<sub>max</sub> = Aω = A × 2πf = 0.1 × 2π × 5 = π m/s ≈ 3.14 m/s.
Q28.
The displacement x = 5 sin(10t + π/6) cm. What is the initial phase?
Answer: A. π/6 rad
Why: In x = A sin(ωt + φ), the initial phase (phase at t=0) is φ = π/6 rad = 30°.
Q29.
The restoring torque on a simple pendulum of mass m and length L displaced by small angle θ is:
Answer: A. -mgL sinθ ≈ -mgLθ
Why: Torque = -mg sinθ × L ≈ -mgLθ (for small θ). This gives α = -(g/L)θ, so ω² = g/L and T = 2π√(L/g).
Q30.
For a spring-mass system, the time period in a lift accelerating upward with acceleration a is:
Answer: A. 2π√(m/(k)) (same as before)
Why: For a spring-mass system, T = 2π√(m/k). This does NOT depend on gravity, so acceleration of lift has no effect. (For a pendulum, effective g changes, but not for spring-mass).
Q31.
A tuning fork of frequency 256 Hz produces SHM. What is the period?
Answer: A. 1/256 s
Why: T = 1/f = 1/256 s ≈ 3.9 ms.
Q32.
In forced oscillations with damping, the steady state amplitude is maximum when:
Answer: A. Driving frequency approaches the natural frequency
Why: Maximum amplitude (resonance) occurs when driving frequency is close to the natural frequency. Higher damping reduces and broadens the resonance peak.
Q33.
A horizontal spring-mass system executes SHM. If the spring is replaced with one that is 4 times stiffer, the period:
Answer: A. Halves
Why: T = 2π√(m/k). If k increases 4 times, T decreases by factor √4 = 2. The period halves.
Q34.
A particle oscillates between x = -5 cm and x = +5 cm. Its amplitude and equilibrium position are:
Answer: A. A = 5 cm, x<sub>eq</sub> = 0
Why: Amplitude = half the range = (5-(-5))/2 = 5 cm. Equilibrium is at the midpoint = 0.
Q35.
The time period of oscillation of a liquid in a U-tube of total length L (liquid length) is:
Answer: A. 2π√(L/2g)
Why: For liquid in U-tube (total column length L), T = 2π√(L/2g). This is similar to a pendulum with effective length L/2.
Q36.
The equation x = 3 sin(πt) + 4 cos(πt) represents SHM. What is the amplitude?
Answer: A. 5 cm
Why: This can be written as A sin(ωt + φ) where A = √(3² + 4²) = √(9+16) = √25 = 5 cm.
Q37.
The acceleration of a SHM particle at displacement x is a = -36x m/s². What is the time period?
Answer: A. π/3 s
Why: Comparing with a = -ω²x: ω² = 36 → ω = 6 rad/s. T = 2π/ω = 2π/6 = π/3 s.
Q38.
A spring stretches by 0.2 m when a mass of 2 kg is hung from it. The time period of vertical oscillation is:
Answer: A. 0.897 s
Why: At equilibrium: kx = mg → k = mg/x = 2×10/0.2 = 100 N/m. T = 2π√(m/k) = 2π√(2/100) = 2π × 0.1414 ≈ 0.889 s.
Q39.
In SHM, the kinetic and potential energies are equal when:
Answer: A. x = A/√2
Why: KE = PE when (1/2)k(A²-x²) = (1/2)kx² → A² = 2x² → x = A/√2.
Q40.
The Q factor (quality factor) of an oscillator measures:
Answer: A. Sharpness of resonance (ratio of resonant frequency to bandwidth)
Why: Q = ω_0/Δω (resonant frequency / bandwidth). A high Q means sharp resonance, low damping, and long decay time.
Q41.
A particle executes SHM with amplitude A. At time t=0, x = A/2 and velocity is positive. The phase constant φ in x = A sin(ωt + φ) is:
Answer: A. π/6
Why: x(0) = A sin(φ) = A/2 → φ = π/6 or 5π/6. Velocity = Aω cos(φ). For positive velocity: cos(φ) > 0, so φ = π/6.
Q42.
Two SHMs x<sub>1</sub> = A sin(ωt) and x<sub>2</sub> = A sin(ωt + π/3) are superposed. The amplitude of the resultant is:
Answer: A. A√3
Why: Resultant amplitude = √(A² + A² + 2A²cos(π/3)) = A√(2 + 2×1/2) = A√3.
Q43.
A mass m is connected to two springs (k<sub>1</sub> = k, k<sub>2</sub> = 3k) in parallel. If one spring is cut, what is the ratio of new period to original?
Answer: A. 2
Why: Original k<sub>eff</sub> = k + 3k = 4k. After cutting (say k<sub>1</sub> remains): k<sub>eff</sub> = k. T ∝ 1/√k<sub>eff</sub>. T<sub>new</sub>/T<sub>old</sub> = √(4k/k) = 2.
Q44.
A pendulum of length L on a trolley that accelerates at a (horizontal) has effective gravity g<sub>eff</sub> = √(g²+a²). Its period is:
Answer: A. 2π√(L/√(g²+a²))
Why: On an accelerating trolley, effective g is the vector sum of g (down) and -a (pseudo-force). g<sub>eff</sub> = √(g²+a²), directed at angle arctan(a/g) from vertical. T = 2π√(L/g<sub>eff</sub>).
Q45.
For a spring-mass system, if the amplitude is halved by damping, the energy becomes:
Answer: A. 1/4 of original
Why: E = (1/2)kA². If A halves to A/2, E = (1/2)k(A/2)² = (1/4)(1/2)kA² = E/4. Energy drops to 1/4.
Q46.
A horizontal disk oscillates as a torsional pendulum. If the wire has torsional constant k<sub>T</sub> and the disk has MI I, the period is:
Answer: A. 2π√(I/k<sub>T</sub>)
Why: For torsional SHM: τ = -k<sub>T</sub> θ, giving Iα = -k<sub>T</sub> θ. ω² = k<sub>T</sub>/I. T = 2π√(I/k<sub>T</sub>).
Q47.
A particle of mass m is in SHM. At time t, its displacement is x = a cos(ωt). The time it takes to go from x=a to x=a/2 is:
Answer: A. T/6
Why: x = a cos(ωt). x = a at t=0. x = a/2 when cos(ωt) = 1/2, so ωt = π/3, giving t = π/3ω = T/6.
Q48.
In forced oscillations, the phase lag of displacement behind the driving force at resonance is:
Answer: A. π/2
Why: At resonance, the displacement lags the driving force by exactly π/2 (90°). Below resonance the lag is less; above resonance it approaches π.
Q49.
Two masses m<sub>1</sub> and m<sub>2</sub> are connected by a spring of constant k. The reduced mass is μ = m<sub>1m</sub>_2/(m<sub>1</sub>+m<sub>2</sub>). The angular frequency of oscillation is:
Answer: A. √(k/μ)
Why: The two-body oscillator reduces to a one-body problem with reduced mass μ. ω = √(k/μ).
Q50.
The time-averaged potential energy of an SHM particle equals:
Answer: A. Half the total energy
Why: Average PE = (1/2) × (1/2)kA² = Total E/2. Similarly, average KE = Total E/2. They are equal and each half the total energy.
Q51.
A particle executes SHM with T = 2 s and amplitude 5 cm. The distance covered by the particle in one full time period starting from extreme is:
Answer: A. 20 cm
Why: In one full period, the particle goes from one extreme to the other and back: 4A = 4 × 5 = 20 cm total distance (not displacement).
Q52.
A block on a spring (k = 200 N/m) is given an initial displacement of 5 cm and initial velocity of 1 m/s. The amplitude of oscillation (mass = 2 kg) is:
Answer: A. 0.1 m
Why: Total E = (1/2)kx_0² + (1/2)mv_0² = (1/2)(200)(0.05)² + (1/2)(2)(1)² = 0.25 + 1 = 1.25 J. E = (1/2)kA² → A = √(2E/k) = √(2.5/200) = √(0.0125) ≈ 0.112 m.
Q53.
A tunnel is drilled through the Earth along a diameter. A ball dropped in undergoes SHM with period:
Answer: A. 84.6 minutes
Why: The restoring force inside Earth = -GMmr/R³ (proportional to r). T = 2π√(R/g) ≈ 2π√(6.4×10⁶/10) ≈ 5078 s ≈ 84.6 minutes.
Q54.
The equation x = A e<sup>-bt/2m</sup> sin(ω_d t + φ) describes:
Answer: A. Underdamped oscillation
Why: Underdamped oscillation: the system still oscillates but with exponentially decreasing amplitude. ω_d = √(ω_0² - b²/4m²) is the damped frequency.
Q55.
At what position in SHM is the KE equal to twice the PE?
Answer: A. x = A/√3
Why: KE = 2 PE: (1/2)k(A²-x²) = 2(1/2)kx² → A² - x² = 2x² → A² = 3x² → x = A/√3.
Q56.
Two springs k<sub>1</sub> and k<sub>2</sub> are attached to a block between two walls. The block is displaced and released. The effective spring constant is:
Answer: A. k<sub>1</sub> + k<sub>2</sub>
Why: When a block is between two walls with springs on each side, both springs act in the same restoring direction when displaced. k<sub>eff</sub> = k<sub>1</sub> + k<sub>2</sub> (parallel combination).
Q57.
A simple pendulum of length L is executing SHM. The tension in the string at the lowest point when the bob has speed v is:
Answer: A. mg + mv²/L
Why: At the lowest point, T - mg = mv²/L (centripetal force equation). So T = mg + mv²/L.
Q58.
The quality factor Q of a damped oscillator with natural frequency f<sub>0</sub> = 1000 Hz and bandwidth Δf = 10 Hz is:
Answer: A. 100
Why: Q = f<sub>0</sub>/Δf = 1000/10 = 100. A Q of 100 means the system rings for about 100 cycles before the amplitude drops significantly.
Q59.
For a seconds pendulum on Earth (T = 2 s, L = 1 m), what must be the length on a planet where g = 4 m/s² for the same period?
Answer: A. 0.4 m
Why: T = 2π√(L/g). For T = 2 s and g = 4 m/s²: 2 = 2π√(L/4) → 1/π = √(L/4) → L/4 = 1/π² ≈ 0.101 → L ≈ 0.405 m.
Q60.
In the equation of SHM, x = 10 sin(πt/2) cm. The acceleration when x = 6 cm is:
Answer: B. -6π²/4 cm/s²
Why: ω = π/2. a = -ω² x = -(π/2)² × 6 = -6π²/4 ≈ -14.8 cm/s².
Q61.
A particle in SHM has amplitude 5 cm and period 2 s. Its maximum speed is:
Answer: B. 0.157 m/s
Why: v<sub>max</sub> = Aω = 0.05·(2π/2) = 0.05π ≈ 0.157 m/s.
Q62.
The length of a simple pendulum is increased to 4 times its original value. Its period becomes:
Answer: C. 2 times
Why: T ∝ √L, so 4× length gives √4 = 2 times the period.
Q63.
Two identical springs each of constant k are connected in parallel and support a mass m. The period of oscillation is:
Answer: A. 2π√(m/2k)
Why: Parallel springs give effective constant 2k, so T = 2π√(m/2k).
Q64.
For a particle in SHM, at displacement x = A/2 the ratio of its kinetic energy to potential energy is:
Answer: C. 3
Why: PE ∝ x² = E/4, KE = 3E/4, so KE/PE = 3.
Q65.
A particle in SHM has angular frequency 10 rad/s and amplitude 0.1 m. Its maximum acceleration is:
Answer: C. 10 m/s²
Why: a<sub>max</sub> = Aω² = 0.1·100 = 10 m/s².
Q66.
When a simple pendulum is in a lift that accelerates upward with acceleration a, its period:
Answer: A. decreases
Why: Effective gravity becomes g + a, so T = 2π√(L/(g+a)) decreases.
Q67.
At the mean (equilibrium) position of a particle executing SHM:
Answer: A. velocity is maximum and acceleration is zero
Why: At the mean position the restoring force (and acceleration) is zero while the speed is maximum.
Q68.
A mass on a spring has period T. If the mass is quadrupled, the new period is:
Answer: C. 2T
Why: T ∝ √m, so 4× mass gives √4 = 2T.