Free Simple Harmonic Motion MCQs with Answers

• A. Halved
• B. Doubled
• C. Quadrupled
• D. None of these

Explanation: The frequency of a simple pendulum is given by: f = 1 / (2π) * √(g/L)where:f is the frequencyg is the acceleration due to gravityL is the…

• A. Total energy
• B. P.F at extreme
• C. K.E at mean
• D. All of these

Explanation: In Simple Harmonic Motion, the total energy of the system remains constant.

• A. Remain Same
• B. Be Doubled
• C. Be Halved
• D. Be Quartered

Explanation: The time period of a spring-mass system is given by:T = 2π√(m/k)where m is the mass and k is the spring constant.If the mass and spring…

• A. At maximum height when a ball is thrown upwards
• B. At extreme position of a Simple Pendulum
• C. At the amplitude of the Mass Spring System
• D. All of these

Explanation: Kinetic energy (KE) is the energy possessed by an object due to its motion.

• A. Decrease
• B. Increase
• C. Remains unchanged
• D. Becomes infinite

Explanation: When the train accelerates horizontally, the pendulum experiences a net effective gravitational force given by geff= √(g² + a²), which is…

• A. 1 Hz
• B. 4π Hz
• C. 1/4 Hz
• D. 4 Hz

Explanation: Compare a + ω2x = 0 to identify ω2 = 4π2, so ω = 2π. Frequency f = ω/(2π) = (2π)/(2π) = 1 Hz. Thus the system oscillates once per second.

• A. 1/2
• B. 1
• C. 1/8
• D. 1/4

Explanation: For SHM potential energy U ∝ x². If x = A/2 then U = ½k(A/2)² = ¼·(½kA²). Total energy is ½kA², so U/E = (¼·½kA²)/(½kA²) = 1/4.

• A. Reasonably large for easy measurement
• B. As small as possible
• C. Without any restriction
• D. Very large because formula T = 2π√(L/g) is derived accordingly

Explanation: The small-angle approximation (sinθ ≈ θ) underlies the standard period formula.

• A. 2.5
• B. 5
• C. 10
• D. 5√2

Explanation: For small amplitudes the period of a simple pendulum is independent of amplitude: T = 2π√(L/g).

• A. When the spring is fully compressed and when the spring is fully extended
• B. When the spring is at its rest length
• C. When the spring is halfway between its rest length and its fully extended or compressed length
• D. The acceleration is constant

Explanation: In SHM acceleration a = −ω² x, so its magnitude is proportional to displacement |x|.

• A. Mean position
• B. Extreme position
• C. Halfway of mean
• D. Everywhere same

Explanation: a = -ω²x → maximum acceleration occurs at maximum x =±A.

• A. 3 Hz
• B. 2 Hz
• C. 4 Hz
• D. 5 Hz

Explanation: Tension provides centripetal force: T = m (2π f)² r = 4π² m r f². Solve for f: f = √(T/(4π² m r)).

• A. Gravitational pull
• B. Surface tension
• C. Speed of rotation
• D. Tension in a spring

Explanation: A tachometer measures rotational speed (revolutions per minute or frequency) of shafts, wheels or engines.

• A. ω²A/2
• B. ω²A
• C. ω²A/4
• D. ω²A√3

Explanation: In SHM, acceleration is a=−ω² x . At displacement x=A/2, the magnitude becomes a=ω² (A/2). So, the acceleration is ω²(A/2).

• A. T/4
• B. T/2
• C. T/8
• D. T

Explanation: One complete oscillation = 4 quarters.Extreme → mean = one quarter.

• A. KE = 0
• B. PE = maximum
• C. KE = maximum
• D. PE = maximum & KE = maximum

Explanation: At x = 0:v = maximumKE maximumPE minimum.

• A. 0
• B. π/2
• C. π
• D. 2π

Explanation: Velocity leads displacement by 90° (π/2).

• A. 50 N/m
• B. 75 N/m
• C. 100 N/m
• D. 150 N/m

Explanation: For springs in series, the equivalent spring constant keq is given by 1/keq = 1/k1 + 1/k2 .

• A. 1.62 m/s2
• B. 9.8 m/s2
• C. Very high
• D. Zero

Explanation: A low Earth satellite experiences nearly the same gravitational acceleration as objects near Earth's surface, approximately g ≈ 9.8 m/s²…

• A. A
• B. √2A
• C. 2A
• D. 2√2A

Explanation: x1 = Asinωt, x2 = Asin(ωt + π/2 ) = Acos ωt. Distance d=|x1 - x2| = A|sinωt- cos ωt|.Maximum value is √2A.