STANDARDPhysics-Oscillations
STANDARD Physics Simple Harmonic Motion MCQ Question
Type: MCQ-conceptual-Medium-Class 11
The displacement of a particle executing simple harmonic motion is given by x = 3 sin(2πt + π/4) where x is in metres and t is in seconds. The amplitude and maximum speed of the particle is
A
3 m, 2π m s⁻¹
B
3 m, 4π m s⁻¹
C
3 m, 6π m s⁻¹
D
3 m, 8π m s⁻¹
Correct Answer
Option C
Detailed Explanation
The given equation of SHM is x = 3 sin(2πt + π/4). Compare with the standard equation x = A sin(ωt + φ), we get A = 3 m, ω = 2π s⁻¹. Maximum speed, v_max = Aω = 3 m × 2π s⁻¹ = 6π m s⁻¹.
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