A particle of mass 250 g executes a simple harmonic motion under a periodic force F = (−25x) N. The particle attains a maximum speed of 4 m/s during its oscillation. The amplitude of the motion is ______ cm.
Physics · Circular Motion
JEE Mains 29 January 2023 Shift 2 Evening Shift

Question

A particle of mass 250 g executes a simple harmonic motion under a periodic force F = (−25x) N. The particle attains a maximum speed of 4 m/s during its oscillation. The amplitude of the motion is ______ cm.

Physics · Circular Motion

G1 Physics · Accepted Answer

For SHM, F = −kx with k = 25 N m⁻¹ and m = 0.25 kg. Therefore ω = √(k/m) = √100 = 10 rad s⁻¹. Since v(max) = ωA = 4 m s⁻¹, A = 4/10 = 0.4 m = 40 cm.