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

JEE Main · 2023 · 29 Jan 2023 · Shift 2 · Elasticity & SHM

A particle of mass 250 g executes a simple harmonic motion under a periodic force

F = (- 25 x) N

. The particle attains a maximum speed of

4 m / s

during its oscillation. The amplitude of the motion is ______ cm.

Answer: 40

The restoring force

F = - 25 x

identifies the force constant as

k = 25

N/m. With mass

m = 0.25

kg, the angular frequency follows from

ω^{2} = \frac{k}{m} = \frac{25}{0.25} = 100

, so

ω = 10

rad/s. The maximum speed in simple harmonic motion is

v_{m a x} = ω A

, hence

A = \frac{v _{m a x}}{ω} = \frac{4}{10} = 0.4

m. The amplitude of the motion is

40

cm.

JEE Main · 2023 · 29 Jan 2023 · Shift 2 · Integer answer