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Q.

A particle of mass 250g 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 oscillations. The amplitude of the motion is ______cm.

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answer is 40.

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Detailed Solution

$Compare with F = - K x; K = 25 N / m$
Angular frequency of oscillation $ω = \sqrt{\frac{K}{M}}$
$ω = \sqrt{\frac{25}{0.25}} = 10 r a d / S$
Maximum speed of particle $V_{\max} = A ω$
$4 = A (10) \Rightarrow A = 0.4 m = 40 c m$

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