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.

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answer is 40.

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Detailed Solution

Given, F = -25 x
Compare with F = - Kx
K = 25 N/m
Angular frequency of oscillation $ω = \sqrt{\frac{K}{M}}$
$ω = \sqrt{\frac{25}{0.25}} = 10 rad / s$
Maximum speed of particle $V_{max} = Aω$
$\Rightarrow 4 = A (10)$

$\Rightarrow A = 0.4 m = 40 cm$

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