A particle of mass 250 g executes a simple harmonic motion under a periodic force $\mathrm{F}=(-25~x)\mathrm{N}$. The particle attains a maximum speed of 4 m/s during its oscillation. The amplitude of the motion is ___________ cm.
Answer (integer)
40
Solution
<p>$F = - 25x$</p>
<p>$.250{{{d^2}x} \over {d{t^2}}} = - 25x$</p>
<p>${{{d^2}x} \over {d{t^2}}} = - 100x$</p>
<p>$\Rightarrow \omega = 10$ rad/sec</p>
<p>& $\omega A = {v_{\max }}$</p>
<p>$10\,A = 4$</p>
<p>$\Rightarrow A = 0.4$ m</p>
<p>$= 40$ cm</p>
About this question
Subject: Physics · Chapter: Oscillations · Topic: Simple Harmonic Motion
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