The displacement $x$ (in metres) of a particle performing simple harmonic motion is related to time $t$ (in seconds) as $x = 0.05\cos \left( {4\,\pi \,t + \frac{\pi }{4}} \right)$. The frequency of the motion will be ..... $Hz$
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(d) From the given equitation $\omega = 2\pi n = 4\pi $
==> $n = 2\,Hz$
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