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(D) The given equation for displacement is $x = 10 \cos \left(2 \pi t + \frac{\pi}{2}\right)$.The velocity $v$ is the derivative of displacement with

Vedclass · Accepted Answer

$31.4$

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(D) The given equation for displacement is $x = 10 \cos \left(2 \pi t + \frac{\pi}{2}\right)$.The velocity $v$ is the derivative of displacement with respect to time: $v = \frac{dx}{dt}$.$v = \frac{d}{dt} \left[10 \cos \left(2 \pi t + \frac{\pi}{2}\right)\right] = -10 \cdot 2 \pi \sin \left(2 \pi t + \frac{\pi}{2}\right)$.$v = -20 \pi \sin \left(2 \pi t + \frac{\pi}{2}\right)$.At $t = \frac{1}{6} \, s$,the velocity is:$v = -20 \pi \sin \left(2 \pi \cdot \frac{1}{6} + \frac{\pi}{2}\right) = -20 \pi \sin \left(\frac{\pi}{3} + \frac{\pi}{2}\right)$.$v = -20 \pi \sin \left(\frac{5\pi}{6}\right)$.Since $\sin \left(\frac{5\pi}{6}\right) = \sin \left(150^\circ\right) = \frac{1}{2}$,we have:$v = -20 \pi \cdot \frac{1}{2} = -10 \pi$.The magnitude of velocity is $|v| = 10 \pi \approx 10 \cdot 3.14 = 31.4 \, cm/s$.

$A$ particle executes $S.H.M.$ according to the equation $x=10 \cos \left[2 \pi t+\frac{\pi}{2}\right] \, cm$,where $t$ is in seconds. The magnitude of the velocity of the particle at $t=\frac{1}{6} \, s$ will be .............. $cm/s$.

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