The displacement of a particle in S.H.M. is g | vedclass

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(A) The displacement of the particle is $x = A \cos(\omega t + \pi/6)$.The velocity $v$ is the derivative of displacement with respect to time: $v = \

Vedclass · Accepted Answer

$\frac{\pi}{3 \omega} 	ext{ s}$

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(A) The displacement of the particle is $x = A \cos(\omega t + \pi/6)$.The velocity $v$ is the derivative of displacement with respect to time: $v = \frac{dx}{dt} = -A \omega \sin(\omega t + \pi/6)$.The speed is maximum when the magnitude of velocity $|v|$ is maximum,which occurs when $|\sin(\omega t + \pi/6)| = 1$.This happens when the argument $(\omega t + \pi/6) = \pi/2$ (or $3\pi/2$,etc.).Setting $\omega t + \pi/6 = \pi/2$,we get $\omega t = \pi/2 - \pi/6 = 3\pi/6 - \pi/6 = 2\pi/6 = \pi/3$.Therefore,$t = \frac{\pi}{3 \omega} 	ext{ s}$.

The displacement of a particle in $S.H.M.$ is given by $x = A \cos(\omega t + \pi/6)$. At what time will its speed be maximum?

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