The displacement of a particle executing S.H. | vedclass

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(B) The displacement of the particle is given by $x = a \sin (\omega t - \phi)$.To find the velocity,we differentiate the displacement with respect to

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$a \omega$

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(B) The displacement of the particle is given by $x = a \sin (\omega t - \phi)$.To find the velocity,we differentiate the displacement with respect to time $t$:$v = \frac{dx}{dt} = \frac{d}{dt} [a \sin (\omega t - \phi)] = a \omega \cos (\omega t - \phi)$.Now,we substitute the given time $t = \frac{\phi}{\omega}$ into the velocity equation:$v = a \omega \cos \left( \omega \left( \frac{\phi}{\omega} ight) - \phi ight)$.$v = a \omega \cos (\phi - \phi) = a \omega \cos (0)$.Since $\cos 0^{\circ} = 1$,we get:$v = a \omega 	imes 1 = a \omega$.

The displacement of a particle executing $S.H.M.$ is $x = a \sin (\omega t - \phi)$. The velocity of the particle at time $t = \frac{\phi}{\omega}$ is (given $\cos 0^{\circ} = 1$):

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