Y = A sin (omega t + phi_{0}) is the time-dis | vedclass

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Question

(C) The displacement equation is $Y = A \sin (\omega t + \phi_{0})$.At $t = 0$,$Y = A \sin(\phi_{0}) = \frac{A}{2}$.This implies $\sin(\phi_{0}) = \fr

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$\frac{5 \pi}{6}$

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(C) The displacement equation is $Y = A \sin (\omega t + \phi_{0})$.At $t = 0$,$Y = A \sin(\phi_{0}) = \frac{A}{2}$.This implies $\sin(\phi_{0}) = \frac{1}{2}$.Thus,$\phi_{0}$ can be $\frac{\pi}{6}$ or $\frac{5 \pi}{6}$.The velocity of the particle is given by $v = \frac{dY}{dt} = A \omega \cos(\omega t + \phi_{0})$.At $t = 0$,$v = A \omega \cos(\phi_{0})$.Since the particle is moving in the negative direction,$v For $\phi_{0} = \frac{\pi}{6}$,$\cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2} > 0$.For $\phi_{0} = \frac{5 \pi}{6}$,$\cos(\frac{5 \pi}{6}) = -\frac{\sqrt{3}}{2} Therefore,the initial phase angle is $\phi_{0} = \frac{5 \pi}{6}$.

$Y = A \sin (\omega t + \phi_{0})$ is the time-displacement equation of a $SHM$. At $t = 0$,the displacement of the particle is $Y = \frac{A}{2}$ and it is moving in the negative direction. Then the initial phase angle $\phi_{0}$ will be ...... .

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