Two S.H.Ms. are represented by equations y_1 | vedclass

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(B) Given equations for displacement are:$y_1 = 0.1 \sin \left(100 \pi t + \frac{\pi}{3} \right)$$y_2 = 0.1 \cos (100 \pi t) = 0.1 \sin \left(100 \pi

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$-\frac{\pi}{6}$

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(B) Given equations for displacement are:$y_1 = 0.1 \sin \left(100 \pi t + \frac{\pi}{3} ight)$$y_2 = 0.1 \cos (100 \pi t) = 0.1 \sin \left(100 \pi t + \frac{\pi}{2} ight)$Velocity is the derivative of displacement with respect to time:$v_1 = \frac{dy_1}{dt} = 0.1 	imes 100 \pi \cos \left(100 \pi t + \frac{\pi}{3} ight) = 10 \pi \sin \left(100 \pi t + \frac{\pi}{3} + \frac{\pi}{2} ight) = 10 \pi \sin \left(100 \pi t + \frac{5\pi}{6} ight)$$v_2 = \frac{dy_2}{dt} = 0.1 	imes 100 \pi \cos \left(100 \pi t + \frac{\pi}{2} ight) = 10 \pi \sin \left(100 \pi t + \frac{\pi}{2} + \frac{\pi}{2} ight) = 10 \pi \sin (100 \pi t + \pi)$The phase of $v_1$ is $\phi_1 = 100 \pi t + \frac{5\pi}{6}$ and the phase of $v_2$ is $\phi_2 = 100 \pi t + \pi$.The phase difference $\Delta \phi = \phi_1 - \phi_2 = \frac{5\pi}{6} - \pi = -\frac{\pi}{6}$.

Two $S$.$H$.Ms. are represented by equations $y_1 = 0.1 \sin \left(100 \pi t + \frac{\pi}{3} \right)$ and $y_2 = 0.1 \cos (100 \pi t)$. The phase difference between the speeds of the two particles is

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