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(D) The given equations for displacement are $y_1 = 4 \sin(10t + \phi)$ and $y_2 = 5 \cos(10t)$.To find the velocity,we differentiate the displacement

Vedclass · Accepted Answer

$\phi - \frac{\pi}{2}$

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(D) The given equations for displacement are $y_1 = 4 \sin(10t + \phi)$ and $y_2 = 5 \cos(10t)$.To find the velocity,we differentiate the displacement with respect to time $t$.For $y_1$,the velocity $v_1 = \frac{dy_1}{dt} = 4 	imes 10 \cos(10t + \phi) = 40 \sin(10t + \phi + \frac{\pi}{2})$.For $y_2$,the velocity $v_2 = \frac{dy_2}{dt} = -5 	imes 10 \sin(10t) = 50 \sin(10t + \pi)$.The phase of velocity $v_1$ is $	heta_1 = 10t + \phi + \frac{\pi}{2}$.The phase of velocity $v_2$ is $	heta_2 = 10t + \pi$.The phase difference $\Delta \phi = 	heta_1 - 	heta_2 = (10t + \phi + \frac{\pi}{2}) - (10t + \pi) = \phi - \frac{\pi}{2}$.

Two simple harmonic motions are represented by equations $y_1 = 4 \sin(10t + \phi)$ and $y_2 = 5 \cos(10t)$. What is the phase difference between their velocities?

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