Two simple harmonic motions are represented b | vedclass

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(C) The velocity of the first particle is given by the derivative of its displacement: ${v_1} = \frac{d{y_1}}{dt} = 0.1 	imes 100\pi \cos(100\pi t +

Vedclass · Accepted Answer

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

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(C) The velocity of the first particle is given by the derivative of its displacement: ${v_1} = \frac{d{y_1}}{dt} = 0.1 	imes 100\pi \cos(100\pi t + \frac{\pi}{3}) = 10\pi \cos(100\pi t + \frac{\pi}{3})$.The velocity of the second particle is given by the derivative of its displacement: ${v_2} = \frac{d{y_2}}{dt} = -0.1\pi \sin(\pi t) = 0.1\pi \cos(\pi t + \frac{\pi}{2})$.The phase of the velocity of the first particle is ${\phi_1} = 100\pi t + \frac{\pi}{3}$,and the phase of the velocity of the second particle is ${\phi_2} = \pi t + \frac{\pi}{2}$.However,the question asks for the phase difference between the two velocities. Since the frequencies are different ($100\pi$ vs $\pi$),the phase difference is time-dependent. Assuming the question implies the difference in initial phases at $t=0$:$\Delta \phi = \phi_1(0) - \phi_2(0) = \frac{\pi}{3} - \frac{\pi}{2} = \frac{2\pi - 3\pi}{6} = -\frac{\pi}{6}$.

Two simple harmonic motions are represented by the equations ${y_1} = 0.1 \sin(100\pi t + \frac{\pi}{3})$ and ${y_2} = 0.1 \cos(\pi t)$. The phase difference of the velocity of particle $1$ with respect to the velocity of particle $2$ is

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