The equations for the displacements of two pa | vedclass

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Question

(C) The displacement of the first particle is $y_1 = 0.1 \sin(100 \pi t + \frac{\pi}{3})$. The velocity $v_1$ is given by $v_1 = \frac{dy_1}{dt} = 0.1

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$\frac{\pi}{6}$

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(C) The displacement of the first particle is $y_1 = 0.1 \sin(100 \pi t + \frac{\pi}{3})$. The velocity $v_1$ is given by $v_1 = \frac{dy_1}{dt} = 0.1 	imes 100 \pi \cos(100 \pi t + \frac{\pi}{3}) = 10 \pi \sin(100 \pi t + \frac{\pi}{3} + \frac{\pi}{2})$.At $t=0$,the phase of $v_1$ is $\phi_1 = \frac{\pi}{3} + \frac{\pi}{2} = \frac{5\pi}{6}$.The displacement of the second particle is $y_2 = 0.1 \cos(\pi t) = 0.1 \sin(\pi t + \frac{\pi}{2})$. The velocity $v_2$ is given by $v_2 = \frac{dy_2}{dt} = 0.1 	imes \pi \cos(\pi t + \frac{\pi}{2}) = 0.1 \pi \sin(\pi t + \frac{\pi}{2} + \frac{\pi}{2}) = 0.1 \pi \sin(\pi t + \pi)$.At $t=0$,the phase of $v_2$ is $\phi_2 = \pi$.The phase difference is $\Delta \phi = |\phi_2 - \phi_1| = |\pi - \frac{5\pi}{6}| = \frac{\pi}{6}$.

The equations for the displacements of two particles in simple harmonic motion are $y_1=0.1 \sin \left(100 \pi t+\frac{\pi}{3}\right)$ and $y_2=0.1 \cos \pi t$ respectively. The phase difference between the velocities of the two particles at a time $t=0$ is

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