The displacements of two particles executing | vedclass

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(C) Given,$y_{1} = 2 \sin (10 t + 	heta)$.The velocity $V_{1} = \frac{dy_{1}}{dt} = 2 	imes 10 \cos (10 t + 	heta) = 20 \cos (10 t + 	heta)$.Given

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$(	heta - \frac{\pi}{2})$

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(C) Given,$y_{1} = 2 \sin (10 t + 	heta)$.The velocity $V_{1} = \frac{dy_{1}}{dt} = 2 	imes 10 \cos (10 t + 	heta) = 20 \cos (10 t + 	heta)$.Given,$y_{2} = 3 \cos 10 t = 3 \sin (10 t + \frac{\pi}{2})$.The velocity $V_{2} = \frac{dy_{2}}{dt} = 3 	imes 10 \cos (10 t + \frac{\pi}{2}) = 30 \cos (10 t + \frac{\pi}{2})$.The phase of $V_{1}$ is $\phi_{1} = 10 t + 	heta$.The phase of $V_{2}$ is $\phi_{2} = 10 t + \frac{\pi}{2}$.The phase difference $\Delta \phi = \phi_{1} - \phi_{2} = (10 t + 	heta) - (10 t + \frac{\pi}{2}) = 	heta - \frac{\pi}{2}$.

The displacements of two particles executing simple harmonic motion are represented as $y_{1} = 2 \sin (10 t + \theta)$ and $y_{2} = 3 \cos 10 t$. The phase difference between the velocities of these waves is

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