Two simple harmonic motions are represented by the equations $y_1 = 0.1 \sin \left( 100 \pi t + \frac{\pi}{3} \right)$ 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
$(a) \frac{-\pi}{3} (b) \frac{\pi}{6} (c) \frac{-\pi}{6} (d) \frac{\pi}{3}$
Text Solution
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$\mathbf{v_1} = \frac{\mathrm{dy_1}}{\mathrm{dt}} = 0.1 \times 100 \pi \cos \left( 100 \pi t + \frac{\pi}{3} \right)$ $\mathbf{v_2} = \frac{\mathrm{dy_2}}{\mathrm{dt}} = -0.1 \pi \sin \pi t = 0.1 \pi \cos \left( \pi t + \frac{\pi}{2} \right)$ Phase difference of velocity of first particle with respect to the velocity of $2^{nd}$ particle at $t=0$ is $\Delta \varphi = \varphi_1 - \varphi_2 = \frac{\pi}{3} - \frac{\pi}{2} = - \frac{\pi}{6}$.
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