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