The displacements of two particles executing | vedclass

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(D) The phase of the first particle is $	heta_1 = \frac{\pi}{2} t + \phi$.At $t = 1 \, s$,$	heta_1 = \frac{\pi}{2}(1) + \phi = \frac{\pi}{2} + \phi$

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$\frac{\pi}{6}$

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(D) The phase of the first particle is $	heta_1 = \frac{\pi}{2} t + \phi$.At $t = 1 \, s$,$	heta_1 = \frac{\pi}{2}(1) + \phi = \frac{\pi}{2} + \phi$.The phase of the second particle is $	heta_2 = \frac{2 \pi}{3} t + \phi$.At $t = 1 \, s$,$	heta_2 = \frac{2 \pi}{3}(1) + \phi = \frac{2 \pi}{3} + \phi$.The phase difference $\Delta 	heta$ is given by $|	heta_2 - 	heta_1|$.$\Delta 	heta = |(\frac{2 \pi}{3} + \phi) - (\frac{\pi}{2} + \phi)| = |\frac{2 \pi}{3} - \frac{\pi}{2}| = |\frac{4 \pi - 3 \pi}{6}| = \frac{\pi}{6}$.

The displacements of two particles executing $S.H.M.$ on the same line are given as $y_1 = a \sin \left(\frac{\pi}{2} t + \phi\right)$ and $y_2 = b \sin \left(\frac{2 \pi}{3} t + \phi\right)$. The phase difference between them at $t = 1 \, s$ is .........

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