Two bodies performing SHM have the same ampli | vedclass

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(A) For the first particle,the displacement from the mean position $O$ is $x_1 = A - \left(\frac{\sqrt{2}-1}{\sqrt{2}}\right)A = A - (1 - \frac{1}{\sq

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$\pi$

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(A) For the first particle,the displacement from the mean position $O$ is $x_1 = A - \left(\frac{\sqrt{2}-1}{\sqrt{2}}\right)A = A - (1 - \frac{1}{\sqrt{2}})A = \frac{A}{\sqrt{2}}$. Since it is moving towards the positive extreme,its phase $\phi_1$ is given by $x_1 = A \sin(\phi_1)$,so $\sin(\phi_1) = \frac{1}{\sqrt{2}}$,which gives $\phi_1 = \frac{\pi}{4}$.For the second particle,the displacement from the mean position $O$ is $x_2 = -\left[A - \left(\frac{\sqrt{2}-1}{\sqrt{2}}\right)A\right] = -\frac{A}{\sqrt{2}}$. Since it is moving towards the mean position (moving in the positive direction),its phase $\phi_2$ is in the third quadrant,so $\phi_2 = \pi + \frac{\pi}{4} = \frac{5\pi}{4}$.The phase difference is $\Delta \phi = \phi_2 - \phi_1 = \frac{5\pi}{4} - \frac{\pi}{4} = \pi$.

Column-$I$	Column-$II$
$(a)$ $t = \frac{T}{8}$	$(i)$ $\theta = \frac{5\pi}{4}$
$(b)$ $t = \frac{5T}{8}$	$(ii)$ $\theta = \frac{3\pi}{2}$
	$(iii)$ $\theta = \frac{\pi}{4}$

Two bodies performing $SHM$ have the same amplitude and frequency. Their positions at a certain instant are as shown in the figure. The phase difference between them is

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