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Question

(A) Let the displacement be $x = A \sin(\omega t + \phi)$.For the first particle,the distance from the mean position $O$ is $x_1 = \frac{2-\sqrt{3}}{2

Vedclass · Accepted Answer

$\frac{2\pi}{3}$

Vedclass · Answer

(A) Let the displacement be $x = A \sin(\omega t + \phi)$.For the first particle,the distance from the mean position $O$ is $x_1 = \frac{2-\sqrt{3}}{2} A$. Since it is moving towards $+x$,its displacement from the mean position is $x_1 = A \sin(\phi_1)$.However,looking at the figure,the distance from the extreme position is $\frac{2-\sqrt{3}}{2} A$. Thus,the displacement from the mean position is $x_1 = A - \frac{2-\sqrt{3}}{2} A = \frac{\sqrt{3}}{2} A$.Since it is moving towards $+x$,$\sin(\phi_1) = \frac{\sqrt{3}}{2}$ and $\cos(\phi_1) > 0$,so $\phi_1 = \frac{\pi}{3}$.For the second particle,the distance from the extreme position $(-A)$ is $\frac{2-\sqrt{3}}{2} A$. Thus,its displacement from the mean position is $x_2 = -A + \frac{2-\sqrt{3}}{2} A = -\frac{\sqrt{3}}{2} A$.Since it is moving towards $+x$,$\sin(\phi_2) = -\frac{\sqrt{3}}{2}$ and $\cos(\phi_2) > 0$,so $\phi_2 = 2\pi - \frac{\pi}{3} = \frac{5\pi}{3}$.The phase difference is $\Delta\phi = \phi_2 - \phi_1 = \frac{5\pi}{3} - \frac{\pi}{3} = \frac{4\pi}{3}$.The equivalent phase difference is $2\pi - \frac{4\pi}{3} = \frac{2\pi}{3}$.

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

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