Two particle executing $S.H.M.$ of same amplitude of $20 \,cm$ with same period along the same line about same equilibrium position. The maximum distance between the two is $20 \,cm$. Their phase difference in radian is equal to
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(a)
$x_1=A \sin \left(\omega t+\phi_1\right)$
$x_2=A \sin \left(\omega t+\phi_2\right)$
$x_1-x_2=A \sin \left(\omega t+\phi_1\right)-A \sin \left(\omega t+\phi_2\right)$
$20=2 \times 20 \sin \left(\frac{\phi_1-\phi_2}{2}\right) \cdot \cos \left[\omega t+\left(\frac{\phi_1+\phi_2}{2}\right)\right]$
$\frac{1}{2}=\sin \left(\frac{\phi_1-\phi_2}{2}\right) \cdot \cos \left(\omega t+\left(\frac{\phi_1+\phi_2}{2}\right)\right) \text { for maximum value. } \Rightarrow \frac{\phi_1-\phi_2}{2}=\frac{\pi}{6} \Rightarrow \phi_1-\phi_2=\frac{\pi}{3}$
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