Two particles execute $SHM$ of same amplitude of $20\, cm$ with same period along the same line about the same equilibrium position. The maximum distance between the two is $20\, cm.$ Their phase difference in radians is
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$\boldsymbol{x}_{1}=20 \sin \omega t$
$\boldsymbol{x}_{2}=20 \sin (\omega t+\phi)$
distance between them is given by
$\left|\boldsymbol{x}_{2}-\boldsymbol{x}_{1}\right|=|20 \sin (\omega t+\phi)-20 \sin \omega t|=20|(\sin (\omega t+\phi)-\sin \omega t)|$
$\Rightarrow\left|\boldsymbol{x}_{2}-\boldsymbol{x}_{1}\right|=40\left|\cos \left(\omega t+\frac{\theta}{2}\right) \sin \left(\frac{\theta}{2}\right)\right|$
The distance between them is maximum when $\cos \left(\omega t+\frac{\theta}{2}\right)=1$ i.e.
$40 \sin \left(\frac{\theta}{2}\right)=20$
$\Rightarrow \sin \left(\frac{\theta}{2}\right)=\frac{1}{2}$
$\Rightarrow \frac{\theta}{2}=\frac{\pi}{\pi}$
$\Rightarrow \theta=\frac{\pi}{3}$
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