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(D) Let the displacements of the two particles be $x_1 = A \sin(\omega t)$ and $x_2 = A \sin(\omega t + \phi)$,where $A = 20 \, cm$.The distance betwe

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$\frac{\pi}{3}$

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(D) Let the displacements of the two particles be $x_1 = A \sin(\omega t)$ and $x_2 = A \sin(\omega t + \phi)$,where $A = 20 \, cm$.The distance between the particles is $d = |x_2 - x_1| = |A \sin(\omega t + \phi) - A \sin(\omega t)|$.Using the trigonometric identity $\sin C - \sin D = 2 \cos(\frac{C+D}{2}) \sin(\frac{C-D}{2})$,we get:$d = |2A \cos(\omega t + \frac{\phi}{2}) \sin(\frac{\phi}{2})|$.The maximum distance occurs when the cosine term is $1$,so $d_{max} = 2A \sin(\frac{\phi}{2})$.Given $d_{max} = 20 \, cm$ and $A = 20 \, cm$,we have:$20 = 2(20) \sin(\frac{\phi}{2})$$\sin(\frac{\phi}{2}) = \frac{1}{2}$.Therefore,$\frac{\phi}{2} = \frac{\pi}{6}$,which gives $\phi = \frac{\pi}{3}$ radians.

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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