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(A) Let the displacements of the two particles be $x_1 = A \sin(\omega t + \phi_1)$ and $x_2 = A \sin(\omega t + \phi_2)$,where $A = 20 \, cm$.The dis

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$\frac{\pi}{3}$

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(A) Let the displacements of the two particles be $x_1 = A \sin(\omega t + \phi_1)$ and $x_2 = A \sin(\omega t + \phi_2)$,where $A = 20 \, cm$.The distance between them is $d = |x_1 - x_2| = |A \sin(\omega t + \phi_1) - A \sin(\omega t + \phi_2)|$.Using the trigonometric identity $\sin C - \sin D = 2 \sin(\frac{C-D}{2}) \cos(\frac{C+D}{2})$,we get:$d = |2A \sin(\frac{\phi_1 - \phi_2}{2}) \cos(\omega t + \frac{\phi_1 + \phi_2}{2})|$.The maximum distance occurs when the cosine term is $1$:$d_{max} = |2A \sin(\frac{\Delta \phi}{2})|$,where $\Delta \phi = \phi_1 - \phi_2$.Given $d_{max} = 20 \, cm$ and $A = 20 \, cm$:$20 = 2 	imes 20 \sin(\frac{\Delta \phi}{2})$$1 = 2 \sin(\frac{\Delta \phi}{2})$$\sin(\frac{\Delta \phi}{2}) = \frac{1}{2}$.Therefore,$\frac{\Delta \phi}{2} = \frac{\pi}{6}$,which gives $\Delta \phi = \frac{\pi}{3}$ radians.

Two particles are executing $S.H.M.$ with the same amplitude of $20 \, cm$ and the 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 equal to

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