Two particles p and q perform SHM with the sa | vedclass

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(B) Let the displacements of the two particles be $x_1 = a \sin(\omega t + \phi_1)$ and $x_2 = a \sin(\omega t + \phi_2)$.The distance between them is

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$\frac{\pi}{2}$

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(B) Let the displacements of the two particles be $x_1 = a \sin(\omega t + \phi_1)$ and $x_2 = a \sin(\omega t + \phi_2)$.The distance between them is $d = |x_1 - x_2| = |a \sin(\omega t + \phi_1) - a \sin(\omega t + \phi_2)|$.Using the formula $\sin C - \sin D = 2 \sin(\frac{C-D}{2}) \cos(\frac{C+D}{2})$,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 $1$,so $d_{max} = |2a \sin(\frac{\Delta \phi}{2})|$,where $\Delta \phi = \phi_1 - \phi_2$ is the phase difference.Given $d_{max} = a\sqrt{2}$,we have $a\sqrt{2} = 2a \sin(\frac{\Delta \phi}{2})$.$\sin(\frac{\Delta \phi}{2}) = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}$.$\frac{\Delta \phi}{2} = 45^{\circ} = \frac{\pi}{4}$ radians.Therefore,$\Delta \phi = 2 	imes \frac{\pi}{4} = \frac{\pi}{2}$ radians.

Two particles $p$ and $q$ perform $SHM$ with the same amplitude $a$ and the same frequency $f$ along a straight line. The maximum distance between the two particles is $a\sqrt{2}$. The initial phase difference between the particles is:

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