Two particles P and Q describe SHM of same am | vedclass

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(B) Let the equations of motion for particles $P$ and $Q$ be $x_1 = a \sin(\omega t)$ and $x_2 = a \sin(\omega t + \phi)$,where $\phi$ is the phase di

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$\frac{\pi}{2}$

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(B) Let the equations of motion for particles $P$ and $Q$ be $x_1 = a \sin(\omega t)$ and $x_2 = a \sin(\omega t + \phi)$,where $\phi$ is the phase difference.The distance between the particles is given by $d = |x_2 - x_1| = |a \sin(\omega t + \phi) - a \sin(\omega t)|$.Using the trigonometric identity $\sin A - \sin B = 2 \sin(\frac{A-B}{2}) \cos(\frac{A+B}{2})$,we get:$d = |2a \sin(\frac{\phi}{2}) \cos(\omega t + \frac{\phi}{2})|$.The maximum value of this distance occurs when the magnitude of the cosine term is $1$.Thus,$d_{max} = 2a \sin(\frac{\phi}{2})$.Given $d_{max} = a \sqrt{2}$,we have $2a \sin(\frac{\phi}{2}) = a \sqrt{2}$.$\sin(\frac{\phi}{2}) = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}$.$\frac{\phi}{2} = \frac{\pi}{4}$,which implies $\phi = \frac{\pi}{2}$.

Two particles $P$ and $Q$ describe $SHM$ of same amplitude $a$ and same frequency $f$ along the same straight line. The maximum distance between the two particles is $a \sqrt{2}$. The phase difference between the particles is

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