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

BITSATBITSAT 2014

Solution:

$\left|x_{1}-x\right|=2 a \sin \left(\omega+\frac{\phi_{1}+\phi_{2}}{2}\right) \cos \left(\frac{\phi_{1}-\phi_{2}}{2}\right)$
To maximize
$\left|x_{1}-x_{2}\right|: \sin \left(\omega t+\frac{\phi_{1}+\phi_{2}}{2}\right)=1$
$\Rightarrow a \sqrt{2}=2 a \times 1 \times \cos \left(\frac{\phi_{1}-\phi_{2}}{2}\right) $
$\Rightarrow \frac{1}{\sqrt{2}}=\cos \left(\frac{\phi_{1}-\phi_{2}}{2}\right)$
$ \Rightarrow \frac{\pi}{4}=\frac{\phi_{1}-\phi_{2}}{2}$
$\Rightarrow \phi_{1}-\phi_{2}=\frac{\pi}{2}$