Question

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

• A. zero
• B. $ \pi /2 $
• C. $ \pi /6 $
• D. $ \pi /3 $

Answer 1

Answer: (B)

$y_{1}=a \sin \omega t$ and
$y_{2}=a \sin (\omega t+\phi)$
$y_{2}-y_{1}=a \sqrt{2}=a \sin (\omega t+\phi)-a \sin \omega t$
$\sqrt{2} a=2 a \cos \left\{\frac{\omega t+\phi+\omega t}{2}\right\} \sin \left(\frac{\phi+\omega t-\omega t}{2}\right)$
$=2 a \cos \left(\omega t+\frac{\phi}{2}\right) \sin \frac{\phi}{2}$
For maximum value, $\cos (\omega t+\phi)=1$
$ 2 \sin \frac{\phi}{2} =\sqrt{2}$ or $ \sin \frac{\phi}{2}=\frac{1}{\sqrt{2}}$
$\frac{\phi}{2} =\frac{\pi}{4} $
$\phi =\frac{\pi}{2}$