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
$y_2=a\sin (\omega t+\varphi )$
$y_2-y_1=a\sqrt{2}=a\sin (\omega t+\varphi )-a\sin \omega t$
$\sqrt{2}a=2a\cos \left\{\frac{\omega t+\varphi +\omega t}{2}\right\}\sin \left(\frac{\varphi +\omega t-\omega t}{2}\right)$
$=2a\cos \left(\omega t+\frac{\varphi }{2}\right)\sin \frac{\varphi }{2}$
For maximum value, $\cos (\omega t+\varphi )=1$
$2\sin \frac{\varphi }{2}=\sqrt{2}$ or $\sin \frac{\varphi }{2}=\frac{1}{\sqrt{2}}$
$\frac{\varphi }{2}=\frac{\pi }{4}$
$\varphi =\frac{\pi }{2}$