Question

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

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

Answer 1

The correct option is B $\pi /2$

$y_1=a\sin \omega t$ and $y_2=a\sin (\omega t+\varphi )$
$y_2-y_1=a\sqrt{2}=a\sin (\omega t+\varphi )-a\sin \omega t$
or
$\sqrt{2}a=2a\cos \left(\frac{\omega t+\varphi +\omega t}{2}\right)\sin \left(\frac{\omega t+\varphi -\omega t}{2}\right)$
$=2a\cos (\omega t+\frac{\varphi }{2})\sin \frac{\varphi }{2}$
For maximum value, $\cos (\omega t+\varphi /2)=1$
$\therefore 2\sin \frac{\varphi }{2}=\sqrt{2}$ or $\sin \frac{\varphi }{2}=\frac{1}{\sqrt{2}}$
or $\frac{\varphi }{2}=\frac{\pi }{4}$ or $\varphi =\frac{\pi }{2}$
Hence, the correct answer is option (b).

Why this question? Tip: For maximum value, we don't always need to use the Double Derivative Test. We know that the maximum value of a sin or cos function is 1.