Question

Two particles are executing SHM in a straight line with same amplitude $A$ and time period $T$. At time $t=0$, one particle is at displacement $x$${}_1$$=+A$ and the other at $x$${}_2$$=A/2$ and they are approaching towards each other. After what time will they cross each other ?

• A. $T/3$
• B. $T/4$
• C. $5T/6$
• D. $T/6$

Answer 1

The correct option is D $T/6$

$Asin\theta =\frac{A}{2}$
$\theta ={30}^o$
$\therefore \frac{\varphi L}{\omega }={90}^o-{30}^o$
$l={60}^o$
$\therefore$ time taken is to corer the phase difference is ${60}^o$
$T$ for ${360}^o$
$t=\frac{{60}^o}{{360}^o}\times T$
$f=\frac{1}{6}T$