Question

Two particles are executing SHM in a straight line with same amplitude A and time period . At $t=0$ one particle is at displacement $x=+A$ and the other at $x=-A/2$ and they are approach towards each other. After what time they cross each other.

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

Answer 1

The correct option is A $T/3$

The displacement equation in S.H.M is given by

$x=Asin\omega t$. . . . . ..(1)

One particle is at $x=+A$, equation (1) becomes,

$A=Asin\omega t_1$

$\omega t_1=si{n}^{-1}\left(1\right)=\frac{\pi }{2}$

$\frac{2\pi }{T}{t}_1=\frac{\pi }{2}$

$t_1=\frac{T}{4}sec$

Second particle at $x=-\frac{A}{2}$

$-\frac{A}{2}=Asin\omega t_2$

$w{t}_2=si{n}^{-1}(-0.5)=-\frac{\pi }{6}$

$\frac{2\pi }{T}{t}_2=-\frac{\pi }{6}$

$t_2=-\frac{T}{12}sec$

They approach towards each other, the at time when they cross each other,

$t=t_1-t_2$

$t=\frac{T}{4}-(-\frac{T}{12})$

$t=\frac{T}{3}sec$

The correct option is A.