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 they cross each other.

• A. $\frac{T}{3}$
• B. $\frac{T}{4}$
• C. $\frac{5T}{6}$
• D. $\frac{T}{6}$

Answer 1

Answer: (A)

$\frac{T}{3}$

Answer 2

The displacement equation in S.H.M is given by $x =A \sin \omega t \ldots \ldots (1)$ One particle is at $x=+A$, equation (1) becomes, $A = A \sin \omega t _{1}$ $\omega t_{1}=\sin ^{-1}(1)=\frac{\pi}{2}$ $\frac{2 \pi}{ T } t _{1}=\frac{\pi}{2}$ $t _{1}=\frac{ T }{4} \sec$ Second particle at $x=-\frac{A}{2}$ $-\frac{ A }{2}= A \sin \omega t _{2}$ $wt _{2}=\sin ^{-1}(-0.5)=-\frac{\pi}{6}$ $\frac{2 \pi}{ T } t _{2}=-\frac{\pi}{6}$ $t _{2}=-\frac{ T }{12} \sec$ They approach towards each other, the at time when they cross each other, $t = t _{1}- t _{2}$ $t =\frac{ T }{4}-\left(-\frac{ T }{12}\right)$ $t =\frac{ T }{3} \sec$