Question

Two particles undergo SHM along same line with same period of $\left(T\right)$ and equal amplitudes $\left(A\right)$. At particular instant one particle is at $x=–A$ and the other is at $x=0$. They move in same direction. They will cross each other at

• A. $t=\frac{3T}{8}$
• B. $t=\frac{4T}{3}$
• C. $x=\frac{A}{\sqrt{2}}$
• D. $x=\frac{A}{2}$

Answer 1

Answer: (A), (C)

$x=\frac{A}{\sqrt{2}}$

$x_1=–A\cos \omega t$

$x_2=–A\sin \omega t$

When the particles meet, $x_1=x_2$

$\Rightarrow \cos \omega t=\sin \omega t$

or $\tan \omega t=–1$

$\omega t=\frac{3\pi }{4}$

$t=\frac{3\pi }{8}$ also $x=A\sin \omega t=A\sin \left(\frac{3\pi }{4}\right)=\frac{A}{\sqrt{2}}$