Question

Two particles undergo SHM along parallel lines with the same time period $\left(T\right)$ and equal amplitudes. At a particular instant, one particle is at its extreme position while the other is at its mean position. They move in the same direction. They will cross each other after a further time of

• A. $\frac{T}{8}$
• B. $\frac{3T}{8}$
• C. $\frac{T}{6}$
• D. $\frac{4T}{3}$

Answer 1

The correct option is B $\frac{3T}{8}$

For a particle executing SHM, the displacement of the particle is given by
$x=a\sin (\omega t+\varphi )$ where, $a$ is the amplitude of SHM.
From the data given in the question, Let us suppose that one of the particle starting from extreme position will start from point $\text{B}$
Let the equation of particle starting from point $\text{B}$ be
$x_1=-a\cos \omega t$
Let the equation of particle starting from point $\text{O'}$ moving towards $\text{A'}$ be
$x_2=a\sin \omega t$
Two particles cross eachother when
$x_1=x_2$
$\therefore -a\cos \omega t=a\sin \omega t$
$\Rightarrow \tan \omega t=-1\Rightarrow \omega t=\frac{3\pi }{4}$
Using $\omega =\frac{2\pi }{T}$, we can say that,
$\frac{2\pi }{T}t=\frac{3\pi }{4}\Rightarrow t=\frac{3T}{8}$
Thus, option (b) is the correct answer.