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

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

Answer 1

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

Given: Same amplitude

This problem is easy to solve with the help of phasor diagram. Firstly, we draw the initial position of both the particles on the phasor as shown in figure.

From above figure, phase difference between both the particles is $\frac{\pi }{2}$.

They will cross each other when their projection from the circle on the horizontal diameter meet at one point.

Let after time $t$, both will reach at $P^{\prime }{Q}^{\prime }$ point having phase difference $\frac{\pi }{2}$ as shown in figure.

Both will meet at $\frac{A}{\sqrt{2}}$

When they meet, angular displacement of $P$ is

$\theta =\frac{\pi }{2}+\frac{\pi }{4}=\frac{3\pi }{4}$

So they will meet after time,

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

$t=\frac{3\pi }{4\times \omega }\times \tau =\frac{3\tau }{8}s$