Question

Time period ($T$) and amplitude ($A$) are same for two particles which undergo $SHM$ along the same line. At one particular instant, one particle is at phase $\frac{3\pi }{2}$ and other is at phase zero. While moving in the same direction. Find the time at which they will cross each other.

• A. $\frac{3T}{7}$
• B. $\frac{3T}{8}$
• C. $\frac{4T}{2}$
• D. $\frac{3T}{4}$

Answer 1

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

Given,
$t=0$,

${\delta }_1=\frac{3\pi }{2}$,

${\delta }_2=0$.

Since particle is already at some initial phase, we can find initial position using standard equation

$x=A\sin (\omega t+\delta )$

At $t=0$

$x_1=A\sin \frac{3\pi }{2}=-A$

Other particle is at mean position ($x_2=0$) So, at $t=0$,

$x_1=-A\cos \omega t$
and
$x_2=-A\sin \omega t\dots \left(i\right)$

Condition for meeting is

$x_1=x_2$

$-\cos \omega t=\sin \omega t$

$\tan \omega t=-1$

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

$\left(\frac{2\pi }{T}\right)t=\frac{3\pi }{4}$

$t=\frac{3T}{8}$