Question

Two particles execute SHM along a straight line with same amplitude and same frequency. Everytime they meet in opposite directions at a displacement equal to half the amplitude. The difference between them is:

• A. $2\pi /3$
• B. $\pi$
• C. $\pi /3$
• D. $\pi /2$

Answer 1

The correct option is A $2\pi /3$

Let the initial phase difference be $\theta$, time of meeting be $t$ and amplitude be $a$.

So for first particle, $asin\omega t=a/2\Rightarrow \omega t=\pi /6-\left(1\right)$

Similarly for the second particle,

$asin(\omega t+\theta )=a/2\Rightarrow \omega t+\theta =\pi -\pi /6=5\pi /6-\left(2\right)$

From $\left(1\right)and\left(2\right)$, $\theta =5\pi /6-\pi /6=2\pi /3$