Question

Two particle of same time period$\left(T\right)$ and amplitude undergo $SHM$ along the same line with initial phase of $\pi /6$. If they start at the same instant and at same point along opposite directions. Find the time after which they will meet again for the first time.

• A. $T/2$
• B. $T/8$
• C. $T$
• D. $T/4$

Answer 1

The correct option is A $T/2$

Let the equation for SHM will be
$x=A\sin (\omega t+\frac{\pi }{6})$ (as initial phase is $\left(\pi /6\right)$)
At $t=0$
$x=A\sin (\omega \times 0+\frac{\pi }{6})=\frac{A}{2}$

Time taken in moving $A/2=T/B$

It is obvious that particle will meet again at $Q$ while having covered a distance
$=4\times \frac{A}{2}=2A$
Time required $=4\times T/8=T/2$

Final answer: (c)