Question

A particle starts its $SHM$ on line at intial phase of $\pi /6$. It reaches again the point of start after time $t$. It crosses yet another point $P$ on the same line at successive intervals $2t$ and $3t$ respectively. Find the amptitude of the motion, if the particle crosses the point of start at speed $2m/s$.

• A. $4t/\pi$
• B. $8t/\pi$
• C. $2t/\pi$
• D. $t/\pi$

Answer 1

The correct option is B $8t/\pi$

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

Let time $t=0$ is considered from point $Q$ where displacement $=A/2$

Given, particle cross point $P$ on the same line at successive interval $2t$ and $3t$ respectively,

Time period
$T=(t_{QA}+t_{AQ})+(t_{QP}+t_{PQ})+(t_{PB}+t_{BP})$$=t+\left[2\right(2t-t)+(2t-t\left)\right]+t=4t$
$\Rightarrow \frac{2\pi }{\omega }=4t$
$\omega =\frac{\pi }{2t}$

Given speed of the particle, $v=2m/s$
$v=A\omega \left(cos(\omega t+\frac{\pi }{6})\right)=2m/s$
$\frac{A\pi }{2t}\left(\cos (\frac{\pi }{2t}t+\frac{\pi }{6})\right)=2$
$\frac{A\pi }{2t}\left(\frac{1}{2}\right)=2$
$A=\frac{8t}{\pi }$

Final answer: (a)