Question

A particle starts its SHM on a line at initial 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 amplitude of the motion, if the particle crosses the point of start at speed $2m/s$ :

• A. $\frac{8t}{\pi }$
• B. $\frac{4t}{\pi }$
• C. $\frac{2t}{\pi }$
• D. $\frac{t}{\pi }$

Answer 1

Answer: (A)

$x=A\sin \left(\omega \times 0+\frac{\pi }{6}\right)=A\sin \frac{\pi }{6}=\frac{A}{2}(t=0$ at initial phase $)$
Let time $t=0$ is considered from point $Q$ where displacement $=A/2$
Time period $T=\left(t_{QA}+t_{AQ}\right)+\left(t_{QP}+t_{PQ}\right)+\left(t_{PB}+t_{BP}\right)$
$=t+[(2t-t)+(2t-t)]+t=4t=2\pi /\omega$
$\omega =\pi /2t;x=A\sin \left(\omega t+\frac{\pi }{6}\right)$
$\frac{dx}{dt}=A\omega \cos \left(\omega t+\frac{\pi }{6}\right)=A\omega \cos \left(\frac{\pi }{2}+\frac{\pi }{6}\right)$
$2=A\times \frac{\pi }{2t}\times \frac{1}{2}$ (sign not considered)
$A=\frac{8t}{\pi }$