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 $2 t$ and $3 t$ respectively. Find the amplitude of the motion, if the particle crosses the point of start at speed $2 m / s$ :

• A. $\frac{8 t}{\pi}$
• B. $\frac{4 t}{\pi}$
• C. $\frac{2 t}{\pi}$
• 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 $T=\left(t_{Q A}+t_{A Q}\right)+\left(t_{Q P}+t_{P Q}\right)+\left(t_{P B}+t_{B P}\right)$
$=t+[(2 t-t)+(2 t-t)]+t=4 t=2 \pi / \omega$
$\omega=\pi / 2 t ; x=A \sin \left(\omega t+\frac{\pi}{6}\right)$
$\frac{d x}{d t}=A \omega \cos \left(\omega t+\frac{\pi}{6}\right)=A \omega \cos \left(\frac{\pi}{2}+\frac{\pi}{6}\right)$
$2=A \times \frac{\pi}{2 t} \times \frac{1}{2} $ (sign not considered)
$A=\frac{8 t}{\pi}$