Question

A particle executes SHM of type $x=A\sin \omega t$. It takes time $t_1$ from $x=0$ to $x=\frac{A}{2}$ and $t_2$ from $x=\frac{A}{2}$ to $x=A.$ The ratio $t_1:t_2$ will be

• A. 1: 1
• B. 1: 2
• C. 1: 3
• D. 2: 1

Answer 1

Answer: (B)

$t_1+t_2=\frac{T}{4}$
or $t_2=\frac{T}{4}-t_1$
At time, $t=t_1,x=\frac{A}{2}$
$\therefore \frac{A}{2}=A\sin \omega t_1$ or $\omega t_1=\frac{\pi }{6}$ or $t_1=\frac{\pi }{6\omega }$
$\therefore t_2=\frac{T}{4}-\frac{\pi }{6\omega }=\frac{2\pi }{4\omega }-\frac{\pi }{6\omega }=\frac{2\pi }{6\omega }\left(\because T=\frac{2\pi }{\omega }\right)$
$\therefore t_1:t_2=1:2$