Question

Displacement-time equation of a particle executing SHM is : $X=Asin(\omega t+\frac{\pi }{6})$. Time taken by the particle to go directly from $X=-\frac{A}{2}$ to $X=+\frac{A}{2}$ is

• A. $\frac{\pi }{3\omega }$
• B. $\frac{\pi }{2\omega }$
• C. $\frac{2\pi }{\omega }$
• D. $\frac{\pi }{\omega }$

Answer 1

The correct option is A $\frac{\pi }{3\omega }$

displacement, $X=Asin(\omega t+\frac{\pi }{6})$

At $X=-\frac{A}{2}$

$-\frac{A}{2}=Asin(\omega t_1+\frac{\pi }{6})$

$\omega t_1+\frac{\pi }{6}=si{n}^{-1}(-0.5)$

$\omega t_1+\frac{\pi }{6}=-\frac{\pi }{6}$

$t_1=-\frac{\pi }{3\omega }$. . . .(1)

At $X=+\frac{A}{2}$

$\frac{A}{2}=Asin(\omega t_2+\frac{\pi }{6})$

$\omega t_2+\frac{\pi }{6}=si{n}^{-1}\left(0.5\right)$

$\omega t_2+\frac{\pi }{6}=\frac{\pi }{6}$

$\omega t_2=0$

$t_2=0$. . . . . . . .(2)

The time taken by the particle,

$T=t_2-t_1$

$T=0-(-\frac{\pi }{3\omega })$

$T=\frac{\pi }{3\omega }$

The correct option is A.