Question

A particle executes a SHM of time period T. The time taken by the particle to go directly from its mean position to half the amplitude.

• A.
• B.
• C.
• D.

Answer 1

The correct option is A

Let the equation of motion be $x=A\sin \omega t$
At t = 0, x = 0 $\therefore$ Particle is at its mean position.
The particle will be at $x=\frac{A}{2}$ at a time t,
Where $\frac{A}{2}=A\sin \omega t$ or $\sin \omega t=\frac{1}{2}$ or $\omega t=\frac{\pi }{6}$
(Minimum value of wt is taken because we want to go directly from x = 0 to $x=\frac{A}{2}$ )
Thus, $t=\frac{\pi }{6\omega }=\frac{\pi }{6\left(\frac{2\pi }{T}\right)}=\frac{T}{12}$