Question

A particle executes simple harmonic oscillation
with an amplitude a. The period of oscillation is
T. The minimum time taken by the particle to
travel half of the amplitude from the
equilibrium position is:

• A. T/2
• B. T/4
• C. T/8
• D. T/12

Answer 1

The correct option is D T/12

$\because X=asin\omega t$
$\therefore \frac{a}{2}=aisn\omega t\Rightarrow \omega t=\frac{\pi }{6}$
$\Rightarrow \left(\frac{2\pi }{T}\right)t=\frac{\pi }{6}\Rightarrow t=\frac{T}{12}$