Question

A point performs simple harmonic oscillation of period $T$ and the equation of motion is given by $x=asin(\omega t+\frac{\pi }{6}).$ After the elapse of what fraction of the time peirod, the velocity of the point will be equal to half of its maximum velocity?

• A. $\frac{T}{8}$
• B. $\frac{T}{6}$
• C. $\frac{T}{3}$
• D. $\frac{T}{12}$

Answer 1

The correct option is D $\frac{T}{12}$

The displacement of the particle is given as

$x=asin(\omega t+\frac{\pi }{6})$

Differentiating w.r.t time, we get velocity

$v=a\omega cos(\omega t+\frac{\pi }{6})$

When $v=\frac{a\omega }{2}$,

$\frac{a\omega }{2}=a\omega cos(\omega t+\frac{\pi }{6})$

$\Rightarrow cos(\omega t+\frac{\pi }{6})=\frac{1}{2}$

$\Rightarrow \omega t+\frac{\pi }{6}=\frac{\pi }{3}$

$\Rightarrow \omega t=\frac{\pi }{6}$

$\Rightarrow t=\frac{\pi }{6\omega }=\frac{T}{12}$
where $T=\frac{2\pi }{\omega }$