Question

The amplitude and time period of simple harmonic oscillator are $a$ and $T$ respectively. The time taken by it in displacing from $x=0$ to $x=a/2$ will be

• A. $T$
• B. $\frac{T}{2}$
• C. $\frac{T}{4}$
• D. $\frac{T}{12}$

Answer 1

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

$x=a\sin \omega t$
$x=0att=0$
When $x=\frac{a}{2}\Rightarrow \frac{a}{2}=a\sin \omega t$ or $\sin \omega t=\frac{1}{2}$
or, $\omega t={30}^0=\frac{\pi }{6}$
This gives, $\frac{2\pi }{T}\times t=\frac{\pi }{6}$ or $t=\frac{T}{12}$