Question

A particle executes simple harmonic oscillation. Its amplitude is $a$. 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. $\frac{T}{8}$
• B. $\frac{T}{12}$
• C. $\frac{T}{2}$
• D. $\frac{T}{4}$

Answer 1

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

We know the general equation of S.H.M. :
$y=a\sin \omega t$
$\frac{a}{2}=a\sin \left(\frac{2\pi t}{T}\right)$
$sin\left(2\pi t/T\right)=sin\left(\pi /6\right)$

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

$t=\frac{T}{12}$