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. $\frac{T}{4}$
• B. $\frac{T}{8}$
• C. $\frac{T}{12}$
• D. $\frac{T}{2}$

Answer 1

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

Equation of a SHM is given by $X=Asin\left(\omega t\right)$

Here $\omega =\frac{2\pi }{T}$

Time to travel half of the amplitude

$\frac{A}{2}=Asin\left(\omega t\right)$

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

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