Question

A particle is executing SHM of periodic time T. The time taken by a particle in moving from mean position to half the maximum displacement is:

(sin ${30}^o=0.5$)

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

Answer 1

Answer: (D)

$\frac{T}{12}$

Let at the instant $t=0$, the particle is at the mean position.

So, the displacement equation of the particle executing $SHM$ is given by, $x=Asin\left(wt\right)$

Where, $A$ and $w$ are the amplitude of oscillation and the frequency respectively.

Let the time taken by the particle in going from mean position to half maximum displacement be $t$.

$\therefore$ $\frac{A}{2}=Asin\left(wt\right)$

$⟹sin\left(wt\right)=0.5$

This gives: $wt=\frac{\pi }{6}$ $\because (\pi /6={30}^o)$

$\therefore$ $\frac{2\pi }{T}t=\frac{\pi }{6}$ $⟹$ $t=\frac{T}{12}$