Question

A particle executes simple harmonic oscillation with an amplitude $A$. If the time period of oscillation is $T$, then 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 $\frac{T}{12}$

Let displacement equation of particle executing SHM is $y=\text{A sin}\omega t$
As particle travels half of the amplitude from the equilibrium position, so $y=\frac{A}{2}$
Therefore , $\frac{A}{2}=\text{A sin}\omega t$
or $\text{sin}\omega t=\frac{1}{2}=\text{sin}\frac{\pi }{6}$
or $\omega t=\frac{\pi }{6}$
or $t=\frac{\pi }{6\omega }$
or $t=\frac{\pi }{6\left(\frac{2\pi }{T}\right)}[\text{as}\omega =\frac{2\pi }{T}]$
or $t=\frac{T}{12}$
Hence, the particle travels half of the amplitude from the equilibrium in $\frac{T}{12}\text{s}$.