Question

A particle executes SHM with a time period of 12 s. Find the time taken by the particle to go directly from its mean position to half of its amplitude.

• A. 6
• B. 1
• C. 7
• D. 8

Answer 1

The correct option is B 1

We know,

Equation of SHM,

$y=A\sin \left(\frac{2\pi t}{T}\right)$

Where, A=Amplitude,T=time period

At $y=0$ mean position,

$0=A\sin \left(\frac{\pi t}{6}\right)$

i.e $t=0$

At $y=\frac{A}{2}$

$\frac{A}{2}=A\sin \left(\frac{\pi t}{6}\right)$

i.e $t=1s$

Difference in times, is the time taken to reach from mean to half of the Amplitude,

i.e $1sec$

Option $\text{B}$ is the correct answer.