Question

A particle executes SHM with amplitude $0.2m$ and time period $24s$. The time required for it to move from the mean position to a point $0.1m$ from the mean position is

• A. $2s$
• B. $3s$
• C. $8s$
• D. $12s$

Answer 1

Answer: (A)

$2s$

Amplitude of particle executing SHM $0.2m$. Time period $T=24s$
Hence
$x=0.2\sin \omega t$
$\omega$ is angular frequency equal to $\frac{2\pi }{24}$
In question $x=0.1m$ and we have to find time $t$
Putting the value of $x$
$0.1=0.2\sin \omega t$
$\sin \omega t=\frac{1}{2}=\sin \frac{\pi }{6}$

$\frac{2\pi }{24}t=\frac{\pi }{6}$
$\Rightarrow t=2s$