Question

A simple harmonic oscillator has an amplitude $A$ and time period $T$. The time required by it to travel from $X=A$ to $A=A / 2$ is

• A. T/6
• B. T/4
• C. T/3
• D. T/2

Answer 1

Answer: (A)

For S.H.M., $x=A \sin \left(\frac{2 \pi}{T} t\right)$
when $x=A, A=A \sin \left(\frac{2 \pi}{T} t\right)$
$\therefore \sin \left(\frac{2 \pi}{T} \cdot t\right)=1$
$ \Rightarrow \sin \left(\frac{2 \pi}{T} \cdot t\right)=\sin \left(\frac{\pi}{2}\right)$
$\Rightarrow t=(T / 4)$
When $x=\frac{A}{2}, \frac{A}{2}=A \sin \left(\frac{2 \pi}{T}, t\right)$
or $\sin \frac{\pi}{6}=\sin \left(\frac{2 \pi}{T} t\right)$
or $t=(T / 12)$
Now, time taken to travel from $x=A$ to
$x=A / 2=T / 4-T / 12=T / 6$