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
$x=A/2=T/4-T/12=T/6$