Question

A particle describes SHM in a straight line about $O$.

If the time period of the motion is $T$ then its kinetic energy at P be half of its peak value at $O$, if the time taken by the particle to travel from $O$ to $P$ is

• A. $\frac{1}{2}T$
• B. $\frac{1}{4}T$
• C. $\frac{1}{2\sqrt{2}}T$
• D. $\frac{1}{8}T$

Answer 1

Answer: (D)

Method-1
$K_{\max }=\frac{1}{2}K{A}^2$
$(KE{)}_P=\frac{1}{2}{K}_{\max }$
$=\frac{1}{2}K\left(A^2-x^2\right)$
$\Rightarrow x=\frac{A}{\sqrt{2}}$ or $\theta =45^{\circ }$
So, time $=\frac{T}{360^{\circ }}\times 45^{\circ }=T/8$
Method-2
$\frac{1}{2}K{A}^2{\cos }^2\omega t$
$=\frac{1}{2}\left(\frac{1}{2}K{A}^2\right)$
${\cos }^2\omega t=\frac{1}{2}$
$\cos \omega t=\frac{1}{\sqrt{2}}$
$\omega t=\frac{\pi }{4}$
$\frac{2\pi t}{T}=\frac{\pi }{4}$
$t=\frac{T}{8}$