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} KA ^{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} KA ^{2} \cos ^{2} \omega t $
$=\frac{1}{2}\left(\frac{1}{2} KA ^{2}\right)$
$\cos ^{2} \omega t =\frac{1}{2}$
$\cos \omega t =\frac{1}{\sqrt{2}}$
$\omega t =\frac{\pi}{4}$
$\frac{2 \pi t }{ T }=\frac{\pi}{4}$
$t =\frac{ T }{8}$