Question

A particle describes SHM in a straight line about O. If the time period of the motion is T then the kinetic energy at P behalf of its peak value at O, if the time taken by the particle to travel from O to P is

A: \[\dfrac{1}{2}T\]
B: \[\dfrac{1}{4}T\]
C: \[\dfrac{1}{2\sqrt{2}}T\]
D: \[\dfrac{1}{8}T\]

Answer 1

Hint: In simple harmonic motion, we can take the displacement of the motion as a sinusoidal function of time and express it accordingly. We also know that at the peak value of oscillation, the particle has the maximum velocity. We can solve the given problem by considering the given criterias.

Formula used:
To find the displacement of a particle in simple harmonic motion,
\[x=A\cos (\omega t)\] or \[A\sin (\omega t)\]
Where x is the required displacement, A is the amplitude, \[\omega \] is the angular velocity and t is the time taken for the oscillation.

Complete step by step answer:
We are given the question that a particle describes SHM in a straight line about O and when the time period of the motion T the kinetic energy at P behalf of its peak value at O.
We have to find the time taken by the particle to travel from O to P in this situation.
If the displacement \[x=A\sin \omega T\]
\[\begin{align}
  & \dfrac{dx}{dt}=A\omega \cos \omega T \\
 & \Rightarrow V=A\omega \cos \omega T \\
\end{align}\]
From this we can obtain that \[{{V}_{\max }}=A\omega \]
Kinetic energy at P is half of its peak value at O.
$\begin{align}
  & \Rightarrow \dfrac{1}{2}m{{v}^{2}}=\dfrac{1}{2}(\dfrac{1}{2}mv_{\max }^{2}) \\
 & \Rightarrow v=\dfrac{{{v}_{\max }}}{\sqrt{2}}=\dfrac{A\omega }{\sqrt{2}} \\
 & \Rightarrow A\omega \cos \omega T=\dfrac{A\omega }{\sqrt{2}} \\
 & \therefore \cos \omega T=\dfrac{1}{\sqrt{2}}\Rightarrow \omega T=\dfrac{\pi }{4} \\
\end{align}$
From this, we can obtain that
$T=\dfrac{\pi }{4\omega }=\dfrac{\pi T}{4(2\pi )}=\dfrac{T}{8}$ (Since $\omega =\dfrac{2\pi }{T}$)
Thus we have obtained the value of the time period as \[\dfrac{1}{8}T\].

So, the correct answer is “Option D”.

Note:
Either sine function or cosine function can be considered as the displacement of the particle in SHM. There will not be any variation in the final answer that is obtained. The time taken to cover the same distance in both motions would be the same as sine and cosine are functions that vary with a phase difference of $\dfrac{\pi }{2}$.