For what instant of time, do potential energy becomes equal to kinetic energy for a particle executing SHM with a time period $T$:
A) $\dfrac{T}{4}$
B) $\dfrac{T}{8}$
C) $\dfrac{T}{2}$
D) $T$

Hint: The question asks us to calculate the time period of the motion when kinetic energy equals potential energy. This can be done, by equating the expression of kinetic and potential energy and thus finding the time period.

Formula Used:
$P.E. = k{A^2}{\sin ^2}\omega t$
where,
$P.E. = $ Potential energy
$k = $Spring constant
$A = $Amplitude
$\omega = $Angular frequency
$t = $Time
$K.E. = k{A^2}{\cos ^2}\omega t$
Where,
$K.E. = $ Kinetic energy
$k = $Spring constant
$A = $Amplitude
$\omega = $Angular frequency
$t = $Time

Complete step by step solution:
We know that in the case of simple harmonic motion, the motion is periodic in nature and the restoring force in the case of an object executing SHM is proportional to the displacement made by it.
Simple harmonic motion is a repetitive motion, therefore, the maximum displacement at one side must be equal to the maximum displacement of the other side.
We know, in the case of SHM, the Potential energy can be written as:
$P.E. = k{A^2}{\sin ^2}\omega t$
where,
\[P.E. = \] Potential energy
$k = $Spring constant
$A = $Amplitude
$\omega = $Angular frequency
$t = $Time
and , Kinetic energy can be written as:
$K.E. = k{A^2}{\cos ^2}\omega t$
where:
$K.E. = $ Kinetic energy
$k = $Spring constant
$A = $Amplitude
$\omega = $Angular frequency
$t = $Time
The value of the spring constant is fixed for a particular SHM.
Now, equating kinetic and potential energy, we get:
$\Rightarrow k{A^2}{\sin ^2}\omega t = k{A^2}{\cos ^2}\omega t$
Thus, on cancelling the identical terms, we get:
$\Rightarrow {\sin ^2}\omega t = {\cos ^2}\omega t$
From the basics of trigonometry, we know that this is true only for:
$\Rightarrow \omega t = \dfrac{\pi }{4}$
Now, we know, $\omega = 2\pi f$
where, $f = $frequency and we also know that:
$f = \dfrac{1}{T}$
where,
$T = $Time period
Thus, we can write:
$\Rightarrow \dfrac{{2\pi t}}{T} = \dfrac{\pi }{4}$
Therefore, or rearranging the equations, we obtain:
$\Rightarrow t = \dfrac{T}{8}$
This is our required solution.

Hence, option (B) is correct.

Note: In simple harmonic motion, since there are no dissipative forces, the total energy remains constant. Therefore, when kinetic energy is maximum, potential energy is zero, similarly, when potential energy is maximum, kinetic energy is zero.