A particle undergoes simple harmonic motion having time period T. the time taken in $\dfrac{3}{8}th$ oscillation is:
$\text{A}\text{. }\dfrac{3}{8}T$
$\text{B}\text{. }\dfrac{5}{8}T$
$\text{C}\text{. }\dfrac{5}{12}T$
$\text{D}\text{. }\dfrac{7}{12}T$

Hint: The displacement of a particle in simple harmonic motion is a sinusoidal function of time. Hence, start with considering the displacement of the particle as a function of cosine or sine of time i.e. $x=A\sin (\omega t)$ or $x=A\cos (\omega t)$. Find the position of the particle at $\dfrac{3}{8}th$ oscillation and then find the time taken to reach this position.

Complete step-by-step answer:
Let the simple harmonic motion of the particle be given by $x=A\sin (\omega t)$ …..(1), where is x is the position of the particle at time t, A is the amplitude of the particle and $\omega $ is the angular frequency of the oscillation.
Let us first find what is meant by $\dfrac{3}{8}th$ oscillation. If the particle starts the oscillations from the mean position and completes one oscillation, it travels a total distance of 4A. When it travels a total distance of 2A, it is called half oscillation and when it travels a total distance of A, it is one-fourth oscillation.
Similarly, when it covers a total distance of $\dfrac{A}{2}$, it is $\dfrac{1}{8}th$ oscillation and therefore, $\dfrac{3}{8}th$ oscillation means that the particle has covered a total distance of $\dfrac{3A}{2}$.
That means the particle started at the mean position travelled to one extreme by covering a distance of A and returned back covering a distance of $\dfrac{A}{2}$. Therefore, the particle is at the position $x=\dfrac{A}{2}$.
Substitute the value of x in equation (1).
$\Rightarrow \dfrac{A}{2}=A\sin (\omega t)\Rightarrow \sin \omega t=\dfrac{1}{2}$
$\Rightarrow \omega t=\dfrac{\pi }{6}\text{ or }\omega t=\pi -\dfrac{\pi }{6}=\dfrac{5\pi }{6}$.
 $\omega t=\dfrac{\pi }{6}$ is when the particle was at the position $x=\dfrac{A}{2}$ for the first time (while going to the extreme).
$\omega t=\dfrac{5\pi }{6}$ is when the particle reached $x=\dfrac{A}{2}$ again. Therefore, we discard $\omega t=\dfrac{\pi }{6}$ and continue with $\omega t=\dfrac{2\pi }{6}$.
Therefore, $t=\dfrac{5\pi }{6\omega }$ and $\omega =\dfrac{2\pi }{T}$
$\Rightarrow t=\dfrac{5\pi }{6\left( \dfrac{2\pi }{T} \right)}=\dfrac{5\pi T}{12\pi }=\dfrac{5T}{12}$
Therefore, the time taken by the particle to complete $\dfrac{3}{8}th$ oscillation is $\dfrac{5T}{12}$.
Hence, the correct option is (C).

Note: You can consider either cosine function or sine function for the simple harmonic motion the answer will not change. Since, sine and cosine functions can be said as the same functions with a phase difference of $\dfrac{\pi }{2}$. The time taken for covering the same distance in both motions will be the same.