A particle is executing SHM with time period T. Starting from the mean position, time taken by it to complete $\dfrac{5}{8}$ oscillations is
\[\begin{align}
  & A.\dfrac{T}{12} \\
 & B.\dfrac{T}{6} \\
 & C.\dfrac{5T}{12} \\
 & D.\dfrac{7T}{12} \\
\end{align}\]

Hint: Start by assuming the time, when the particle will be at the mean position .Then using the SHM equation for displacement of particle $x=A\sin(\omega t)$, find the time for the particle to complete $\dfrac{5}{8}$ oscillations
Formula: $x=A\sin(\omega t)$

Complete answer:
SHM or simple harmonic motion is the motion caused by the restoring force; it is directly proportional to the displacement of the object from its mean position. Clearly, this is a reactive force, which is always directed towards the mean. Let us assume the equation for displacement of particle to be given as $x=A\sin(\omega t)$, then the acceleration of the particle is given by, $a(t)=-\omega^{2}x(t)$. We know that $\omega$ is the angular velocity and is given as $\omega=\dfrac{2\pi}{T}$, where $T$ is the time period of the motion.
Given that the time period of the particle is $T$. Let the total distance covered by the particle during the time period T be $4\;A$. Then the distance covered during $\dfrac{5}{8}$ can be written as $\dfrac{5}{8}\times 4A=\dfrac{5A}{2}$
 Also, $\dfrac{5A}{2}$ can be written in terms of $\dfrac{A}{2}$ as $\dfrac{5A}{2}=2A+\dfrac{A}{2}$
Now, from the equation of the particle, we can find the time $t$ taken due to $\dfrac{A}{2}$ distance. Consider the equation$\dfrac{A}{2}=A sin\omega t$
$\implies \dfrac{A}{2}=A sin \dfrac{2\pi}{T}t$
$\implies \dfrac{1}{2}=sin\dfrac{2\pi}{T}t$
$\implies sin\dfrac{\pi}{6}=sin\dfrac{2\pi}{T}t$
$\implies t=\dfrac{T}{12}$
Similarly, the time $t\prime$ taken to cover $2\;A$ is given as, $2A=Asin\dfrac{2\pi}{T}t\prime$
$\implies 2\times sin \dfrac{\pi}{2}=sin\dfrac{2\pi}{T}t\prime$
$\implies t\prime=\dfrac{T}{2}$
Then total time take to complete $\dfrac{5}{8}$ oscillations is given as $T\prime=t+t\prime$
$\implies T\prime=\dfrac{T}{12}+\dfrac{T}{2}$
$\therefore T\prime=\dfrac{7T}{12}$

Thus the correct answer is option \[D.\dfrac{7T}{12}\]

Note:
Remember SHM motions are sinusoidal in nature. Here, we are assuming, the particle is at mean when, $t=0$. This makes the further steps easier. Then, it will take time $T$ for the particle to cover one oscillation. For simplification, we are finding the time taken due to the small parts of the oscillation