The velocity of a particle executing a simple harmonic motion is $13m{{s}^{-1}}$, when its distance from the equilibrium position (Q) is 3m and its velocity is $12m{{s}^{-1}}$, when it is 5m away from Q. The frequency of the simple harmonic motion is:
$\begin{align}
  & \text{A}\text{.}\dfrac{5\pi }{8} \\
 & \text{B}\text{.}\dfrac{5}{8\pi } \\
 & \text{C}\text{.}\dfrac{8\pi }{5} \\
 & \text{D}\text{.}\dfrac{8}{5\pi } \\
\end{align}$

Hint: We have a particle in simple harmonic motion. We have to find the particle’s frequency. To find frequency we have the equation for frequency. We first solve for $\omega $ by using the expression for velocity of simple harmonic motion and then solve for frequency using $\omega $.

Formula Used:
Frequency, $f=\dfrac{1}{T}=\dfrac{\omega }{2\pi }$
Velocity of a particle in SHM, $v=\omega \sqrt{{{A}^{2}}-{{x}^{2}}}$

Complete step by step answer:
In the question we have a particle executing simple harmonic motion.
The equilibrium position of the particle is said to be ‘Q”
The velocity of the particle when it is 3m away from Q is given.
Let this distance be $'{{x}_{1}}'$.
Let this velocity be \['{{v}_{1}}'\].
It is given that,
${{x}_{1}}=3m$
${{v}_{1}}=13m{{s}^{-1}}$
When the particle is at a distance 5 m from the equilibrium,
${{x}_{2}}=5m$
The velocity of the particle is,
${{v}_{2}}=12m{{s}^{-1}}$
We have to find the frequency of this simple harmonic motion.
We know that for a simple harmonic motion, velocity $'v'$ is given as
$v=\omega \left( \sqrt{{{A}^{2}}-{{x}^{2}}} \right)$, Where $'\omega '$ is angular velocity, $'A'$ is amplitude and $'x'$ is the distance or position.
Squaring on both sides, we get
$\Rightarrow $ ${{v}^{2}}={{\omega }^{2}}\left( {{A}^{2}}-{{x}^{2}} \right)$
$\Rightarrow $ $\dfrac{{{v}^{2}}}{{{\omega }^{2}}}={{A}^{2}}-{{x}^{2}}$
Here, for ${{v}_{1}}=13m{{s}^{-1}}$ and${{x}_{1}}=3m$
$\Rightarrow $ $\dfrac{{{v}_{1}}^{2}}{{{\omega }^{2}}}={{A}^{2}}-{{x}_{1}}^{2}$
$\Rightarrow $ $\left( \dfrac{{{13}^{2}}}{\omega } \right)={{A}^{2}}-\left( {{3}^{2}} \right)$
For ${{v}_{2}}=12m{{s}^{-1}}$ and${{x}_{2}}=5m$
$\Rightarrow $ $\dfrac{{{v}_{2}}^{2}}{{{\omega }^{2}}}={{A}^{2}}-{{x}_{2}}^{2}$
$\Rightarrow $ $\left( \dfrac{{{12}^{2}}}{\omega } \right)={{A}^{2}}-\left( {{5}^{2}} \right)$
By equating both these equations, we get
${{\left( \dfrac{13}{\omega } \right)}^{2}}-{{\left( \dfrac{12}{\omega } \right)}^{2}}={{5}^{2}}-{{3}^{2}}$
$\begin{align}
 & {{\omega }^{2}}=\left( \dfrac{\left( {{13}^{2}}-{{12}^{2}} \right)}{25-9} \right) \\
& {{\omega }^{2}}=\dfrac{169-144}{16} \\
& {{\omega }^{2}}=\dfrac{25}{16} \\
\end{align}$
Therefore,
$\omega =\dfrac{5}{4}$
We need to find frequency.
Frequency of a simple harmonic motion is given by the equation
$f=\dfrac{\omega }{2\pi }$
By substituting the value of ω in the above equation, we get
$\begin{align}
  & f=\dfrac{5}{4\times 2\pi } \\
 & f=\dfrac{5}{8\pi } \\
\end{align}$
Therefore, the frequency of the simple harmonic motion is, $f=\dfrac{5}{8\pi }$ .

So, the correct answer is “Option B”.

Note:
Simple harmonic motion is a type of periodic motion where the restoring force on the moving object is directly proportional to the object's displacement magnitude and acts towards the object's equilibrium position.
Frequency is the number of oscillations per unit time.
Angular frequency $\left( \omega \right)$ is the measure of angular displacement per unit time in a simple harmonic oscillation.