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**Hint:**In this solution, we will use the formula of displacement of a simple harmonic oscillator to determine the equation of velocity and acceleration of the same. Since they are both equal, we will use the condition to determine the time period of oscillation.

**Formula used:**In this solution, we will use the following formula:

$x = A\sin (\omega t + \phi )$ where $x$ is the displacement of the oscillator, $\omega $ is the angular velocity, $t$ is the time, and $\phi $ is the phase.

**Complete step by step answer:**

We’ve been given that the velocity and acceleration of a simple harmonic oscillator are the same.

Now the velocity of the harmonic oscillator can be calculated using:

$v = \dfrac{{dx}}{{dt}}$

Substituting the value of $x = A\sin (\omega t + \phi )$ in the above equation, and taking the derivative, we can write

$v = A\omega \cos (\omega t + \phi )$

Similarly, the acceleration can be determined as

$a = \dfrac{{dv}}{{dt}}$

Substituting $v = A\omega \cos (\omega t + \phi )$ in the above equation, and taking the derivative, we get

$a = - A{\omega ^2}\sin (\omega t + \phi )$

Now, since the maximum velocity and the maximum acceleration are the same for our case, we can write

${v_{max}} = {a_{max}}$

$ \Rightarrow A\omega = A{\omega ^2}$

Dividing both sides, by $A\omega $, we get

$\omega = 1$

Since $\omega = \dfrac{{2\pi }}{T}$, we can write

$T = 2\pi = 6..28\,\sec $

**Hence the correct choice is option (C).**

**Note:**The maximum values of velocities of acceleration will be achieved at different positions of the simple harmonic oscillator. However, that is not of consequence to us since we are only focused on the magnitude of the maximum velocity and acceleration. Hence, we do not need to worry about the phase of the simple harmonic oscillator as well. Both the sine and the cosine functions have a maximum value of 1 which is why it corresponds to the maximum value of velocity and acceleration.

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