If displacement `x` and velocity `v` related as
`4v^(2) = 25 - x^(2)m` in a `SHM` Then time period of given `SHM` is (consider SI unit)

Options

A. `pi`
B. `2pi`
C. `4pi`
D. `6pi`

Explanation

To find the time period of the simple harmonic motion (SHM) given the relationship between displacement \( x \) and velocity \( v \) as \( 4v^2 = 25 - x^2 \), we can follow these steps:

### Step 1: Differentiate the given equation
We start with the equation:
\[ 4v^2 = 25 - x^2 \]
Differentiating both sides with respect to time \( t \):
\[ \frac{d}{dt}(4v^2) = \frac{d}{dt}(25 - x^2) \]
This gives us:
\[ 4 \cdot 2v \frac{dv}{dt} = 0 - 2x \frac{dx}{dt} \]
Simplifying, we have:
\[ 8v \frac{dv}{dt} = -2x \frac{dx}{dt} \]

### Step 2: Substitute acceleration and velocity
We know that:
- \( \frac{dv}{dt} \) is the acceleration \( a \)
- \( \frac{dx}{dt} \) is the velocity \( v \)

Substituting these into the equation gives:
\[ 8v a = -2x v \]
If we divide both sides by \( v \) (assuming \( v \neq 0 \)):
\[ 8a = -2x \]

### Step 3: Express acceleration in terms of displacement
Rearranging the equation:
\[ a = -\frac{2}{8} x = -\frac{1}{4} x \]
This shows that the acceleration is proportional to the displacement \( x \) and is directed opposite to it, which is characteristic of SHM.

### Step 4: Relate to the standard SHM equation
In SHM, the acceleration can also be expressed as:
\[ a = -\omega^2 x \]
Comparing the two expressions for acceleration:
\[ -\frac{1}{4} x = -\omega^2 x \]
This implies:
\[ \omega^2 = \frac{1}{4} \]
Taking the square root gives:
\[ \omega = \frac{1}{2} \]

### Step 5: Calculate the time period
The time period \( T \) of SHM is given by the formula:
\[ T = \frac{2\pi}{\omega} \]
Substituting the value of \( \omega \):
\[ T = \frac{2\pi}{\frac{1}{2}} = 4\pi \]

### Conclusion
Thus, the time period of the given SHM is:
\[ \boxed{4\pi} \]