If the displacement `x` and velocity `v` of a aprticle executing `SHM` are related as `4v^(2) = 25 - x^(2)`. Then its maximum displacement in metre `(x, v` are in `SI`) is

Options

A. `1`
B. `2`
C. `5`
D. `6`

Explanation

To solve the problem, we need to find the maximum displacement (amplitude) of a particle executing simple harmonic motion (SHM) given the relationship between its displacement \( x \) and velocity \( v \) as:

\[ 4v^2 = 25 - x^2 \]

### Step 1: Rearranging the equation
We start by rearranging the given equation to express \( v \) in terms of \( x \):

\[ 4v^2 = 25 - x^2 \]

Dividing both sides by 4, we get:

\[ v^2 = \frac{25 - x^2}{4} \]

Taking the square root of both sides gives:

\[ v = \frac{1}{2} \sqrt{25 - x^2} \]

### Step 2: Understanding the relationship in SHM
In SHM, the velocity \( v \) can also be expressed in terms of the angular frequency \( \omega \) and the amplitude \( A \):

\[ v = \omega \sqrt{A^2 - x^2} \]

### Step 3: Equating the two expressions for velocity
Now we can equate the two expressions for \( v \):

\[ \frac{1}{2} \sqrt{25 - x^2} = \omega \sqrt{A^2 - x^2} \]

### Step 4: Finding the maximum displacement
To find the maximum displacement (amplitude \( A \)), we need to find the value of \( A \) when \( x = 0 \) (the maximum displacement occurs when the particle is at the extreme position).

Substituting \( x = 0 \):

\[ v = \frac{1}{2} \sqrt{25 - 0^2} = \frac{1}{2} \sqrt{25} = \frac{1}{2} \times 5 = \frac{5}{2} \]

### Step 5: Relating \( \omega \) and \( A \)
From the previous equation, we can also express \( \omega \):

\[ \omega = \frac{1}{2} \quad \text{(from the earlier equation)} \]

Now we know that:

\[ \frac{5}{2} = \frac{1}{2} \sqrt{A^2 - 0^2} \]

This simplifies to:

\[ \frac{5}{2} = \frac{1}{2} A \]

Multiplying both sides by 2:

\[ 5 = A \]

### Conclusion
Thus, the maximum displacement (amplitude) \( A \) is:

\[ \boxed{5} \text{ meters} \]