For a particle in `S.H.M.` if the maximum acceleration is `a` and maximum velocity is `v` then amplitude is

Options

A. `va`
B. `(v^(2))/(a)`
C. `(v^(2))/(2a)`
D. `(v)/(a)`

Explanation

To find the amplitude of a particle in simple harmonic motion (SHM) given the maximum acceleration \( a \) and maximum velocity \( v \), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the formulas for maximum velocity and acceleration in SHM:**
- The maximum velocity \( V_{\text{max}} \) of a particle in SHM is given by:
\[ V_{\text{max}} = \omega A \]
- The maximum acceleration \( A_{\text{max}} \) is given by:
\[ A_{\text{max}} = \omega^2 A \]

2. **Set the maximum acceleration and maximum velocity equal to the given values:**
- We know from the problem that:
\[ A_{\text{max}} = a \quad \text{and} \quad V_{\text{max}} = v \]

3. **Express \( \omega \) in terms of \( A \):**
- From the equation for maximum velocity:
\[ v = \omega A \quad \Rightarrow \quad \omega = \frac{v}{A} \quad \text{(Equation 1)} \]

4. **Substitute \( \omega \) into the equation for maximum acceleration:**
- Substitute \( \omega \) from Equation 1 into the equation for maximum acceleration:
\[ a = \omega^2 A \]
- This becomes:
\[ a = \left(\frac{v}{A}\right)^2 A \]

5. **Simplify the equation:**
- Rearranging gives:
\[ a = \frac{v^2}{A} \]

6. **Solve for amplitude \( A \):**
- Rearranging the equation to solve for \( A \):
\[ A = \frac{v^2}{a} \]

### Final Answer:
The amplitude \( A \) is given by:
\[ A = \frac{v^2}{a} \]