A particle is executing a simple harmonic motion its maximum acceleration is a and maximum velocity is `beta` .Then its time of vibration will be

Options

A. `(2pi beta)/(alpha)`
B. `(beta^(2))/(alpha^(2))`
C. `(alpha)/(beta)`
D. `(beta^(2))/(alpha)`

Explanation

To solve the problem, we need to find the time period of a particle executing simple harmonic motion (SHM) given its maximum acceleration \( a \) and maximum velocity \( \beta \).

### Step-by-Step Solution:

1. **Understand the Relationships**:
- The maximum velocity \( V_{\text{max}} \) in SHM is given by:
\[ V_{\text{max}} = A \omega \]
- The maximum acceleration \( a_{\text{max}} \) in SHM is given by:
\[ a_{\text{max}} = A \omega^2 \]
Here, \( A \) is the amplitude and \( \omega \) is the angular frequency.

2. **Express Amplitude in Terms of Given Quantities**:
- From the equation for maximum velocity:
\[ A = \frac{V_{\text{max}}}{\omega} = \frac{\beta}{\omega} \]
- From the equation for maximum acceleration:
\[ A = \frac{a_{\text{max}}}{\omega^2} = \frac{a}{\omega^2} \]

3. **Set the Two Expressions for Amplitude Equal**:
- Since both expressions equal \( A \), we can set them equal to each other:
\[ \frac{\beta}{\omega} = \frac{a}{\omega^2} \]

4. **Cross-Multiply to Solve for \( \omega \)**:
- Rearranging gives:
\[ \beta \omega = a \quad \Rightarrow \quad \omega = \frac{a}{\beta} \]

5. **Relate Angular Frequency to Time Period**:
- The angular frequency \( \omega \) is related to the time period \( T \) by:
\[ \omega = \frac{2\pi}{T} \]
- Substituting for \( \omega \):
\[ \frac{2\pi}{T} = \frac{a}{\beta} \]

6. **Solve for the Time Period \( T \)**:
- Rearranging gives:
\[ T = \frac{2\pi \beta}{a} \]

### Final Answer:
The time period \( T \) of the vibration is given by:
\[ T = \frac{2\pi \beta}{a} \]

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