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

Options

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

Answer

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

### Step-by-Step Solution:

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

2. **Express Amplitude in Terms of Given Quantities**:
- From the maximum velocity equation, we can express the amplitude \( A \):
\[ A = \frac{v_{\text{max}}}{\omega} = \frac{\beta}{\omega} \]

3. **Substitute Amplitude into the Acceleration Equation**:
- Substitute \( A \) into the maximum acceleration equation:
\[ a_{\text{max}} = \left(\frac{\beta}{\omega}\right) \omega^2 = \beta \omega \]
- Since \( a_{\text{max}} = \alpha \), we can set up the equation:
\[ \alpha = \beta \omega \]

4. **Solve for Angular Frequency \( \omega \)**:
- Rearranging the equation gives:
\[ \omega = \frac{\alpha}{\beta} \]

5. **Find the Time Period \( T \)**:
- The time period \( T \) is related to the angular frequency \( \omega \) by the formula:
\[ T = \frac{2\pi}{\omega} \]
- Substitute \( \omega \) into this equation:
\[ T = \frac{2\pi}{\frac{\alpha}{\beta}} = \frac{2\pi \beta}{\alpha} \]

6. **Final Answer**:
- Therefore, the time period of vibration is:
\[ T = \frac{2\pi \beta}{\alpha} \]