If the maximum velocity and acceleration of a particle executing SHM are equal in magnitude, the time period will be

Options

A. 1.57
B. 3.17
C. 6.28
D. 12.56

Explanation

To solve the problem, we need to find the time period of a particle executing simple harmonic motion (SHM) when its maximum velocity and maximum acceleration are equal in magnitude.

### 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:
\[ V_{\text{max}} = A \omega \]
where \( A \) is the amplitude and \( \omega \) is the angular frequency.

- The maximum acceleration \( a_{\text{max}} \) is given by:
\[ a_{\text{max}} = A \omega^2 \]

2. **Set the magnitudes of maximum velocity and acceleration equal**:
- According to the problem, we have:
\[ V_{\text{max}} = a_{\text{max}} \]
- Substituting the formulas, we get:
\[ A \omega = A \omega^2 \]

3. **Simplify the equation**:
- Since \( A \) (amplitude) is non-zero, we can divide both sides by \( A \):
\[ \omega = \omega^2 \]

4. **Rearrange the equation**:
- Rearranging gives:
\[ \omega^2 - \omega = 0 \]
- Factoring out \( \omega \):
\[ \omega(\omega - 1) = 0 \]

5. **Find the solutions for \( \omega \)**:
- This gives us two solutions:
\[ \omega = 0 \quad \text{or} \quad \omega = 1 \]
- Since \( \omega = 0 \) does not represent a physical situation in SHM, we take \( \omega = 1 \).

6. **Calculate the time period**:
- The time period \( T \) of SHM is given by:
\[ T = \frac{2\pi}{\omega} \]
- Substituting \( \omega = 1 \):
\[ T = \frac{2\pi}{1} = 2\pi \]

7. **Convert to decimal**:
- The value of \( 2\pi \) is approximately:
\[ T \approx 6.28 \text{ seconds} \]

### Conclusion:
The time period when the maximum velocity and acceleration of a particle executing SHM are equal in magnitude is \( 6.28 \) seconds.