A particle executing SHM has a maximum speed of 30 cm `s^(-1)` and a maximum acceleration of 60 cm `s^(-1)`. The period of oscillation is

Options

A. `pis`
B. `(pi)/(2)s`
C. `2pis`
D. `(pi)/(t)s`

Explanation

To find the period of oscillation for a particle executing simple harmonic motion (SHM) given its maximum speed and maximum acceleration, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the given values**:
- Maximum speed \( V_{\text{max}} = 30 \, \text{cm/s} \)
- Maximum acceleration \( A_{\text{max}} = 60 \, \text{cm/s}^2 \)

2. **Use the formulas for maximum speed and maximum acceleration**:
- The maximum speed in SHM is given by:
\[ V_{\text{max}} = A \omega \]
- The maximum acceleration in SHM is given by:
\[ A_{\text{max}} = A \omega^2 \]

3. **Divide the equations**:
- To eliminate \( A \), divide the equation for maximum speed by the equation for maximum acceleration:
\[ \frac{V_{\text{max}}}{A_{\text{max}}} = \frac{A \omega}{A \omega^2} = \frac{1}{\omega} \]
- Substituting the known values:
\[ \frac{30 \, \text{cm/s}}{60 \, \text{cm/s}^2} = \frac{1}{\omega} \]

4. **Calculate \( \omega \)**:
- Simplifying the left side:
\[ \frac{30}{60} = \frac{1}{2} \]
- Therefore:
\[ \frac{1}{\omega} = \frac{1}{2} \implies \omega = 2 \, \text{rad/s} \]

5. **Relate angular frequency to the period**:
- The relationship between angular frequency \( \omega \) and the period \( T \) is given by:
\[ \omega = \frac{2\pi}{T} \]
- Rearranging this gives:
\[ T = \frac{2\pi}{\omega} \]

6. **Substitute \( \omega \) into the period formula**:
- Now substituting \( \omega = 2 \, \text{rad/s} \):
\[ T = \frac{2\pi}{2} = \pi \, \text{s} \]

7. **Final answer**:
- The period of oscillation is:
\[ T = \pi \, \text{s} \]

### Summary:
The period of oscillation for the particle executing SHM is \( \pi \) seconds.