A particle executing S.H.M. has a maximum speed of 30`cm//s` and a maximum acceleration of `60cm//s^(2)`. The period of oscillation is

Options

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

Explanation

To solve the problem, we need to find the period of oscillation of a particle executing Simple Harmonic Motion (S.H.M.) given its maximum speed and maximum acceleration.

### Step-by-Step Solution:

1. **Identify the given values**:
- Maximum speed (\(V_{max}\)) = 30 cm/s
- Maximum acceleration (\(A_{max}\)) = 60 cm/s²

2. **Relate maximum speed and maximum acceleration to angular frequency (\(\omega\)) and amplitude (A)**:
- The maximum speed in S.H.M. is given by the formula:
\[ V_{max} = \omega A \]
- The maximum acceleration in S.H.M. is given by the formula:
\[ A_{max} = \omega^2 A \]

3. **Express amplitude (A) in terms of \(V_{max}\) and \(\omega\)**:
- From the first equation, we can express \(A\) as:
\[ A = \frac{V_{max}}{\omega} \]

4. **Substitute \(A\) into the second equation**:
- Substitute \(A\) from the previous step into the equation for maximum acceleration:
\[ A_{max} = \omega^2 \left(\frac{V_{max}}{\omega}\right) \]
- This simplifies to:
\[ A_{max} = \omega V_{max} \]

5. **Solve for \(\omega\)**:
- Rearranging the equation gives:
\[ \omega = \frac{A_{max}}{V_{max}} \]
- Substituting the given values:
\[ \omega = \frac{60 \, \text{cm/s}^2}{30 \, \text{cm/s}} = 2 \, \text{rad/s} \]

6. **Calculate the period (T)**:
- The period of oscillation is given by the formula:
\[ T = \frac{2\pi}{\omega} \]
- Substituting the value of \(\omega\):
\[ T = \frac{2\pi}{2} = \pi \, \text{seconds} \]

### Final Answer:
The period of oscillation is \(T = \pi\) seconds.