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

Options

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

Explanation

To solve the problem, we need to find the period of oscillation for a particle executing simple harmonic motion (SHM) given its maximum speed and maximum acceleration.

### 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 Relationship for Maximum Speed:**
The maximum speed in SHM is given by the formula:
\[ V_{\text{max}} = A \omega \]
where \( A \) is the amplitude and \( \omega \) is the angular frequency.

3. **Use the Relationship for Maximum Acceleration:**
The maximum acceleration in SHM is given by the formula:
\[ a_{\text{max}} = A \omega^2 \]

4. **Set Up the Equations:**
From the maximum speed equation:
\[ A \omega = 30 \, \text{(Equation 1)} \]
From the maximum acceleration equation:
\[ A \omega^2 = 60 \, \text{(Equation 2)} \]

5. **Divide Equation 2 by Equation 1:**
To eliminate \( A \), divide Equation 2 by Equation 1:
\[ \frac{A \omega^2}{A \omega} = \frac{60}{30} \]
Simplifying gives:
\[ \omega = \frac{60}{30} = 2 \, \text{radians/second} \]

6. **Calculate the Period of Oscillation:**
The period \( T \) of SHM is related to the angular frequency \( \omega \) by the formula:
\[ T = \frac{2\pi}{\omega} \]
Substituting \( \omega = 2 \):
\[ T = \frac{2\pi}{2} = \pi \, \text{seconds} \]

7. **Final Answer:**
The period of oscillation is \( T = \pi \, \text{seconds} \).

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