THE magnitude of maximum accesleration is `pi` times that of maximum velocity of a simble harmonic oscillator The time period of the oscillator. The time period of the oscillator in second is,

Options

A. 4
B. 2
C. 1
D. 0.5

Answer

To solve the problem, we need to find the time period of a simple harmonic oscillator (SHO) given that the magnitude of maximum acceleration is π times that of maximum velocity.

### Step-by-Step Solution:

1. **Understand Maximum Velocity and Acceleration in SHM:**
- The maximum velocity \( V_{\text{max}} \) of a simple harmonic oscillator 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:
\[ a_{\text{max}} = \omega^2 A \]

2. **Set Up the Relationship Given in the Problem:**
- According to the problem, the maximum acceleration is π times the maximum velocity:
\[ a_{\text{max}} = \pi V_{\text{max}} \]

3. **Substitute the Expressions for Maximum Velocity and Acceleration:**
- Substitute the expressions we have for \( a_{\text{max}} \) and \( V_{\text{max}} \):
\[ \omega^2 A = \pi (A \omega) \]

4. **Simplify the Equation:**
- We can cancel \( A \) from both sides (assuming \( A \neq 0 \)):
\[ \omega^2 = \pi \omega \]

5. **Rearranging the Equation:**
- Rearranging gives:
\[ \omega^2 - \pi \omega = 0 \]
- Factor out \( \omega \):
\[ \omega(\omega - \pi) = 0 \]

6. **Solve for \( \omega \):**
- This gives us two solutions:
- \( \omega = 0 \) (not physically meaningful for an oscillator)
- \( \omega = \pi \)

7. **Find the Time Period:**
- The time period \( T \) of the oscillator is related to the angular frequency \( \omega \) by:
\[ T = \frac{2\pi}{\omega} \]
- Substituting \( \omega = \pi \):
\[ T = \frac{2\pi}{\pi} = 2 \text{ seconds} \]

### Conclusion:
The time period of the oscillator is **2 seconds**.