The amplitude of a particle executing S.H.M. with frequency of 60 Hz is 0.01 m . The maximum value of the acceleration of the particle is

Options

A. `144pi^(2) m// sec^(2)`
B. `144 m// sec^(2)`
C. `144/(pi^(2)) m// sec^(2)`
D. `288pi^(2) m// sec^(2)`

Answer

To find the maximum value of the acceleration of a particle executing Simple Harmonic Motion (S.H.M.), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values**:
- Frequency (f) = 60 Hz
- Amplitude (A) = 0.01 m

2. **Understand the Relationship**:
- The maximum acceleration (a_max) in S.H.M. can be calculated using the formula:
\[ a_{\text{max}} = A \cdot \omega^2 \]
- Where \( \omega \) is the angular frequency.

3. **Relate Frequency to Angular Frequency**:
- The angular frequency \( \omega \) is related to the frequency \( f \) by the formula:
\[ \omega = 2\pi f \]

4. **Substitute the Frequency into the Angular Frequency Formula**:
- Substitute \( f = 60 \, \text{Hz} \):
\[ \omega = 2\pi \cdot 60 \]
\[ \omega = 120\pi \, \text{rad/s} \]

5. **Calculate \( \omega^2 \)**:
- Now, calculate \( \omega^2 \):
\[ \omega^2 = (120\pi)^2 = 14400\pi^2 \]

6. **Substitute \( A \) and \( \omega^2 \) into the Maximum Acceleration Formula**:
- Now, substitute \( A = 0.01 \, \text{m} \) and \( \omega^2 = 14400\pi^2 \):
\[ a_{\text{max}} = 0.01 \cdot 14400\pi^2 \]

7. **Simplify the Expression**:
- Calculate the maximum acceleration:
\[ a_{\text{max}} = 144 \pi^2 \]

8. **Final Result**:
- The maximum value of the acceleration of the particle is:
\[ a_{\text{max}} = 144\pi^2 \, \text{m/s}^2 \]