A ball hung from a vertical spring oscillates in simple harmonic motion with an angular frequency of 2.6 rad/s and an amplitude of 0.075m. What is the maximum acceleration of the ball?

Options

A. `0.13m//s^(2)`
B. `0.20m//s^(2)`
C. `0.51m//s^(2)`
D. `2.6m//s^(2)`

Answer

To find the maximum acceleration of a ball oscillating in simple harmonic motion, we can use the following steps:

### Step-by-Step Solution:

1. **Understand the Formula for Maximum Acceleration**:
The maximum acceleration \( a_{max} \) in simple harmonic motion (SHM) is given by the formula:
\[ a_{max} = \omega^2 A \]
where \( \omega \) is the angular frequency and \( A \) is the amplitude of the motion.

2. **Identify Given Values**:
From the problem, we have:
- Angular frequency \( \omega = 2.6 \, \text{rad/s} \)
- Amplitude \( A = 0.075 \, \text{m} \)

3. **Substitute the Values into the Formula**:
Now, substituting the values into the formula for maximum acceleration:
\[ a_{max} = (2.6)^2 \times 0.075 \]

4. **Calculate \( \omega^2 \)**:
First, calculate \( \omega^2 \):
\[ \omega^2 = 2.6 \times 2.6 = 6.76 \, \text{rad}^2/\text{s}^2 \]

5. **Calculate Maximum Acceleration**:
Now substitute \( \omega^2 \) back into the equation:
\[ a_{max} = 6.76 \times 0.075 \]
\[ a_{max} = 0.507 \, \text{m/s}^2 \]

6. **Final Result**:
The maximum acceleration of the ball is:
\[ a_{max} = 0.507 \, \text{m/s}^2 \]

### Conclusion:
Thus, the maximum acceleration of the ball is \( 0.507 \, \text{m/s}^2 \).

---