A body executing simple harmonic motion has a maximum acceleration equal to `24 metres// sec ^(2)` metres and maximum velocity equal to 16 metres/ sec . The amplitude of the simple harmonic motion is

Options

A. `32/3` metres
B. `3/32 ` metres
C. `1024/ 9` metres
D. `64/ 9` metres

Answer

To solve the problem, we need to find the amplitude of a body executing simple harmonic motion (SHM) given its maximum acceleration and maximum velocity.

### Step-by-Step Solution:

1. **Identify the given values:**
- Maximum acceleration, \( A_{max} = 24 \, \text{m/s}^2 \)
- Maximum velocity, \( V_{max} = 16 \, \text{m/s} \)

2. **Use the formulas for maximum velocity and maximum acceleration in SHM:**
- The maximum velocity in SHM is given by:
\[ V_{max} = A \cdot \omega \]
- The maximum acceleration in SHM is given by:
\[ A_{max} = A \cdot \omega^2 \]

3. **Set up the equations:**
- From the maximum velocity equation:
\[ V_{max} = A \cdot \omega \quad \text{(1)} \]
- From the maximum acceleration equation:
\[ A_{max} = A \cdot \omega^2 \quad \text{(2)} \]

4. **Divide equation (2) by equation (1):**
\[ \frac{A_{max}}{V_{max}} = \frac{A \cdot \omega^2}{A \cdot \omega} \]
This simplifies to:
\[ \frac{A_{max}}{V_{max}} = \omega \]

5. **Substitute the known values:**
\[ \frac{24 \, \text{m/s}^2}{16 \, \text{m/s}} = \omega \]
Simplifying this gives:
\[ \omega = \frac{24}{16} = \frac{3}{2} \, \text{rad/s} \]

6. **Substitute \( \omega \) back into equation (1) to find \( A \):**
\[ 16 = A \cdot \frac{3}{2} \]
Rearranging gives:
\[ A = \frac{16}{\frac{3}{2}} = 16 \cdot \frac{2}{3} = \frac{32}{3} \, \text{m} \]

7. **Final answer:**
The amplitude \( A \) of the simple harmonic motion is:
\[ A = \frac{32}{3} \, \text{m} \]