A particle executing simple harmonic motion has an amplitude of 6 cm . Its acceleration at a distance of 2 cm from the mean position is `8 cm/s^(2)` The maximum speed of the particle is

Options

A. 8 cm/s
B. 12 cm /s
C. 16 cm/s
D. 24 cm/s

Answer

To find the maximum speed of a particle executing simple harmonic motion (SHM), we can use the relationship between acceleration, displacement, and maximum speed.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Amplitude (A) = 6 cm
- Displacement from mean position (x) = 2 cm
- Acceleration (a) at x = 2 cm = 8 cm/s²

2. **Use the formula for acceleration in SHM:**
The acceleration (a) of a particle in SHM is given by the formula:
\[ a = -\omega^2 x \]
where:
- \( \omega \) is the angular frequency,
- \( x \) is the displacement from the mean position.

3. **Rearranging the formula to find \( \omega \):**
Since we know \( a \) and \( x \), we can rearrange the formula to solve for \( \omega \):
\[ \omega^2 = \frac{-a}{x} \]
Substituting the known values:
\[ 8 = \omega^2 \cdot 2 \]
\[ \omega^2 = \frac{8}{2} = 4 \]
\[ \omega = \sqrt{4} = 2 \, \text{rad/s} \]

4. **Calculate Maximum Speed (V_max):**
The maximum speed (V_max) in SHM is given by the formula:
\[ V_{max} = A \cdot \omega \]
Substituting the values for amplitude and angular frequency:
\[ V_{max} = 6 \, \text{cm} \cdot 2 \, \text{rad/s} = 12 \, \text{cm/s} \]

5. **Final Answer:**
The maximum speed of the particle is:
\[ V_{max} = 12 \, \text{cm/s} \]