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 cms^(-1)`
B. `12 cms^(-1)`
C. `16 cms^(-1)`
D. `24 cms^(-1)`

Explanation

To solve the problem step by step, we will use the formulas related to simple harmonic motion (SHM).

### Step 1: Understand the given information
- Amplitude (A) = 6 cm
- Distance from the mean position (x) = 2 cm
- Acceleration (a) at x = 2 cm = 8 cm/s²

### Step 2: Use the formula for acceleration in SHM
The acceleration \( a \) of a particle in simple harmonic motion is given by the formula:
\[ a = -\omega^2 x \]
For the magnitude, we can write:
\[ a = \omega^2 x \]

### Step 3: Substitute the known values into the acceleration formula
We know that at \( x = 2 \) cm, the acceleration \( a = 8 \) cm/s². Thus, we can write:
\[ 8 = \omega^2 \cdot 2 \]

### Step 4: Solve for \( \omega^2 \)
Rearranging the equation gives:
\[ \omega^2 = \frac{8}{2} = 4 \]

### Step 5: Find \( \omega \)
Taking the square root of both sides:
\[ \omega = \sqrt{4} = 2 \text{ rad/s} \]

### Step 6: Use the formula for maximum speed in SHM
The maximum speed \( V_{max} \) in simple harmonic motion is given by the formula:
\[ V_{max} = A \cdot \omega \]
where \( A \) is the amplitude.

### Step 7: Substitute the values of \( A \) and \( \omega \)
Substituting the values we have:
\[ V_{max} = 6 \cdot 2 = 12 \text{ cm/s} \]

### Step 8: Conclusion
Thus, the maximum speed of the particle is:
\[ \boxed{12 \text{ cm/s}} \]