The amplitude of a executing `SHM` is `4cm` At the mean position the speed of the particle is `16 cm//s` The distance of the particle from the mean position at which the speed the particle becomes `8 sqrt(3)cm//s` will be

Options

A. `2sqrt(3) cm`
B. `sqrt(3) cm`
C. `1 cm`
D. `2 cm`

Explanation

To solve the problem step by step, we will use the principles of Simple Harmonic Motion (SHM).

### Given:
- Amplitude (A) = 4 cm
- Speed at mean position (V_max) = 16 cm/s
- Speed at distance y from mean position (V) = 8√3 cm/s

### Step 1: Find the angular frequency (ω)
The maximum speed (V_max) in SHM is given by the formula:
\[ V_{max} = A \cdot \omega \]
Substituting the known values:
\[ 16 = 4 \cdot \omega \]
To find ω, rearrange the equation:
\[ \omega = \frac{16}{4} = 4 \, \text{rad/s} \]

### Step 2: Use the formula for speed at a distance y from the mean position
The speed (V) at a distance y from the mean position in SHM is given by:
\[ V = \omega \sqrt{A^2 - y^2} \]
Substituting the known values:
\[ 8\sqrt{3} = 4 \sqrt{4^2 - y^2} \]

### Step 3: Simplify the equation
Square both sides to eliminate the square root:
\[ (8\sqrt{3})^2 = (4^2 - y^2) \cdot 4^2 \]
This simplifies to:
\[ 192 = 16(16 - y^2) \]
Divide both sides by 16:
\[ 12 = 16 - y^2 \]

### Step 4: Solve for y^2
Rearranging gives:
\[ y^2 = 16 - 12 \]
\[ y^2 = 4 \]

### Step 5: Find y
Taking the square root of both sides:
\[ y = \sqrt{4} = 2 \, \text{cm} \]

### Final Answer:
The distance of the particle from the mean position at which the speed becomes \( 8\sqrt{3} \, \text{cm/s} \) is **2 cm**.

---