The amplitude of a particle executing SHM is 4 cm. 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 of the particle becomes `8sqrt(3)` cm/sec will be

Options

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

Explanation

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

### Step 1: Identify the given values
- Amplitude (A) = 4 cm
- Speed at mean position (v_max) = 16 cm/s
- Speed at the position we want to find (v) = 8√3 cm/s

### Step 2: Relate speed and angular frequency
In SHM, the maximum speed (v_max) is given by the formula:
\[ v_{max} = \omega A \]
Where:
- \( \omega \) is the angular frequency.

Substituting the known values:
\[ 16 = \omega \times 4 \]

### Step 3: Solve for angular frequency (ω)
Rearranging the equation gives:
\[ \omega = \frac{16}{4} = 4 \, \text{rad/s} \]

### Step 4: Use the speed formula in SHM
The speed (v) at a distance (x) from the mean position is given by:
\[ v = \omega \sqrt{A^2 - x^2} \]

Substituting the known values:
\[ 8\sqrt{3} = 4 \sqrt{4^2 - x^2} \]

### Step 5: Simplify the equation
First, divide both sides by 4:
\[ 2\sqrt{3} = \sqrt{16 - x^2} \]

### Step 6: Square both sides to eliminate the square root
Squaring both sides gives:
\[ (2\sqrt{3})^2 = 16 - x^2 \]
\[ 4 \times 3 = 16 - x^2 \]
\[ 12 = 16 - x^2 \]

### Step 7: Rearrange to find x²
Rearranging gives:
\[ x^2 = 16 - 12 \]
\[ x^2 = 4 \]

### Step 8: Solve for x
Taking the square root of both sides:
\[ x = \pm 2 \, \text{cm} \]

### Step 9: Determine the distance from the mean position
Since we are looking for the distance from the mean position, we take the absolute value:
\[ \text{Distance from mean position} = 2 \, \text{cm} \]

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

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