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 concepts of Simple Harmonic Motion (SHM) and the formulas related to it.

### Step 1: Understand the Given Information
We are given:
- Amplitude (A) = 4 cm
- Speed at mean position (v) = 16 cm/s
- Speed at a certain position (v') = 8√3 cm/s

### Step 2: Relate the Speed at Mean Position to Angular Frequency
At the mean position (x = 0), the speed is given by the formula:
\[ v = A \omega \]
Where:
- \( v \) is the speed at the mean position,
- \( A \) is the amplitude,
- \( \omega \) is the angular frequency.

Substituting the known values:
\[ 16 = 4 \omega \]
From this, we can solve for \( \omega \):
\[ \omega = \frac{16}{4} = 4 \, \text{rad/s} \]

### Step 3: Use the Velocity Formula for SHM
The velocity at a distance \( x \) from the mean position is given by:
\[ v' = \omega \sqrt{A^2 - x^2} \]
We need to find \( x \) when \( v' = 8\sqrt{3} \) cm/s.

### Step 4: Substitute Known Values into the Velocity Formula
Substituting \( v' = 8\sqrt{3} \), \( \omega = 4 \), and \( A = 4 \) into the formula:
\[ 8\sqrt{3} = 4 \sqrt{4^2 - x^2} \]

### Step 5: Simplify the Equation
Dividing 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:
\[ (2\sqrt{3})^2 = 16 - x^2 \]
\[ 12 = 16 - x^2 \]

### Step 7: Solve for \( x^2 \)
Rearranging the equation:
\[ x^2 = 16 - 12 \]
\[ x^2 = 4 \]

### Step 8: Find \( x \)
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 positive value:
\[ \text{Distance} = 2 \, \text{cm} \]

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

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