A particle executes simple harmonic motion with an amplitude of `4cm` At the mean position the velocity of tge particle is `10` earth distance of the particle from the mean position when its speed `5` point is

Options

A. `sqrt(3)cm`
B. `sqrt(5)cm`
C. `21(sqrt(3))cm`
D. `21(sqrt(5))cm`

Explanation

To solve the problem step by step, we will follow the principles of simple harmonic motion (SHM).

### Step 1: Identify the given values
- Amplitude (A) = 4 cm
- Maximum velocity (V_max) = 10 cm/s
- Speed at a certain position (V) = 5 cm/s

### Step 2: Relate maximum velocity to angular frequency
The maximum velocity in SHM is given by the formula:
\[ V_{max} = A \cdot \omega \]
Where:
- \( \omega \) is the angular frequency.

Substituting the known values:
\[ 10 = 4 \cdot \omega \]

### Step 3: Solve for angular frequency (ω)
Rearranging the equation to find \( \omega \):
\[ \omega = \frac{10}{4} = 2.5 \, \text{rad/s} \]

### Step 4: Use the formula for velocity in SHM
The velocity at any position \( x \) in SHM is given by:
\[ V = A \cdot \omega \cdot \cos(\omega t) \]

### Step 5: Substitute the known values into the velocity equation
We know:
- \( V = 5 \, \text{cm/s} \)
- \( A = 4 \, \text{cm} \)
- \( \omega = 2.5 \, \text{rad/s} \)

Substituting these values into the velocity equation:
\[ 5 = 4 \cdot 2.5 \cdot \cos(\omega t) \]

### Step 6: Simplify the equation
Calculating \( 4 \cdot 2.5 \):
\[ 5 = 10 \cdot \cos(\omega t) \]

### Step 7: Solve for cos(ωt)
Rearranging gives:
\[ \cos(\omega t) = \frac{5}{10} = \frac{1}{2} \]

### Step 8: Find the angle ωt
The cosine of \( \frac{1}{2} \) corresponds to:
\[ \omega t = \frac{\pi}{3} \, \text{or} \, \omega t = \frac{5\pi}{3} \]

### Step 9: Calculate the distance from the mean position
The position \( x \) in SHM is given by:
\[ x = A \cdot \sin(\omega t) \]

Using \( A = 4 \, \text{cm} \) and \( \omega t = \frac{\pi}{3} \):
\[ x = 4 \cdot \sin\left(\frac{\pi}{3}\right) \]

### Step 10: Calculate sin(π/3)
We know:
\[ \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2} \]

Substituting this into the equation:
\[ x = 4 \cdot \frac{\sqrt{3}}{2} = 2\sqrt{3} \, \text{cm} \]

### Final Answer
The distance of the particle from the mean position when its speed is 5 cm/s is \( 2\sqrt{3} \, \text{cm} \) (approximately 3.46 cm).

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