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

Options

A. `sqrt(3)` cm
B. `sqrt(5)` cm
C. `2sqrt(3)` cm
D. `2sqrt(5)` cm

Explanation

To solve the problem step by step, we will use the formula for velocity in simple harmonic motion (SHM) and the given data.

### Step 1: Identify the given values
- Amplitude (A) = 4 cm
- Velocity at mean position (V_max) = 10 cm/s
- Speed when we need to find the distance (V) = 5 cm/s

### Step 2: Use the formula for maximum velocity
In SHM, the maximum velocity (V_max) is given by:
\[ V_{max} = \omega A \]
Where:
- \( \omega \) is the angular frequency.

Since we know \( V_{max} = 10 \) cm/s and \( A = 4 \) cm, we can find \( \omega \):
\[ 10 = \omega \times 4 \]
\[ \omega = \frac{10}{4} = 2.5 \text{ rad/s} \]

### Step 3: Use the velocity formula in SHM
The velocity at any position \( x \) in SHM is given by:
\[ V = \omega \sqrt{A^2 - x^2} \]
We need to find \( x \) when \( V = 5 \) cm/s.

### Step 4: Substitute known values into the velocity formula
Substituting \( V = 5 \) cm/s and \( \omega = 2.5 \) rad/s into the equation:
\[ 5 = 2.5 \sqrt{4^2 - x^2} \]
\[ 5 = 2.5 \sqrt{16 - x^2} \]

### Step 5: Solve for \( \sqrt{16 - x^2} \)
Dividing both sides by 2.5:
\[ \frac{5}{2.5} = \sqrt{16 - x^2} \]
\[ 2 = \sqrt{16 - x^2} \]

### Step 6: Square both sides
Now, squaring both sides to eliminate the square root:
\[ 2^2 = 16 - x^2 \]
\[ 4 = 16 - x^2 \]

### Step 7: Rearranging the equation
Rearranging gives:
\[ x^2 = 16 - 4 \]
\[ x^2 = 12 \]

### Step 8: Taking the square root
Taking the square root of both sides:
\[ x = \pm \sqrt{12} \]
\[ x = \pm 2\sqrt{3} \]

### Step 9: Determine the distance from the mean position
Since distance cannot be negative, we take the positive value:
\[ \text{Distance} = 2\sqrt{3} \text{ cm} \]

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

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