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

Options

A. `sqrt(3)` cm
B. `sqrt5` cm
C. `2(sqrt3)` cm
D. `2(sqrt5)` cm

Answer

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

### Given Data:
- Amplitude (A) = 4 cm
- Maximum Velocity (v_max) = 10 cm/s
- Speed at which we need to find the distance (v) = 5 cm/s

### Step 1: Find the Angular Frequency (ω)
The maximum velocity in SHM is given by the formula:
\[ v_{max} = \omega A \]

We can rearrange this to find the angular frequency (ω):
\[ \omega = \frac{v_{max}}{A} \]

Substituting the known values:
\[ \omega = \frac{10 \, \text{cm/s}}{4 \, \text{cm}} = 2.5 \, \text{rad/s} \]

### Step 2: Use the Velocity-Displacement Relationship
The velocity of a particle in SHM can also be expressed in terms of its displacement (y) from the mean position:
\[ v = \omega \sqrt{A^2 - y^2} \]

### Step 3: Rearranging the Equation
We need to find y when v = 5 cm/s. Rearranging the equation gives:
\[ v^2 = \omega^2 (A^2 - y^2) \]

### Step 4: Substitute Known Values
Substituting the known values into the rearranged equation:
\[ (5 \, \text{cm/s})^2 = (2.5 \, \text{rad/s})^2 (4^2 - y^2) \]

Calculating the left side:
\[ 25 = (6.25)(16 - y^2) \]

### Step 5: Solve for y^2
Expanding the equation:
\[ 25 = 100 - 6.25y^2 \]

Rearranging gives:
\[ 6.25y^2 = 100 - 25 \]
\[ 6.25y^2 = 75 \]

Dividing both sides by 6.25:
\[ y^2 = \frac{75}{6.25} = 12 \]

### Step 6: Find y
Taking the square root gives:
\[ y = \sqrt{12} = 2\sqrt{3} \, \text{cm} \]

### Conclusion
The distance of the particle from the mean position when its speed becomes 5 cm/s is:
\[ y = 2\sqrt{3} \, \text{cm} \]