A particle executes SHM on a line 8 cm long . Its KE and PE will be equal when its distance from the mean position is

Options

A. 4 cm
B. 2 cm
C. `2sqrt(2)`
D. `sqrt(2)` cm

Explanation

To solve the problem of finding the distance from the mean position where the kinetic energy (KE) and potential energy (PE) of a particle executing simple harmonic motion (SHM) are equal, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Amplitude**:
The total length of the line is given as 8 cm. In SHM, the amplitude (A) is half of the total length.
\[ A = \frac{8 \, \text{cm}}{2} = 4 \, \text{cm} \]

2. **Write the Formulas for KE and PE**:
The kinetic energy (KE) and potential energy (PE) in SHM are given by the following formulas:
- Kinetic Energy:
\[ KE = \frac{1}{2} k (A^2 - x^2) \]
- Potential Energy:
\[ PE = \frac{1}{2} k x^2 \]

3. **Set KE Equal to PE**:
To find the position where KE equals PE, we set the two equations equal to each other:
\[ \frac{1}{2} k (A^2 - x^2) = \frac{1}{2} k x^2 \]

4. **Cancel Common Terms**:
Since \(\frac{1}{2} k\) is common on both sides, we can cancel it out:
\[ A^2 - x^2 = x^2 \]

5. **Rearrange the Equation**:
Rearranging gives us:
\[ A^2 = 2x^2 \]

6. **Solve for x**:
Now, we can solve for \(x\):
\[ x^2 = \frac{A^2}{2} \]
Taking the square root:
\[ x = \pm \frac{A}{\sqrt{2}} \]

7. **Substitute the Amplitude**:
Substitute \(A = 4 \, \text{cm}\):
\[ x = \pm \frac{4 \, \text{cm}}{\sqrt{2}} = \pm 2\sqrt{2} \, \text{cm} \]

8. **Determine the Distance**:
Since we are asked for the distance from the mean position, we take the positive value:
\[ x = 2\sqrt{2} \, \text{cm} \]

### Final Answer:
The distance from the mean position at which the kinetic energy and potential energy are equal is \(2\sqrt{2} \, \text{cm}\).