A particle executes simple harmonic motion of ampliltude A. At what distance from the mean position is its kinetic energy equal to its potential energy?

Options

A. 0.51 A
B. 0.61 A
C. 0.71 A
D. 0.81 A

Explanation

To solve the problem of finding the distance from the mean position at which the kinetic energy (KE) is equal to the potential energy (PE) for a particle executing simple harmonic motion (SHM) with amplitude A, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Energies in SHM**:
- The total mechanical energy (E) in SHM is given by the sum of kinetic energy (KE) and potential energy (PE):
\[ E = KE + PE \]
- The expressions for KE and PE are:
\[ KE = \frac{1}{2} mv^2 \]
\[ PE = \frac{1}{2} kx^2 \]

2. **Setting KE Equal to PE**:
- We need to find the position \( x \) where:
\[ KE = PE \]
- This implies:
\[ \frac{1}{2} mv^2 = \frac{1}{2} kx^2 \]
- We can cancel the \( \frac{1}{2} \) from both sides:
\[ mv^2 = kx^2 \]

3. **Relating Velocity to Position**:
- The velocity \( v \) of a particle in SHM can be expressed as:
\[ v = \omega \sqrt{A^2 - x^2} \]
- Substituting this into the equation gives:
\[ m(\omega^2 (A^2 - x^2)) = kx^2 \]

4. **Using the Relationship Between k and m**:
- We know that \( k = m\omega^2 \). Substituting this into the equation:
\[ m\omega^2 (A^2 - x^2) = m\omega^2 x^2 \]
- We can cancel \( m\omega^2 \) from both sides (assuming \( m \neq 0 \) and \( \omega \neq 0 \)):
\[ A^2 - x^2 = x^2 \]

5. **Rearranging the Equation**:
- Rearranging gives:
\[ A^2 = 2x^2 \]
- Dividing both sides by 2:
\[ x^2 = \frac{A^2}{2} \]

6. **Finding the Distance x**:
- Taking the square root gives:
\[ x = \frac{A}{\sqrt{2}} \quad \text{or} \quad x = \pm \frac{A}{\sqrt{2}} \]

7. **Final Answer**:
- The distance from the mean position where the kinetic energy equals the potential energy is:
\[ x = \frac{A}{\sqrt{2}} \quad \text{(approximately } 0.707A\text{)} \]