A particle executes simple harmonic motion with an amplitude of 10 cm. At what distance from the mean position are the kinetic and potential energies equal?

Explanation

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

### Step-by-Step Solution:

1. **Understand the formulas for kinetic and potential energy in SHM**:
- The potential energy (PE) of a particle in SHM is given by:
\[ PE = \frac{1}{2} m \omega^2 x^2 \]
- The kinetic energy (KE) of a particle in SHM is given by:
\[ KE = \frac{1}{2} m \omega^2 (A^2 - x^2) \]
where \( A \) is the amplitude, \( x \) is the displacement from the mean position, and \( m \) is the mass of the particle.

2. **Set the kinetic energy equal to the potential energy**:
\[ \frac{1}{2} m \omega^2 x^2 = \frac{1}{2} m \omega^2 (A^2 - x^2) \]

3. **Cancel out common terms**:
Since \( \frac{1}{2} m \omega^2 \) is common on both sides, we can cancel it out:
\[ x^2 = A^2 - x^2 \]

4. **Rearrange the equation**:
\[ x^2 + x^2 = A^2 \]
\[ 2x^2 = A^2 \]

5. **Solve for \( x \)**:
\[ x^2 = \frac{A^2}{2} \]
\[ x = \sqrt{\frac{A^2}{2}} = \frac{A}{\sqrt{2}} \]

6. **Substitute the given amplitude**:
Given that the amplitude \( A = 10 \) cm:
\[ x = \frac{10}{\sqrt{2}} = 5\sqrt{2} \text{ cm} \]

7. **Conclusion**:
The distance from the mean position where the kinetic and potential energies are equal is:
\[ x = 5\sqrt{2} \text{ cm} \]

### Final Answer:
The distance from the mean position where the kinetic and potential energies are equal is \( 5\sqrt{2} \) cm.

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