A particle executes shm of amplitude 'a'. (i) At what distance from the mean position is its kinetic energy equal to its potential energy? (ii) At what points is its speed half the maximum speed ?

Explanation

To solve the problem step by step, we will break it down into two parts as per the questions asked.

### Part (i): Finding the distance from the mean position where kinetic energy equals potential energy.

1. **Understanding the Energy in SHM**:
- In simple harmonic motion (SHM), the total mechanical energy (E) is the sum of kinetic energy (KE) and potential energy (PE).
- The total energy is given by:
\[ E = \frac{1}{2} k A^2 \]
- Where \( k \) is the spring constant and \( A \) is the amplitude.

2. **Kinetic Energy (KE)**:
- The kinetic energy at a distance \( x \) from the mean position is:
\[ KE = \frac{1}{2} m v^2 \]
- The velocity \( v \) at distance \( x \) is given by:
\[ v = \omega \sqrt{A^2 - x^2} \]
- Therefore, substituting for \( v \):
\[ KE = \frac{1}{2} m \left(\omega \sqrt{A^2 - x^2}\right)^2 = \frac{1}{2} m \omega^2 (A^2 - x^2) \]

3. **Potential Energy (PE)**:
- The potential energy at distance \( x \) is:
\[ PE = \frac{1}{2} k x^2 \]
- Using \( k = m \omega^2 \):
\[ PE = \frac{1}{2} m \omega^2 x^2 \]

4. **Setting KE equal to PE**:
- We set the expressions for KE and PE equal to each other:
\[ \frac{1}{2} m \omega^2 (A^2 - x^2) = \frac{1}{2} m \omega^2 x^2 \]
- Canceling \( \frac{1}{2} m \omega^2 \) from both sides gives:
\[ A^2 - x^2 = x^2 \]
- Rearranging leads to:
\[ A^2 = 2x^2 \implies x^2 = \frac{A^2}{2} \implies x = \frac{A}{\sqrt{2}} \]

### Part (ii): Finding the points where speed is half of the maximum speed.

1. **Maximum Speed**:
- The maximum speed \( v_{max} \) occurs at the mean position (when \( x = 0 \)):
\[ v_{max} = \omega A \]

2. **Speed at Distance \( x \)**:
- We want to find \( x \) such that:
\[ v = \frac{1}{2} v_{max} = \frac{1}{2} \omega A \]
- Using the expression for speed in SHM:
\[ v = \omega \sqrt{A^2 - x^2} \]
- Setting this equal to \( \frac{1}{2} \omega A \):
\[ \omega \sqrt{A^2 - x^2} = \frac{1}{2} \omega A \]
- Dividing both sides by \( \omega \) (assuming \( \omega \neq 0 \)):
\[ \sqrt{A^2 - x^2} = \frac{1}{2} A \]

3. **Squaring Both Sides**:
- Squaring both sides gives:
\[ A^2 - x^2 = \frac{1}{4} A^2 \]
- Rearranging leads to:
\[ A^2 - \frac{1}{4} A^2 = x^2 \implies \frac{3}{4} A^2 = x^2 \implies x = \frac{\sqrt{3}}{2} A \]

### Final Answers:
- (i) The distance from the mean position where kinetic energy equals potential energy is \( x = \frac{A}{\sqrt{2}} \).
- (ii) The points where the speed is half of the maximum speed are \( x = \frac{\sqrt{3}}{2} A \) and \( x = -\frac{\sqrt{3}}{2} A \).