A particle executes SHM.
(a) What fraction of total energy is kinetic and what fraction is potential when displacement is one half of the amplitude?
(b) At what value of displacement are the kinetic and potential energies equal?

Explanation

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

### Part (a): Finding the fractions of kinetic and potential energy when displacement is one half of the amplitude.

1. **Understanding the total energy in SHM**:
The total mechanical energy (E) in Simple Harmonic Motion (SHM) is constant and is given by the formula:
\[ E = \frac{1}{2} m \omega^2 A^2 \]
where \( A \) is the amplitude, \( m \) is the mass, and \( \omega \) is the angular frequency.

2. **Kinetic Energy (KE) at displacement \( x = \frac{A}{2} \)**:
The kinetic energy at a displacement \( x \) is given by:
\[ KE = \frac{1}{2} m \omega^2 (A^2 - x^2) \]
Substituting \( x = \frac{A}{2} \):
\[ KE = \frac{1}{2} m \omega^2 \left( A^2 - \left( \frac{A}{2} \right)^2 \right) = \frac{1}{2} m \omega^2 \left( A^2 - \frac{A^2}{4} \right) = \frac{1}{2} m \omega^2 \left( \frac{3A^2}{4} \right) = \frac{3}{8} m \omega^2 A^2 \]

3. **Potential Energy (PE) at displacement \( x = \frac{A}{2} \)**:
The potential energy at a displacement \( x \) is given by:
\[ PE = \frac{1}{2} m \omega^2 x^2 \]
Substituting \( x = \frac{A}{2} \):
\[ PE = \frac{1}{2} m \omega^2 \left( \frac{A}{2} \right)^2 = \frac{1}{2} m \omega^2 \left( \frac{A^2}{4} \right) = \frac{1}{8} m \omega^2 A^2 \]

4. **Finding the fractions of total energy**:
The total energy \( E \) is:
\[ E = KE + PE = \frac{3}{8} m \omega^2 A^2 + \frac{1}{8} m \omega^2 A^2 = \frac{4}{8} m \omega^2 A^2 = \frac{1}{2} m \omega^2 A^2 \]
- Fraction of kinetic energy:
\[ \text{Fraction of } KE = \frac{KE}{E} = \frac{\frac{3}{8} m \omega^2 A^2}{\frac{1}{2} m \omega^2 A^2} = \frac{3/8}{1/2} = \frac{3}{4} \]
- Fraction of potential energy:
\[ \text{Fraction of } PE = \frac{PE}{E} = \frac{\frac{1}{8} m \omega^2 A^2}{\frac{1}{2} m \omega^2 A^2} = \frac{1/8}{1/2} = \frac{1}{4} \]

### Part (b): Finding the displacement where kinetic and potential energies are equal.

1. **Setting KE equal to PE**:
We know:
\[ KE = \frac{1}{2} m \omega^2 (A^2 - x^2) \]
\[ PE = \frac{1}{2} m \omega^2 x^2 \]
Setting them equal:
\[ \frac{1}{2} m \omega^2 (A^2 - x^2) = \frac{1}{2} m \omega^2 x^2 \]

2. **Simplifying the equation**:
Cancel \( \frac{1}{2} m \omega^2 \) from both sides:
\[ A^2 - x^2 = x^2 \]
Rearranging gives:
\[ A^2 = 2x^2 \]
Thus:
\[ x^2 = \frac{A^2}{2} \]
Taking the square root:
\[ x = \frac{A}{\sqrt{2}} \]

### Final Answers:
(a) The fraction of total energy that is kinetic is \( \frac{3}{4} \) and the fraction that is potential is \( \frac{1}{4} \).

(b) The displacement at which kinetic and potential energies are equal is \( x = \frac{A}{\sqrt{2}} \).