The average energy in one time period in simple harmonic motion is

Options

A. `1/2momega^2A^2`
B. `1/4 m omega^2A^2`
C. `m omega^2A^2`
D. zero

Answer

To find the average energy in one time period in simple harmonic motion (SHM), we need to consider both the kinetic and potential energy of the system.

### Step-by-Step Solution:

1. **Understand the Energy Components**:
- In SHM, the total mechanical energy (E) is the sum of kinetic energy (KE) and potential energy (PE).
- The formulas for these energies are:
- Kinetic Energy (KE): \( KE = \frac{1}{2} m \omega^2 (A^2 - x^2) \)
- Potential Energy (PE): \( PE = \frac{1}{2} m \omega^2 x^2 \)

2. **Total Energy Calculation**:
- The total energy (E) in SHM can be expressed as:
\[ E = KE + PE \]
- Substituting the expressions for KE and PE:
\[ E = \frac{1}{2} m \omega^2 (A^2 - x^2) + \frac{1}{2} m \omega^2 x^2 \]

3. **Simplifying the Total Energy**:
- Combine the two terms:
\[ E = \frac{1}{2} m \omega^2 A^2 - \frac{1}{2} m \omega^2 x^2 + \frac{1}{2} m \omega^2 x^2 \]
- The \( -\frac{1}{2} m \omega^2 x^2 \) and \( +\frac{1}{2} m \omega^2 x^2 \) cancel each other out:
\[ E = \frac{1}{2} m \omega^2 A^2 \]

4. **Average Energy Over One Time Period**:
- Since the total energy \( E = \frac{1}{2} m \omega^2 A^2 \) is constant and does not change with time, the average energy over one complete cycle (one time period) is simply equal to the total energy:
\[ \text{Average Energy} = \frac{1}{2} m \omega^2 A^2 \]

5. **Conclusion**:
- Therefore, the average energy in one time period in simple harmonic motion is:
\[ \frac{1}{2} m \omega^2 A^2 \]
- This corresponds to option 1 in the given choices.