The average kinetic energy of a simple harmonic oscillator is `2` joule and its total energy is `5` joule. Its minimum potential energy is

Options

A. `1J`
B. `1.5J`
C. `2J`
D. `3J`

Explanation

To find the minimum potential energy of a simple harmonic oscillator given the average kinetic energy and total energy, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Given Values:**
- Average kinetic energy (KE_avg) = 2 joules
- Total energy (E_total) = 5 joules

2. **Understand the Relationship:**
- The total energy (E_total) in a simple harmonic oscillator is the sum of the maximum kinetic energy (KE_max) and the maximum potential energy (PE_max).
- At the mean position, the potential energy is at its minimum (PE_min), and the kinetic energy is at its maximum (KE_max).
- Therefore, we can write:
\[ E_{total} = KE_{max} + PE_{min} \]

3. **Calculate Maximum Kinetic Energy:**
- The average kinetic energy is given by:
\[ KE_{avg} = \frac{KE_{max} + KE_{min}}{2} \]
- Since the minimum kinetic energy (KE_min) in SHM can be zero, we have:
\[ 2 = \frac{KE_{max} + 0}{2} \]
- Solving for KE_max gives:
\[ KE_{max} = 4 \text{ joules} \]

4. **Substitute into Total Energy Equation:**
- Now substitute KE_max into the total energy equation:
\[ E_{total} = KE_{max} + PE_{min} \]
- Plugging in the values:
\[ 5 = 4 + PE_{min} \]

5. **Solve for Minimum Potential Energy:**
- Rearranging the equation gives:
\[ PE_{min} = 5 - 4 = 1 \text{ joule} \]

### Final Answer:
The minimum potential energy is **1 joule**.

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