In simple harmonic motion of a particle, maximum kinetic energy is 40 J and maximum potential energy is 60 J. then

Options

A. minimum potential energy will be 20 J
B. potential energy at half the displacement will be 30J
C. kinetic energy at half the displacement is 40 J
D. potential energy or kinetic energy at some intermediate position cannot be found the given data

Answer

To solve the problem, we need to analyze the simple harmonic motion (SHM) of a particle with the given maximum kinetic energy (KE) and maximum potential energy (PE).

### Step-by-Step Solution:

1. **Identify Given Values**:
- Maximum Kinetic Energy (KE_max) = 40 J
- Maximum Potential Energy (PE_max) = 60 J

2. **Calculate Total Mechanical Energy (E)**:
- In SHM, the total mechanical energy (E) is the sum of maximum kinetic energy and maximum potential energy.
\[ E = KE_{max} + PE_{max} = 40 \, \text{J} + 60 \, \text{J} = 100 \, \text{J} \]

3. **Determine Minimum Potential Energy (PE_min)**:
- At the mean position, the potential energy is minimum (PE_min) and the kinetic energy is maximum.
- Since total mechanical energy is conserved:
\[ PE_{min} = E - KE_{max} = 100 \, \text{J} - 40 \, \text{J} = 60 \, \text{J} \]

4. **Calculate Potential Energy at Half the Displacement (X = A/2)**:
- At half the amplitude (X = A/2), we can find the kinetic energy (KE) and potential energy (PE) using the conservation of energy.
- The potential energy at this position can be calculated as:
\[ PE = E - KE \]
- To find KE at X = A/2, we use the relationship:
\[ KE = \frac{1}{2} m \omega^2 (A^2 - X^2) \]
- Substituting \(X = \frac{A}{2}\):
\[ KE = \frac{1}{2} m \omega^2 \left(A^2 - \left(\frac{A}{2}\right)^2\right) = \frac{1}{2} m \omega^2 \left(A^2 - \frac{A^2}{4}\right) = \frac{1}{2} m \omega^2 \left(\frac{3A^2}{4}\right) \]
- Therefore, the kinetic energy at half the displacement is:
\[ KE = \frac{3}{4} KE_{max} = \frac{3}{4} \times 40 \, \text{J} = 30 \, \text{J} \]

5. **Calculate Potential Energy at Half the Displacement**:
- Now using the total mechanical energy:
\[ PE = E - KE = 100 \, \text{J} - 30 \, \text{J} = 70 \, \text{J} \]

6. **Summarize the Results**:
- Minimum Potential Energy (PE_min) = 60 J
- Potential Energy at half the displacement = 70 J
- Kinetic Energy at half the displacement = 30 J

### Final Answers:
- Minimum Potential Energy = 60 J
- Potential Energy at half the displacement = 70 J
- Kinetic Energy at half the displacement = 30 J