A linear harmonic oscillator of force constant `6 xx 10^(5)` N/m and amplitude 4cm, has a total energy 600J. Select the correct statement.

Options

A. Maximum potential energy is 600J
B. Maximum kinetic energy is 480 J
C. Minimum potential energy is 120J
D. All of these

Answer

To solve the problem, we need to analyze the properties of a linear harmonic oscillator. We are given the force constant (k), amplitude (A), and total energy (E) of the oscillator. We will find the maximum kinetic energy (K.E.) and minimum potential energy (P.E.) and check the statements provided.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Force constant, \( k = 6 \times 10^5 \, \text{N/m} \)
- Amplitude, \( A = 4 \, \text{cm} = 0.04 \, \text{m} \)
- Total energy, \( E = 600 \, \text{J} \)

2. **Calculate Maximum Kinetic Energy:**
The maximum kinetic energy in a harmonic oscillator occurs at the mean position (x = 0) and is given by the formula:
\[ K.E._{\text{max}} = \frac{1}{2} k A^2 \]
Substituting the known values:
\[ K.E._{\text{max}} = \frac{1}{2} \times (6 \times 10^5) \times (0.04)^2 \]
\[ = \frac{1}{2} \times (6 \times 10^5) \times (0.0016) \]
\[ = \frac{1}{2} \times 960 \, \text{J} \]
\[ = 480 \, \text{J} \]

3. **Calculate Minimum Potential Energy:**
The total energy in a harmonic oscillator is conserved and can be expressed as:
\[ E = K.E. + P.E. \]
At the maximum displacement (x = A), all energy is potential, and at the mean position (x = 0), all energy is kinetic. Thus, we can write:
\[ E = K.E._{\text{max}} + P.E._{\text{min}} \]
Rearranging gives:
\[ P.E._{\text{min}} = E - K.E._{\text{max}} \]
Substituting the known values:
\[ P.E._{\text{min}} = 600 \, \text{J} - 480 \, \text{J} = 120 \, \text{J} \]

4. **Calculate Maximum Potential Energy:**
The maximum potential energy occurs at maximum displacement (x = A):
\[ P.E._{\text{max}} = E \]
Therefore, the maximum potential energy is:
\[ P.E._{\text{max}} = 600 \, \text{J} \]

5. **Conclusion:**
- Maximum Kinetic Energy = 480 J
- Minimum Potential Energy = 120 J
- Maximum Potential Energy = 600 J

### Final Answer:
All the statements provided in the question are correct:
1. Maximum potential energy = 600 J
2. Maximum kinetic energy = 480 J
3. Minimum potential energy = 120 J