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 will follow these steps:

### Step 1: Understand the given parameters
We are given:
- Force constant \( k = 6 \times 10^5 \, \text{N/m} \)
- Amplitude \( A = 4 \, \text{cm} = 0.04 \, \text{m} \)
- Total energy \( E = 600 \, \text{J} \)

### Step 2: Calculate the maximum kinetic energy
The maximum kinetic energy \( K_{\text{max}} \) in a simple harmonic oscillator is given by the formula:
\[ K_{\text{max}} = \frac{1}{2} k A^2 \]
Substituting the values:
\[ K_{\text{max}} = \frac{1}{2} \times (6 \times 10^5) \times (0.04)^2 \]
Calculating \( (0.04)^2 = 0.0016 \):
\[ K_{\text{max}} = \frac{1}{2} \times (6 \times 10^5) \times 0.0016 \]
\[ K_{\text{max}} = 3 \times 10^5 \times 0.0016 = 480 \, \text{J} \]

### Step 3: Relate total energy to kinetic and potential energy
In a harmonic oscillator, the total energy \( E \) is the sum of maximum kinetic energy and maximum potential energy:
\[ E = K_{\text{max}} + U_{\text{max}} \]
Since at maximum displacement (amplitude), the kinetic energy is zero and potential energy is maximum:
\[ E = U_{\text{max}} \quad \text{(when kinetic energy is minimum)} \]
Thus, we can find the maximum potential energy:
\[ U_{\text{max}} = E = 600 \, \text{J} \]

### Step 4: Calculate the minimum potential energy
The minimum potential energy occurs when the kinetic energy is maximum, which we calculated as \( K_{\text{max}} = 480 \, \text{J} \). Therefore, the minimum potential energy \( U_{\text{min}} \) can be calculated as:
\[ U_{\text{min}} = E - K_{\text{max}} = 600 \, \text{J} - 480 \, \text{J} = 120 \, \text{J} \]

### Conclusion
From the calculations:
- Maximum potential energy \( U_{\text{max}} = 600 \, \text{J} \)
- Minimum potential energy \( U_{\text{min}} = 120 \, \text{J} \)

Thus, the correct statement regarding the energies in the harmonic oscillator is that the maximum potential energy is \( 600 \, \text{J} \) and the minimum potential energy is \( 120 \, \text{J} \).

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