A linear harmoinc oscillator of force constant `10^(6)N/m` and amplitude 1cm has a total mechanical energy of 80j . The its maximum

Options

A. potential energy is 50J
B. potential energy is 80J
C. kinetic eneregy is 80J
D. Kineteic energy is 50J

Explanation

To solve the problem, we need to find the maximum potential energy (PE_max) and maximum kinetic energy (KE_max) of a linear harmonic oscillator given the force constant (k), amplitude (A), and total mechanical energy (E).

### Given:
- Force constant, \( k = 10^6 \, \text{N/m} \)
- Amplitude, \( A = 1 \, \text{cm} = 0.01 \, \text{m} \)
- Total mechanical energy, \( E = 80 \, \text{J} \)

### Step 1: Calculate Maximum Potential Energy (PE_max)
The maximum potential energy in a simple harmonic oscillator is equal to the total mechanical energy. Therefore, we have:
\[ PE_{\text{max}} = E \]
Substituting the given total mechanical energy:
\[ PE_{\text{max}} = 80 \, \text{J} \]

### Step 2: Calculate Maximum Kinetic Energy (KE_max)
The maximum kinetic energy in a simple harmonic oscillator can be calculated using the relationship:
\[ KE_{\text{max}} = E - PE_{\text{max}} \]
Since we already found \( PE_{\text{max}} = 80 \, \text{J} \) and the total mechanical energy is also \( 80 \, \text{J} \), we can substitute:
\[ KE_{\text{max}} = E - PE_{\text{max}} = 80 \, \text{J} - 80 \, \text{J} = 0 \, \text{J} \]

### Step 3: Verify with the formula for KE_max
Alternatively, we can calculate the maximum kinetic energy using the formula:
\[ KE_{\text{max}} = \frac{1}{2} k A^2 \]
Substituting the values:
\[ KE_{\text{max}} = \frac{1}{2} \times 10^6 \, \text{N/m} \times (0.01 \, \text{m})^2 \]
Calculating:
\[ KE_{\text{max}} = \frac{1}{2} \times 10^6 \times 0.0001 = \frac{1}{2} \times 100 = 50 \, \text{J} \]

### Final Results:
- Maximum Potential Energy, \( PE_{\text{max}} = 80 \, \text{J} \)
- Maximum Kinetic Energy, \( KE_{\text{max}} = 50 \, \text{J} \)

### Summary:
- Maximum Potential Energy: \( 80 \, \text{J} \)
- Maximum Kinetic Energy: \( 50 \, \text{J} \)