A linear harmonic oscillator of force constant `2 xx 10^(6)N//m` and amplitude `0.01 m` has a total mechanical energy of `160 J`. Its

Options

A. maximum potential energy is 100 J
B. maximum kinetic energy is 100 J
C. maximum potential energy is 160 J
D. minimum potential energy is zero

Explanation

To solve the problem step by step, we need to find the maximum potential energy and the maximum kinetic energy of a linear harmonic oscillator given the force constant, amplitude, and total mechanical energy.

### Given Data:
- Force constant, \( k = 2 \times 10^6 \, \text{N/m} \)
- Amplitude, \( A = 0.01 \, \text{m} \)
- Total mechanical energy, \( E = 160 \, \text{J} \)

### Step 1: Determine Maximum Potential Energy
The maximum potential energy \( U_{\text{max}} \) in a harmonic oscillator is equal to the total mechanical energy \( E \).

\[ U_{\text{max}} = E = 160 \, \text{J} \]

### Step 2: Determine Maximum Kinetic Energy
The maximum kinetic energy \( K_{\text{max}} \) can be calculated using the formula:

\[ K_{\text{max}} = \frac{1}{2} k A^2 \]

### Step 3: Substitute Values into the Kinetic Energy Formula
First, we need to square the amplitude:

\[ A^2 = (0.01 \, \text{m})^2 = 0.0001 \, \text{m}^2 = 10^{-4} \, \text{m}^2 \]

Now substitute \( k \) and \( A^2 \) into the kinetic energy formula:

\[ K_{\text{max}} = \frac{1}{2} \times (2 \times 10^6 \, \text{N/m}) \times (10^{-4} \, \text{m}^2) \]

### Step 4: Calculate the Kinetic Energy
Now calculate:

\[ K_{\text{max}} = \frac{1}{2} \times 2 \times 10^6 \times 10^{-4} \]

\[ K_{\text{max}} = \frac{1}{2} \times 2 \times 10^{2} \, \text{J} \]

\[ K_{\text{max}} = 100 \, \text{J} \]

### Final Results:
- Maximum Potential Energy \( U_{\text{max}} = 160 \, \text{J} \)
- Maximum Kinetic Energy \( K_{\text{max}} = 100 \, \text{J} \)

### Summary:
- Maximum Potential Energy: \( 160 \, \text{J} \)
- Maximum Kinetic Energy: \( 100 \, \text{J} \)

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