The force constant of a simple harmonic oscillator is `3 xx 10^(6) N//m`, amplitude 0.02 m has a total energy of 1250 J

Options

A. Maximum potential energy is 650 J
B. Maximum kinetic energy is 600 J
C. Maximum kinetic energy is 700 J
D. Maximum potential energy is 450 J

Explanation

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

### Step-by-Step Solution:

1. **Identify the given values**:
- Force constant (k) = \(3 \times 10^6 \, \text{N/m}\)
- Amplitude (A) = \(0.02 \, \text{m}\)
- Total energy (E) = \(1250 \, \text{J}\)

2. **Calculate the maximum kinetic energy (KE_max)**:
The formula for maximum kinetic energy in a simple harmonic oscillator is given by:
\[ KE_{\text{max}} = \frac{1}{2} k A^2 \]
Substitute the values of k and A into the formula:
\[ KE_{\text{max}} = \frac{1}{2} \times (3 \times 10^6) \times (0.02)^2 \]
Calculate \(A^2\):
\[ A^2 = (0.02)^2 = 0.0004 \, \text{m}^2 \]
Now substitute this back into the equation:
\[ KE_{\text{max}} = \frac{1}{2} \times (3 \times 10^6) \times 0.0004 \]
\[ KE_{\text{max}} = \frac{1}{2} \times 1200 = 600 \, \text{J} \]

3. **Determine the maximum potential energy (PE_max)**:
The total energy in a simple harmonic oscillator is the sum of maximum kinetic energy and maximum potential energy:
\[ E = KE_{\text{max}} + PE_{\text{max}} \]
Rearranging for PE_max gives:
\[ PE_{\text{max}} = E - KE_{\text{max}} \]
Substitute the known values:
\[ PE_{\text{max}} = 1250 \, \text{J} - 600 \, \text{J} \]
\[ PE_{\text{max}} = 650 \, \text{J} \]

4. **Conclusion**:
- Maximum Kinetic Energy (KE_max) = \(600 \, \text{J}\)
- Maximum Potential Energy (PE_max) = \(650 \, \text{J}\)