A linear harmonic oscillator of force constant 2 × 106 N/m and displacement 0.01 m has a total mechanical energy of 160 J. Its

A linear harmonic oscillator of force constant \( 2 \times 10^6 \, \text{N/m} \) and displacement \( 0.01 \, \text{m} \) has a total mechanical energy of \( 160 \, \text{J} \). Its:

1. P.E. is 160 J
2. P.E. is zero
3. P.E. is 100 J
4. P.E. is 120 J

Detailed Answer:

The total energy of a linear harmonic oscillator is given by: \[ E_{\text{total}} = \frac{1}{2} k A^2 \] where: - \( k = 2 \times 10^6 \, \text{N/m} \) is the force constant, - \( A \) is the amplitude (maximum displacement). The potential energy (P.E.) at any displacement \( x \) is: \[ U = \frac{1}{2} k x^2 \] Substituting \( k = 2 \times 10^6 \, \text{N/m} \) and \( x = 0.01 \, \text{m} \): \[ U = \frac{1}{2} (2 \times 10^6) (0.01)^2 \] \[ U = \frac{1}{2} (2 \times 10^6) (0.0001) \] \[ U = 100 \, \text{J} \] Therefore, the potential energy at the given displacement is: \[ \boxed{100 \, \text{J}} \] Thus, the correct answer is: (c) P.E. is 100 J.