) A simple harmonic oscillator of amplitude A has a total energy E. Determine a) the kinetic energy nd b) the potential energy when the position is one third the amplitude. c) For what values of the position does the kinetic energy equal one half the potential energy? d) Are there any values of the position where the kinetic energy is greater than the maximum potential energy? Explain. (a: 8E/9, : E/9, c: A = Aar V2/3 , d: not giving this one)

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### Simple Harmonic Oscillator Energy Analysis

**Problem Statement:**
A simple harmonic oscillator of amplitude \( A \) has a total energy \( E \). Determine the following:
1. (a) The kinetic energy and (b) the potential energy when the position is one-third the amplitude.
2. (c) For what values of the position does the kinetic energy equal one-half the potential energy?
3. (d) Are there any values of the position where the kinetic energy is greater than the maximum potential energy? Explain.

**Answers:**
- **(a)** When the position is one-third the amplitude: 
  - Kinetic Energy: \( \frac{8E}{9} \)
- **(b)** When the position is one-third the amplitude: 
  - Potential Energy: \( \frac{E}{9} \)
- **(c)** For the position at which kinetic energy equals half the potential energy:
  - Position: \( A = A_{\text{max}} \sqrt{\frac{2}{3}} \)
- **(d)** Not provided.

### Explanation of Concepts:
- Simple harmonic oscillators (SHOs) involve sinusoidal motion where potential and kinetic energies interchange but sum up to a constant total energy.
- At maximum amplitude \( A \), the potential energy is maximized, and kinetic energy is zero.
- At equilibrium (zero position), potential energy is zero, and kinetic energy is maximized.

### Detailed Calculations:
1. **For part (a) and (b):**
   - **Potential Energy \( U \):** \( U = \frac{1}{2}k x^2 \)
   - **Kinetic Energy \( K \):** \( K = E - U \)
   - If the position \( x \) is \( \frac{A}{3} \):
     - \( U = \frac{1}{2} k \left(\frac{A}{3}\right)^2 = \frac{1}{2} k \frac{A^2}{9} = \frac{k A^2}{18} \)
     - Since \( E = \frac{1}{2} k A^2 \):
       - \( U = \frac{E}{9} \)
     - Thus, \( K = E - U = E - \frac{E}{9} = \frac{8E}{9} \)

2
Transcribed Image Text:### Simple Harmonic Oscillator Energy Analysis **Problem Statement:** A simple harmonic oscillator of amplitude \( A \) has a total energy \( E \). Determine the following: 1. (a) The kinetic energy and (b) the potential energy when the position is one-third the amplitude. 2. (c) For what values of the position does the kinetic energy equal one-half the potential energy? 3. (d) Are there any values of the position where the kinetic energy is greater than the maximum potential energy? Explain. **Answers:** - **(a)** When the position is one-third the amplitude: - Kinetic Energy: \( \frac{8E}{9} \) - **(b)** When the position is one-third the amplitude: - Potential Energy: \( \frac{E}{9} \) - **(c)** For the position at which kinetic energy equals half the potential energy: - Position: \( A = A_{\text{max}} \sqrt{\frac{2}{3}} \) - **(d)** Not provided. ### Explanation of Concepts: - Simple harmonic oscillators (SHOs) involve sinusoidal motion where potential and kinetic energies interchange but sum up to a constant total energy. - At maximum amplitude \( A \), the potential energy is maximized, and kinetic energy is zero. - At equilibrium (zero position), potential energy is zero, and kinetic energy is maximized. ### Detailed Calculations: 1. **For part (a) and (b):** - **Potential Energy \( U \):** \( U = \frac{1}{2}k x^2 \) - **Kinetic Energy \( K \):** \( K = E - U \) - If the position \( x \) is \( \frac{A}{3} \): - \( U = \frac{1}{2} k \left(\frac{A}{3}\right)^2 = \frac{1}{2} k \frac{A^2}{9} = \frac{k A^2}{18} \) - Since \( E = \frac{1}{2} k A^2 \): - \( U = \frac{E}{9} \) - Thus, \( K = E - U = E - \frac{E}{9} = \frac{8E}{9} \) 2
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