### 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

Name: ### Simple Harmonic Oscillator Energy Analysis**Problem Statement:**A
Uploaded: 2024-03-20T06:58:04.000Z
Channel: Bartleby
Description: ### 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