AP Physics 1- 7.4 Energy of Simple Harmonic Oscillators - Study Notes- New Syllabus
AP Physics 1-7.4 Energy of Simple Harmonic Oscillators – Study Notes
AP Physics 1-7.4 Energy of Simple Harmonic Oscillators – Study Notes -AP Physics 1 – per latest Syllabus.
Key Concepts:
- Energy of Simple Harmonic Oscillators
Energy of Simple Harmonic Oscillators
In SHM, energy continuously transforms between kinetic energy (KE) and potential energy (PE), but the total mechanical energy (E) of the system remains constant (for ideal, frictionless oscillators).
Kinetic Energy:
The instantaneous kinetic energy of a mass-spring system is given by:
\( KE = \dfrac{1}{2} m v^2 = \dfrac{1}{2} m \omega^2 (A^2 – x^2) \)
Maximum kinetic energy occurs at equilibrium position (\( x = 0 \)): \( KE_{max} = \dfrac{1}{2} m \omega^2 A^2 \)
Potential Energy:
The instantaneous potential energy stored in the spring is:
\( PE = \dfrac{1}{2} k x^2 = \dfrac{1}{2} m \omega^2 x^2 \)
Maximum potential energy occurs at maximum displacement (\( x = \pm A \)): \( PE_{max} = \dfrac{1}{2} k A^2 \)
Total Mechanical Energy:
The total energy of the system remains constant:
\( E = KE + PE = \dfrac{1}{2} m \omega^2 A^2 \)
- Independent of time and position
- Energy oscillates between KE and PE as the object moves
Key Points:
- Energy in SHM is a scalar quantity.
- At maximum displacement: \( PE = E \), \( KE = 0 \)
- At equilibrium: \( KE = E \), \( PE = 0 \)
- At intermediate positions: both KE and PE are non-zero but sum equals total energy.
Example :
A mass-spring system has mass \( m = 0.5 \, \text{kg} \), spring constant \( k = 200 \, \text{N/m} \), and amplitude \( A = 0.1 \, \text{m} \). Find total energy, kinetic energy, and potential energy at \( x = 0.05 \, \text{m} \).
▶️Answer/Explanation
Angular frequency: \( \omega = \sqrt{k/m} = \sqrt{200/0.5} = 20 \, \text{rad/s} \)
Total energy: \( E = \dfrac{1}{2} m \omega^2 A^2 = 0.5 \cdot 20^2 \cdot 0.1^2 / 2 \approx 1 \, \text{J} \)
Potential energy at \( x = 0.05 \): \( PE = \dfrac{1}{2} m \omega^2 x^2 = 0.5 \cdot 20^2 \cdot 0.05^2 / 2 \approx 0.25 \, \text{J} \)
Kinetic energy: \( KE = E – PE = 1 – 0.25 = 0.75 \, \text{J} \)
Example :
A simple pendulum of length \( L = 0.8 \, \text{m} \) and mass \( m = 0.3 \, \text{kg} \) oscillates with amplitude \( \theta_{max} = 0.1 \, \text{rad} \). Find total mechanical energy and energy at maximum displacement.
▶️Answer/Explanation
Potential energy at maximum displacement: \( PE_{max} = m g L (1 – \cos\theta_{max}) \approx 0.3 \cdot 9.8 \cdot 0.8 (1 – \cos0.1) \approx 0.012 \, \text{J} \)
Total energy \( E = PE_{max} \approx 0.012 \, \text{J} \) (KE = 0 at maximum displacement)
Example:
A mass-spring system has mass \( m = 0.5 \, \text{kg} \), spring constant \( k = 50 \, \text{N/m} \), and amplitude \( A = 0.2 \, \text{m} \).
Find the displacement where kinetic energy equals potential energy, and the total mechanical energy at that position.
▶️Answer/Explanation
Total energy in SHM is constant: \( E = \dfrac{1}{2} k A^2 = \dfrac{1}{2} \cdot 50 \cdot (0.2)^2 = 1 \, \text{J} \)
Let displacement be \( x \). Potential energy: \( PE = \dfrac{1}{2} k x^2 \)
Since \( KE = E – PE \) and KE = PE, we have:
\( PE = KE = \dfrac{E}{2} = 0.5 \, \text{J} \)
\( \dfrac{1}{2} k x^2 = 0.5 \Rightarrow 25 x^2 = 0.5 \Rightarrow x^2 = 0.02 \Rightarrow x = \pm 0.141 \, \text{m} \)
Total energy: \( E = KE + PE = 0.5 + 0.5 = 1 \, \text{J} \)