Section 6.3: Energy in SHM
In simple harmonic motion, energy is continuously exchanged between kinetic energy and potential energy. The total mechanical energy remains constant if no damping exists.
Formulas:
- Displacement: \( x(t) = A \cos(\omega t + \phi) \)
- Velocity: \( v(t) = -\omega A \sin(\omega t + \phi) \)
- Kinetic Energy: \( KE = \frac{1}{2} m v^2 = \frac{1}{2} m \omega^2 (A^2 - x^2) \)
- Potential Energy: \( PE = \frac{1}{2} k x^2 = \frac{1}{2} m \omega^2 x^2 \)
- Total Energy: \( E = KE + PE = \frac{1}{2} k A^2 = \frac{1}{2} m \omega^2 A^2 \)
Example 1
A 0.5 kg mass oscillates on a spring with \( k = 200 \, \text{N/m} \) and amplitude 0.1 m. Find the total energy, kinetic energy at \( x = 0.05 \, \text{m} \), and potential energy at the same point.
Total energy: \( E = \frac{1}{2} k A^2 = 0.5 \cdot 200 \cdot 0.1^2 = 1 \, \text{J} \)
Potential energy: \( PE = \frac{1}{2} k x^2 = 0.5 \cdot 200 \cdot 0.05^2 = 0.25 \, \text{J} \)
Kinetic energy: \( KE = E - PE = 1 - 0.25 = 0.75 \, \text{J} \)
Practice Problems
- A 0.2 kg mass oscillates with amplitude 0.15 m. Spring constant is 50 N/m. Find total energy.
- For a spring-mass system, find KE when displacement is half the amplitude.
- A pendulum of length 1 m has maximum PE of 2 J. Find KE at 60° displacement.
- Mass 0.3 kg, \( k = 80 \, \text{N/m} \), amplitude 0.2 m. Find KE at \( x = 0.1 \, \text{m} \).
- Show that \( KE_\text{max} = PE_\text{max} \) at displacement \( x = A/\sqrt{2} \).