Energy of Simple Harmonic Oscillators
Introduced| English |
|---|
| potential energy/pəˈtenʃl ˈenədʒi/ |
| kinetic energy/kɪˈnetɪk ˈenədʒi/ |
The endless trade
- A swing slows to a stop at the top, then rushes fastest through the bottom.
- Nothing is lost -- the motion just keeps swapping form.
- Stretch is stored; speed is spent; back and forth forever.
- Two stores hand the same total to and fro.
Two stores in a spring
- The spring holds potential energy 势能 $U = \tfrac{1}{2}kx^2$.
- The mass carries kinetic energy 动能 $K = \tfrac{1}{2}mv^2$.
- As it oscillates, energy pours from one into the other.
A spring with $k = 100\ \text{N/m}$ is displaced $x = 0.1\ \text{m}$. Its stored potential energy (in J)?
$U = \tfrac{1}{2}kx^2 = \tfrac{1}{2}(100)(0.1)^2 = 0.5\ \text{J}$.
The total never changes
- Add them and the total stays fixed:
- At the ends it is all potential (speed zero); at the middle it is all kinetic.
- The amplitude alone sets the total.
A spring with $k = 100\ \text{N/m}$ oscillates with amplitude $A = 0.2\ \text{m}$. Its total energy (in J)?
$E = \tfrac{1}{2}kA^2 = \tfrac{1}{2}(100)(0.2)^2 = 2\ \text{J}$.
At which point is the oscillator's energy entirely kinetic?
At equilibrium $x = 0$ so $U = 0$ and all the energy is kinetic.
Cashing energy for speed
- At equilibrium every bit is kinetic, so:
- This recovers $v_{max} = A\omega$, since $\omega^2 = k/m$.
- Double the amplitude and you get four times the energy, twice the top speed.
Energy of an oscillator
Over a cycle, energy shifts between kinetic and potential while the total stays fixed.
A spring with $k = 200\ \text{N/m}$ oscillates with amplitude $A = 0.05\ \text{m}$.
- Total energy: $E = \tfrac{1}{2}kA^2 = \tfrac{1}{2}(200)(0.05)^2 = 0.25\ \text{J}$.
- It is all kinetic at the middle and all potential at the ends.
If you double the amplitude of an oscillator, its total energy...
$E = \tfrac{1}{2}kA^2 \propto A^2$, so $2^2 = 4$ times the energy.
Kinetic and potential energy can both reach their maximum at the same instant.
Each is largest where the other is zero -- they are never both maximal.
In a real oscillator with friction, over time the amplitude...
Friction drains energy, so the amplitude decays -- this is damping.
The energy goes as amplitude squared ($E = \tfrac{1}{2}kA^2$), so doubling $A$ quadruples the energy, not doubles it. Kinetic and potential energy are each largest where the other is zero, never both at once. And "the total stays constant" assumes no friction or drag -- a real oscillator slowly loses energy and its amplitude decays (damping).
An oscillator's energy trades between potential energy $U = \tfrac{1}{2}kx^2$ and kinetic energy $K = \tfrac{1}{2}mv^2$, with a fixed total $E = \tfrac{1}{2}kA^2$. It is all potential at the extremes and all kinetic at equilibrium, giving $\tfrac{1}{2}mv_{max}^2 = \tfrac{1}{2}kA^2$. Because $E \propto A^2$, doubling the amplitude quadruples the energy.