Energy in SHM
A short lesson with diagrams, a worked example, a problem to try, and a quick check. Then quiz yourself on 6 practice questions.
Without friction, an oscillator’s total energy stays constant: E = ½kA². At the ends it’s all spring potential energy; at equilibrium it’s all kinetic energy. In between, it’s a mix: ½kx² + ½mv² = ½kA².
1. Energy back and forth
2. Amplitude squared
Energy depends on A². Double the amplitude and the energy is four times as large; the maximum speed doubles.
The period, though, doesn’t change with amplitude.
Because U = ½kx² grows with x², halfway out (x = A/2) the spring holds only ¼ of the energy. The other ¾ is still kinetic, so the mass is still moving fast.
3. Example, then try it
A 0.20 kg mass on a spring (k = 80 N/m) oscillates with amplitude 0.25 m. Find its total energy and maximum speed.
E = ½kA² = ½(80)(0.25)² = 2.5 J.
At equilibrium it’s all kinetic: ½(0.20)v² = 2.5 → v² = 25 → vmax = 5.0 m/s. (Check: A√(k/m) = 0.25 × 20 = 5.0.)
Same oscillator: m = 0.20 kg, k = 80 N/m, A = 0.25 m, total energy 2.5 J. How fast is the mass moving when it is 0.15 m from equilibrium?
Spring energy at x = 0.15 m: U = ½(80)(0.15)² = 0.90 J.
Kinetic energy: K = E − U = 2.5 − 0.90 = 1.6 J.
Speed: ½(0.20)v² = 1.6 → v² = 16 → v = 4.0 m/s.
Answer: 4.0 m/s, still most of its 5.0 m/s top speed.
4. Common mistakes
A mass on a spring is halfway between equilibrium and its maximum displacement. What fraction of its total energy is kinetic?
U = ½kx² at x = A/2 is ½k(A/2)² = ¼ of ½kA². The remaining ¾ is kinetic.
Ready to practice?
6 questions on Energy in SHM, each with a hint and an explanation for every wrong answer.