Energy in SHM
A short lesson with diagrams, a worked example, a problem to try, and a quick check. Then quiz yourself on 5 practice questions.
With no friction, the oscillator’s total energy E = ½kA² stays constant, trading back and forth between spring potential energy and kinetic energy: ½kx² + ½mv² = ½kA². That gives the speed at any position: v = ω√(A² − x²).
1. Energy over time
2. Energy facts
Energy depends on A²: double the amplitude, four times the energy, twice the maximum speed.
At x = A/2, U = ¼E, so K = ¾E and v = (√3/2)vmax ≈ 0.87vmax.
Since x ∝ cos ωt, U ∝ cos²ωt = ½(1 + cos 2ωt): the energies oscillate at 2ω, twice the frequency of the motion.
For a vertical spring, measure x from the hanging equilibrium and use ½kx² for the combined spring + gravity potential energy.
3. Example, then try it
A 0.50 kg block on a spring (k = 50 N/m) oscillates with amplitude 0.20 m. Find its total energy and maximum speed.
E = ½(50)(0.20)² = 1.0 J.
ω = √(50/0.50) = 10 rad/s, so vmax = Aω = 2.0 m/s (check: ½(0.50)(2.0)² = 1.0 J ✓).
For that oscillator (k = 50 N/m, m = 0.50 kg, A = 0.20 m), find its speed at x = 0.10 m, where K = U, and its total energy if the amplitude is doubled.
At x = 0.10 m: v = ω√(A² − x²) = 10√(0.040 − 0.010) = 10(0.173) ≈ 1.7 m/s.
K = U: each is E/2, so ½kx² = ½(½kA²) → x = A/√2 = 0.20 ÷ 1.414 ≈ 0.14 m.
Double amplitude: E ∝ A², so E = 4(1.0) = 4.0 J.
Answer: 1.7 m/s at 0.10 m; K = U at 0.14 m; 4.0 J at double amplitude.
4. Common mistakes
A mass-spring oscillator is at x = A/2. What fraction of its total energy is kinetic?
U = ½k(A/2)² = ¼ of ½kA². The other ¾ is kinetic.
Ready to practice?
5 questions on Energy in SHM, each with a hint and an explanation for every wrong answer.