Energy in SHM
- Describe the continuous exchange between kinetic and potential energy during SHM
- Apply the total-energy relation E = ½kA² for a spring oscillator
- Compute the maximum speed and the speed at a given displacement
Energy sloshes between two forms
In SHM the total mechanical energy is constant, but it continuously trades between kinetic and potential. At the extremes (x = ±A) the oscillator is momentarily at rest, so all the energy is potential: E = ½kA². At the equilibrium point (x = 0) the spring is relaxed, so all the energy is kinetic: E = ½mv_max². In between, the two shares add to the same constant total.
Speed at any displacement
Setting ½kA² = ½kx² + ½mv² and solving for speed gives v = ω√(A² − x²). The speed is largest at x = 0 (v_max = Aω) and drops to zero at x = ±A, where the motion reverses. This single energy statement lets you find the speed anywhere in the cycle without solving the differential equation for x(t).
A 0.5 kg mass on a spring with k = 200 N/m oscillates with amplitude A = 0.1 m. Find the total energy and the maximum speed.
- 1.Total energy is set by the amplitude: E = ½kA² = ½(200)(0.1²) = ½(200)(0.01) = 1 J.
- 2.At equilibrium all the energy is kinetic: E = ½mv_max².
- 3.Solve: 1 = ½(0.5)v_max² = 0.25 v_max², so v_max² = 4.
- 4.Therefore v_max = 2 m/s (equivalently v_max = Aω = 0.1 × √(200/0.5) = 0.1 × 20 = 2 m/s).
A spring oscillator has spring constant k = 100 N/m and amplitude A = 0.2 m. What is its total mechanical energy?
During simple harmonic motion, at what point is the oscillator's speed greatest?
Anchor SHM energy problems on E = ½kA². Set it equal to ½kx² + ½mv² to find the speed at any position, and equal to ½mv_max² to get the maximum speed. The energy is fixed by the amplitude and never changes as the motion proceeds.
Answer the 2 checkpoints as you read.
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