AP® Physics 1: Algebra-Based review sheet from Aim for Five (aimforfive.com/physics/units/7/7-4)
Unit 7 · Topic 7.4
7.4 Energy of Simple Harmonic Oscillators
Without friction, the total mechanical energy of an oscillator stays constant while it shifts back and forth between kinetic and potential energy. For a block on a spring, E = ½kA²: all potential at the turning points and all kinetic at equilibrium, so doubling the amplitude makes the total energy four times as large.
Key terms
- mechanical energy
- kinetic energy
- spring potential energy
- amplitude
- conservation of energy
Energy sloshes back and forth
The total energy of an oscillating system is the sum of its kinetic and potential energies: E = K + U. With no friction or other outside work, that total stays constant (3.4). What changes is how it's split.
At a turning point the object is momentarily at rest, so K = 0 and the potential energy is at its maximum. At equilibrium the potential energy is at its minimum and the kinetic energy is at its maximum. In between, the energy is partly both.
The lowest the kinetic energy ever gets is zero, at the turning points.
Spring–block energy
For a block on a horizontal spring, U = ½kx². At x = ±A, all of the energy is potential, so E = ½kA².
At equilibrium all of it is kinetic: ½mv_max² = ½kA², which gives v_max = A√(k/m).
At any position in between, ½mv² + ½kx² = ½kA². Solving gives v = √((k/m)(A² − x²)).
Because energy depends on x², the split isn't even. Halfway out (x = A/2), U is only a quarter of the total and K is three quarters. Kinetic and potential are equal at x = A/√2 ≈ 0.71A.
For a vertical spring, both spring and gravitational energy change, but measured from the hanging equilibrium their changes combine to act like ½kx², so the same results hold.
Amplitude and energy
Changing the amplitude changes the maximum potential energy, and so the total energy. Since E = ½kA², doubling the amplitude makes the energy four times as large and doubles the maximum speed.
The period stays the same (7.2). The extra energy shows up as faster motion, not slower cycles.
| Change (same k and m) | Total energy | Max speed | Period |
|---|---|---|---|
| A × 2 | × 4 | × 2 | same |
| A × 3 | × 9 | × 3 | same |
| A ÷ 2 | × ¼ | × ½ | same |
Pendulum energy
For a pendulum, the potential energy is gravitational: U = mgh, where h is the bob's height above its lowest point. If the string of length L makes angle θ with the vertical, h = L(1 − cos θ). Energy conservation gives the speed at the bottom: v = √(2gh). This works at any angle, small or not.
Energy graphs
Against position, U = ½kx² is an upward parabola, K is an upside-down parabola that touches zero at x = ±A, and the total E is a horizontal line above them. At every x, K and U add up to E.
Against time, both K and U wiggle between zero and E but never go negative. They peak twice per oscillation, so their graphs repeat every half period. Real oscillators lose a little energy to friction each cycle, so their amplitude slowly shrinks; on the exam, assume no losses unless told otherwise.
Worked examples
Try each one yourself first, then open the solution.
- Example 1Calculator allowed
Energy and speed of a spring oscillator
A 0.20 kg block on a frictionless surface is attached to a spring with k = 80 N/m and oscillates with amplitude 0.050 m. Find (a) the total energy, (b) the maximum speed and (c) the speed when the block is 0.030 m from equilibrium.
Show the solutionHide the solution
- Step 1: (a) E = ½kA² = ½(80)(0.050)² = 0.10 J.
- Step 2: (b) At equilibrium all of it is kinetic: ½mv_max² = 0.10 J, so v_max = √(2(0.10)/0.20) = 1.0 m/s.
- Step 3: (c) At x = 0.030 m, U = ½(80)(0.030)² = 0.036 J, so K = 0.10 − 0.036 = 0.064 J.
- Step 4: v = √(2K/m) = √(2(0.064)/0.20) = √0.64 = 0.80 m/s.
Answer: (a) 0.10 J (b) 1.0 m/s (c) 0.80 m/s
- Example 2Calculator allowed
Speed of a pendulum bob
A 0.40 kg pendulum bob on a 1.5 m string is released from rest at 20° from the vertical. Use g = 9.8 m/s². How fast is it moving at the bottom of its swing?
Show the solutionHide the solution
- Step 1: Height of the release point above the bottom: h = L(1 − cos θ) = 1.5(1 − cos 20°) = 1.5(1 − 0.940) ≈ 0.090 m.
- Step 2: The string's tension is always perpendicular to the bob's motion, so it does no work, and mechanical energy is conserved: mgh = ½mv².
- Step 3: v = √(2gh) = √(2(9.8)(0.090)) ≈ 1.3 m/s. The mass cancels, so the 0.40 kg wasn't needed.
Answer: About 1.3 m/s
- Example 3Calculator allowed
Doubling the amplitude (classic trap)
A block oscillates on a spring with amplitude A and total energy E. It is restarted with amplitude 2A. By what factor do the total energy, the maximum speed and the period change? Then: at what position are the kinetic and potential energies equal?
Show the solutionHide the solution
- Step 1: E = ½kA², so with 2A the energy is ½k(2A)² = 4 × ½kA². Total energy × 4.
- Step 2: ½mv_max² = E, so v_max ∝ √E ∝ A. Maximum speed × 2.
- Step 3: T = 2π√(m/k) has no A in it. Period unchanged.
- Step 4: Equal energies: ½kx² = ½E = ¼kA², so x² = A²/2 and x = A/√2 ≈ 0.71A. The trap is answering x = A/2; there U is only a quarter of E, because U depends on x².
Answer: Energy × 4, maximum speed × 2, period unchanged; K = U at x = ±A/√2 ≈ ±0.71A
Common mistakes
- Assuming energy is split evenly halfway to the turning point. At x = A/2 the potential energy is only E/4.
- Thinking a larger amplitude means a longer period. It means more energy and higher speeds, with the same period.
- Drawing kinetic or potential energy graphs that dip below zero. Both are never negative.
- Using the string length L as the height in mgh for a pendulum. The height is L(1 − cos θ).
On the exam
- Energy bar charts drawn at a turning point, at equilibrium and in between are a common way to show energy in SHM. Keep the total bar height the same at every position.
- Free-response questions often combine 7.2 and 7.4: a change in amplitude changes the energy and maximum speed but not the period. State each effect separately and justify it with the equation.
Connected topics
Videos
Check yourself
4 questions on 7.4 Energy of Simple Harmonic Oscillators. Pick an answer to see if you got it, and why.
A block on an ideal spring with spring constant 200 N/m oscillates with an amplitude of 0.050 m. What is the total mechanical energy of the block–spring system?
A block oscillates on an ideal spring with amplitude A. What fraction of the system's total energy is kinetic energy when the block is at x = A/2?
The amplitude of a mass–spring oscillator is doubled. How does the total mechanical energy of the system change?
A pendulum 1.0 m long is pulled aside so that its bob is 0.10 m higher than its lowest point, then released from rest. What is the bob's speed at the lowest point? Use g = 9.8 m/s².
0 of 4 answered