Unit 7 · Topic 7.4

7.4 Energy of Simple Harmonic Oscillators

In SHM, energy keeps changing form while the total stays constant. Kinetic energy is largest at equilibrium and potential energy is largest at the turning points. For a spring–object system, $E = \tfrac{1}{2}kA^2$ , so a bigger amplitude means more energy but the same period.

Key terms

kinetic energy
spring potential energy
total mechanical energy
amplitude
conservation of energy

What changed in the 2024 update

Before the Fall 2024 course update, oscillations were Unit 6. They're now Unit 7, so older videos and practice may call this unit "Unit 6."

Energy back and forth

For a mass on a spring with no friction, the total mechanical energy is constant: $E = K + U = \tfrac{1}{2}mv^2 + \tfrac{1}{2}kx^2$

At a turning point (x = ±A), the object is momentarily at rest. All the energy is potential: $E = \tfrac{1}{2}kA^2$ . At equilibrium (x = 0), U = 0 and all the energy is kinetic: $E = \tfrac{1}{2}mv_{\max}^2$ .

Setting those equal gives $v_{\max} = A\sqrt{k/m} = A\omega$ , the same result as 7.3. At any position, $v = \omega\sqrt{A^2 - x^2}$

Amplitude and energy

Total energy depends on the square of the amplitude. Double A and the energy is four times as large. The maximum speed doubles. The period stays the same.

The minimum kinetic energy is always zero, at the turning points. The minimum potential energy is zero at equilibrium, if you measure x from there.

Energy also lets you find the amplitude from a single snapshot. If you know the position x and speed v at one moment, then $\tfrac{1}{2}kA^2 = \tfrac{1}{2}mv^2 + \tfrac{1}{2}kx^2$ , which you can solve for A.

Energy graphs

Against position: U = ½kx² is a parabola with its minimum at x = 0. The total energy E is a horizontal line at height ½kA². K is the gap between them, an upside-down parabola that's largest at x = 0 and zero at x = ±A. The object can only be where U ≤ E, which is why it turns around at ±A.

Against time: K and U each swing between 0 and E, and they always add to E. They repeat twice per oscillation, because K is at its maximum each time the object passes through equilibrium, going either way. So K and U have period T/2, half the motion's period.

Other oscillators

The same pattern holds for a pendulum: gravitational potential energy mgh is greatest at the ends of the swing and kinetic energy is greatest at the bottom.

For a vertical spring, measure y from the hanging equilibrium position. Then the spring's and gravity's potential energies together act like ½ky², plus a constant you can ignore. So the energy analysis works just like the horizontal case.

If friction is present, the total energy slowly drops and so does the amplitude. The ideal SHM results assume no energy is lost.

Worked examples

Try each one yourself first, then open the solution.

Example 1Calculator allowed
Energy and speed of a spring oscillator
A 0.50 kg block on a spring (k = 50 N/m) oscillates with amplitude 0.10 m. Find the total energy, the maximum speed and the speed when x = 0.050 m.
Show the solutionHide the solution
1. Step 1: $E = \tfrac{1}{2}kA^2 = \tfrac{1}{2}(50)(0.10)^2 = 0.25$ J.
2. Step 2: At equilibrium all of it is kinetic: $0.25 = \tfrac{1}{2}(0.50)v_{\max}^2$ , so $v_{\max} = 1.0$ m/s.
3. Step 3: At x = 0.050 m: $U = \tfrac{1}{2}(50)(0.050)^2 = 0.0625$ J, so K = 0.25 − 0.0625 = 0.1875 J and $v = \sqrt{\dfrac{2(0.1875)}{0.50}} \approx 0.87$ m/s.
4. Step 4: Check: $\omega = \sqrt{50/0.50} = 10$ rad/s and $\omega\sqrt{A^2 - x^2} = 10\sqrt{0.0075} \approx 0.87$ m/s.
Answer: E = 0.25 J, v_max = 1.0 m/s, v ≈ 0.87 m/s at x = 0.050 m
Example 2Calculator allowed
Where are K and U equal? (classic trap)
For the oscillator above, at what position are the kinetic and potential energies equal?
Show the solutionHide the solution
1. Step 1: The trap is answering x = A/2. At A/2, U = ½k(A/2)² = ¼ of E, so K is ¾ of E. Not equal.
2. Step 2: K = U means each is half of E: $\tfrac{1}{2}kx^2 = \tfrac{1}{2}\cdot\tfrac{1}{2}kA^2$ , so $x = \dfrac{A}{\sqrt{2}}$ .
3. Step 3: $x = \dfrac{0.10}{\sqrt{2}} \approx \pm 0.071$ m.
Answer: x = ±A/√2 ≈ ±0.071 m
Example 3
Changing the amplitude
The same block is set oscillating with amplitude 0.20 m instead of 0.10 m. By what factor does each of these change: total energy, maximum speed, maximum acceleration, period?
Show the solutionHide the solution
1. Step 1: Energy goes as A²: ×4, to 1.0 J.
2. Step 2: Maximum speed Aω goes as A: ×2, to 2.0 m/s.
3. Step 3: Maximum acceleration Aω² goes as A: ×2.
4. Step 4: Period 2π√(m/k) has no A in it: no change.
Answer: Energy ×4, v_max ×2, a_max ×2, period unchanged

Common mistakes

Assuming K = U at x = A/2. They're equal at x = A/√2.
Thinking a larger amplitude means a longer period. It means more energy and a higher maximum speed, with the same period.
Giving K and U vs. time graphs the same period as the motion. They repeat twice per cycle.
Drawing a total energy bar that changes in a frictionless system. It stays the same height.

On the exam

Energy bar charts at the turning point and at equilibrium are a favorite. Make the total the same height in both, and show which bar is zero at each point.
When asked how a change affects energy, speed and period, answer each separately, with the dependence (like E ∝ A²) as your reason.

Connected topics

Videos

Topic 7.4 - Energy of SHO
Lessons With LondotWatch on YouTube (opens in a new tab)
Energy of spring-mass oscillators | AP Physics | Khan Academy
Khan AcademyWatch on YouTube (opens in a new tab)
Total Mechanical Energy in Simple Harmonic Motion
Flipping PhysicsWatch on YouTube (opens in a new tab)
Energy In a Simple Harmonic Oscillator - Maximum Velocity & Acceleration Calculations
The Organic Chemistry TutorWatch on YouTube (opens in a new tab)
Energy in Simple Harmonic Motion
Physics with Professor Matt AndersonWatch on YouTube (opens in a new tab)

Check yourself

4 questions on 7.4 Energy of Simple Harmonic Oscillators. Pick an answer to see if you got it, and why.

Question 1 of 4Calculator allowed

A block on a spring with spring constant 200 N/m oscillates with amplitude 0.050 m on a frictionless surface. What is the total mechanical energy of the block–spring system?

Question 2 of 4Calculator allowed

A block on a spring is restarted with twice its original amplitude. Which correctly describes the new total energy and maximum speed?

Question 3 of 4Calculator allowed

A block on a spring has total energy E. What is its kinetic energy when it is halfway between equilibrium and a turning point?

Question 4 of 4Calculator allowed

A block oscillates on a spring with amplitude A. At what distance from equilibrium are the kinetic energy and the spring potential energy equal?

0 of 4 answered