7.4 Energy of simple harmonic oscillators

Four little picture stories

Press Next (or Play) to walk through each story one small step at a time. The numbers come last.

1. Stretch energy and motion energy take turns

2. Halfway out is not half the energy

Where does a block on a spring move fastest?

3. The trampoline

Read the story as text

A gymnast bounces on a trampoline. At the bottom of each bounce the springs are stretched hard and the gymnast stops for an instant. A moment later the springs fling them up, fastest just as the bed passes its flat position. The energy did not disappear at the bottom. It was hiding in the stretched springs.

The question: how is energy shared between motion and the spring during an oscillation, and how do we use that to find speeds and amplitudes?

At the very bottom of a trampoline bounce, where is most of the gymnast's energy?

4. Check yourself

Think of your answer first, then tap to see it.

a) A block on a spring is at the very end of its swing. Where is all the energy?

Show answer

All in the spring. The block is stopped for an instant (no motion energy) and the spring is stretched (or squeezed) the most.

b) The block passes x = A/2, halfway out. What fraction of the energy is motion energy?

Show answer

Three quarters. Spring energy goes with x², so at half the stretch the spring holds (½)² = ¼ of the energy. The other ¾ is motion.

c) You pull the block out twice as far before letting go. What happens to the total energy?

Show answer

4 times as much: E = ½kA², and 2² = 4. (The top speed only doubles, and the period does not change.)

Already know this?

Get all three right on the first try and you can skip ahead.

A spring (k = 100 N/m) oscillates with amplitude 0.10 m. What is its total energy?

At x = A/2, what fraction of the energy is kinetic?

You double the amplitude. The total energy becomes:

An oscillator keeps passing the same energy back and forth: stored in the spring at the ends, moving with the block in the middle. The total stays the same.

The idea: one fixed total, two places to keep it

In pictures

  • At the ends, the block is stopped: all the energy is in the stretched (or squeezed) spring.
  • In the middle, the spring is relaxed: all the energy is motion, and the block is fastest.
  • In between, the energy is shared. Halfway out, the spring holds only a quarter.
  • The total never changes: it just keeps changing places.

Below is the same idea in words, and then with numbers.

For a block on a spring with no friction, the system (block + spring) has two kinds of energy:

K = ½ m v²     Us = ½ k x²

No outside force does work on the system, so the total mechanical energy stays constant:

E = K + Us = constant

Setting the two ends equal gives the top speed: ½kA² = ½mvmax², so vmax = A√(k/m).

Pendulum too: a pendulum swaps kinetic energy and gravitational potential energy (mgh). At the ends it is highest and stopped; at the bottom it is lowest and fastest: mghmax = ½mvmax².

Doubling the amplitude makes the energy 4 times bigger (A²) and the top speed 2 times bigger, but the period does not change.

With numbers

m = 0.40 kg, k = 40 N/m, pulled to A = 0.10 m and released.

Show all steps as text
  1. Total energy: E = ½kA² = ½(40)(0.10)² = ½(40)(0.010) = 0.20 J. At release v = 0, so all the energy is in the spring.
  2. Top speed: ½mvmax² = 0.20 J → vmax = √(2 × 0.20/0.40) = √1.0 = 1.0 m/s. At x = 0 all of E is kinetic.
  3. At x = 0.050 m (A/2): Us = ½(40)(0.050)² = 0.050 J. That is ¼ of 0.20 J, as expected from x².
  4. So K = 0.20 − 0.050 = 0.15 J, and v = √(2 × 0.15/0.40) = √0.75 = 0.87 m/s. Halfway out, it still has 87% of its top speed.

Play: live energy bars

Green is kinetic energy K, blue is spring energy Us, purple is the total E. Step frame by frame: the green and blue bars always add to the purple one. Double the amplitude and watch the total.

kinetic Kspring Ustotal E

Worked examples

Energy from the amplitude basic

A spring with k = 200 N/m oscillates with amplitude 0.050 m. What is its total energy?

Show solution
Show all steps as text
  1. E = ½kA² = ½(200)(0.050)² = ½(200)(0.0025) = 0.25 J.The mass is not needed for E.

The A/2 trap basic

A block oscillates with total energy E. What fraction of E is kinetic when x = A/2?

Show solution
Show all steps as text
  1. Us = ½k(A/2)² = ¼ · ½kA² = ¼E.Spring energy goes with x squared.
  2. K = E − ¼E = ¾E.Not half! Halfway in distance is not halfway in energy.

Amplitude from top speed medium

A 0.50 kg block on a spring with k = 50 N/m has a top speed of 2.0 m/s. Find the amplitude.

Show solution
Show all steps as text
  1. E = ½mvmax² = ½(0.50)(2.0)² = 1.0 J.Top speed happens at x = 0 where all energy is kinetic.
  2. ½kA² = 1.0 J → A² = 2(1.0)/50 = 0.040 m² → A = 0.20 m.At the ends all energy is in the spring.

Double the amplitude medium

The same block is pulled out twice as far. How do E, vmax, amax and T change?

Show solution
Show all steps as text
  1. E = ½kA²: × 4.A is squared.
  2. vmax = A√(k/m): × 2.4 times the energy, and v goes with √E.
  3. amax = kA/m: × 2.Twice the biggest stretch, twice the biggest force.
  4. T = 2π√(m/k): unchanged.Faster, but further to go.

Pendulum energy AP

A pendulum bob is released from rest 0.10 m above its lowest point. Find its speed at the bottom. Does the mass matter? Does the length?

Show solution
Show all steps as text
  1. System: bob + Earth. Only gravity does work (tension is perpendicular to the motion), so K + Ug is constant.AP wants the system and the reason energy is conserved.
  2. mgh = ½mv² → v = √(2gh) = √(2 × 9.8 × 0.10) = √1.96 = 1.4 m/s.Mass cancels.
  3. Neither the mass nor the length matters for this speed; only the drop height h.The length changes the period, not the energy balance for a given h.

Practice

  1. A block on a spring passes through equilibrium (x = 0). At that moment the energy of the block-spring system is

    Show answer

    (A). At x = 0, Us = ½k(0)² = 0, so all of E is kinetic. That is why the speed is greatest there.

  2. The amplitude of a spring oscillator is doubled. The total energy

    Show answer

    (C). E = ½kA², and (2A)² = 4A².

  3. A spring with k = 100 N/m oscillates with amplitude 0.20 m. What is the total energy?

    Show answer

    (D). E = ½(100)(0.20)² = ½(100)(0.040) = 2.0 J. (C) forgets the ½; (B) forgets to square.

  4. What fraction of the total energy is kinetic when the block is at x = A/2?

    Show answer

    (C). Us = ½k(A/2)² = ¼E, so K = ¾E.

  5. A 0.25 kg block on a spring with k = 100 N/m has amplitude 0.10 m. Its top speed is

    Show answer

    (C). E = ½(100)(0.10)² = 0.50 J. ½(0.25)v² = 0.50 → v² = 4.0 → v = 2.0 m/s. Or vmax = A√(k/m) = 0.10 × √400 = 2.0 m/s.

  6. At what position is the kinetic energy equal to the spring potential energy?

    Show answer

    (B). K = Us means Us = ½E: ½kx² = ½ · ½kA² → x² = A²/2 → x = A/√2.

  7. Short answer. Describe the graphs of K, Us and E against position x (from −A to +A) for a frictionless spring oscillator, and explain each shape.

    Show answer

    Us is a U-shaped parabola (½kx²): zero at x = 0, rising to ½kA² at both ends. K is an upside-down parabola: largest (½kA²) at x = 0, zero at both ends. E is a flat horizontal line at ½kA², because with no friction no energy leaves the system. At every x the K and Us heights add up to E.

    Scoring idea: 1 point each for the Us and K shapes with correct zero and max positions; 1 point for the flat E with the reason (no work by outside forces / no friction).

  8. Short answer. In a real lab, a little friction acts on the block. Over many cycles, what happens to (a) the total mechanical energy, (b) the amplitude, (c) the period? Explain.

    Show answer

    (a) Friction does negative work, turning mechanical energy into thermal energy, so E slowly decreases. (b) E = ½kA², so as E falls the amplitude gets smaller. (c) The period stays (very nearly) the same, because T = 2π√(m/k) does not depend on amplitude. This is why a clock pendulum keeps good time even as its swings shrink.

AP question types for this topic

Here are four short free response questions, one of each AP Physics 1 type, all about this topic. Try each one on paper first, then open the worked answer.

1. Mathematical Routines

Nadia's 0.50 kg block sits on a frictionless table, attached to a horizontal spring with k = 200 N/m. She pulls it 0.060 m from equilibrium and lets go.

  1. Calculate the total mechanical energy and the block's largest speed.
  2. Calculate the block's speed when it is 0.030 m from equilibrium.
  3. Derive an expression, in terms of A, for the position where the kinetic energy equals the spring energy, and evaluate it.
Show worked answer and scoring

(a)

  1. At the turning point v = 0, so all the energy is in the spring: E = ½kA² = ½(200)(0.060)² = 0.36 J.
  2. At equilibrium it is all kinetic: ½mvmax² = ½kA², so vmax = A√(k/m) = 0.060√(200/0.50) = 1.2 m/s.

(b)

  1. Energy is shared: ½mv² + ½kx² = ½kA², so v = √((k/m)(A² − x²)).
  2. v = √((200/0.50)(0.060² − 0.030²)) = 1.04 m/s.Halfway out, the speed is still about 87% of vmax, not half.

(c)

  1. K = Us means each is half of E: ½kx² = ½(½kA²), so x² = A²/2 and x = ±A/√2.
  2. x = 0.060/√2 ≈ ±0.042 m (about 71% of the way out, not halfway).

AP-style scoring

  • 1 point for E = ½kA² ≈ 0.36 J, and 1 point for vmax = A√(k/m) ≈ 1.2 m/s.
  • 1 point for the energy equation with both terms and v ≈ 1.04 m/s.
  • 1 point for x = ±A/√2 ≈ ±0.042 m.

2. Translation Between Representations

Tomas's cart (m = 0.20 kg) oscillates on a spring (k = 80 N/m) with amplitude A = 0.10 m on a frictionless track. The energy bar chart shows one moment of the motion; each grid line is the same amount of energy.

Energy bars at one instant, then four possible graphs of kinetic energy K against position x (from −A to +A):

KUsE
xK
(A) highest at the middle, curved
xK
(B) zero at the middle, curved
xK
(C) highest at the middle, straight sides
xK
(D) flat
  1. At this instant, where is the cart? Give x as a fraction of A and in metres.
  2. Find the cart's speed at this instant, both as a fraction of vmax and in m/s.
  3. Which graph, A to D, shows K against x? Justify using the bars.
Show worked answer and scoring

(a)

  1. Us/E = ¼, and Us/E = (½kx²)/(½kA²) = (x/A)². So (x/A)² = ¼ and x = ±A/2.
  2. x = ±0.10/2 = ±0.050 m. (Check: E = ½kA² = 0.40 J, Us = ½(80)(0.050)² = 0.10 J, one quarter. ✓)

(b)

  1. K/E = ¾ = (v/vmax)², so v = (√3/2)vmax ≈ 0.87 vmax.
  2. vmax = A√(k/m) = 0.10√(80/0.20) = 2.0 m/s, so v = 1.73 m/s.

(c)

  1. Graph A. K = E − ½kx²: E is fixed, and the spring part grows as x², so K is largest (= E) at x = 0 and falls to zero at ±A along a curve.
  2. The bars confirm the curve: halfway out (x = A/2) K is still ¾E, not ½E, which a straight-sided graph (C) would give. B has K = 0 at the middle, and D ignores the energy shifting into the spring.

AP-style scoring

  • 1 point for (x/A)² = ¼, and 1 point for x = ±A/2 = ±0.050 m.
  • 1 point for v = (√3/2)vmax with vmax = 2.0 m/s, giving ≈ 1.7 m/s.
  • 1 point for graph A, with the bars (¾E at A/2) used to rule out the straight-sided graph.

3. Experimental Design and Analysis

Priya and Jun have a 0.40 kg cart on a horizontal spring, a low-friction track, a motion sensor that draws x-t and v-t graphs, and a metre stick. They want to find the spring constant k using energy, without hanging masses on the spring.

Their data:

Amplitude A (m)Largest speed vmax (m/s)
0.0400.45
0.0600.66
0.0800.90
0.1001.11
0.1201.35
  1. Describe a procedure to get A and vmax for each trial.
  2. Using energy conservation, what should they graph to get a straight line, and how is k found from it?
  3. Use the data to find k.
Show worked answer and scoring

(a)

  1. Pull the cart to several different distances (measured with the metre stick and checked on the x-t graph, half of peak-to-valley) and release from rest.
  2. Read vmax as the peak height on the v-t graph (where x crosses the middle). Average the first few peaks, since friction slowly shrinks them.

(b)

  1. All spring energy at the end becomes kinetic energy at the middle: ½mvmax² = ½kA².
  2. Graph Kmax = ½mvmax² against A²: a line through the origin with slope ½k, so k = 2 × slope.

(c)

  1. Computed values:
    A² (×10⁻⁴ m²)Kmax = ½mvmax² (J)
    160.0405
    360.0871
    640.1620
    1000.2464
    1440.3645
  2. Best-fit slope ≈ 25.3 J/m² (intercept ≈ -0.0018 J, essentially zero), so k = 2 × 25.3 ≈ 51 N/m.

AP-style scoring

  • 1 point for several amplitudes measured, and 1 point for vmax from v-t peaks with repeats or averaging.
  • 1 point for Kmax (or vmax²) against A² with the slope linked to k.
  • 1 point for k ≈ 50 N/m from a best-fit line.

4. Qualitative/Quantitative Translation

Two identical springs on a frictionless table. Lena attaches a block of mass m, Omar a block of mass 4m, and both pull their blocks out the same distance A before releasing. Omar claims: "Both systems have the same total energy, but my block's top speed is only half of Lena's."

  1. Without equations, explain why the claim is correct.
  2. Derive vmax for each block from energy conservation and show the ratio.
  3. Check with k = 120 N/m, A = 0.050 m and m = 0.30 kg. Which part of the derivation matches each part of the claim?
Show worked answer and scoring

(a)

  1. The energy is put in by stretching the spring, and the stretch is the same for both, so the stored energy is the same: the mass on the end does not change how hard the spring is to pull.
  2. At the middle that same energy is all kinetic. A heavier block needs less speed to carry the same kinetic energy, and since speed is squared, four times the mass needs only half the speed.

(b)

  1. E = ½kA² for both (no m in it). At x = 0: ½mvmax² = ½kA², so vmax = A√(k/m).
  2. vOmar/vLena = √(m/4m) = √(¼) = ½.

(c)

  1. E = ½(120)(0.050)² = 0.15 J for both. Lena: v = 0.050√(120/0.30) = 1.0 m/s; Omar: 0.050√(120/1.20) = 0.5 m/s. ✓
  2. “Same energy” is E = ½kA² with no m in it; “half the speed” is the 1/√m in vmax = A√(k/m).

AP-style scoring

  • 1 point for same stretch so same stored energy, and 1 point for heavier mass needing less speed for the same K.
  • 1 point for vmax = A√(k/m) and the ratio ½.
  • 1 point for the numbers (0.15 J, 1.0 m/s, 0.5 m/s) linked back to the claim.

Common mistakes

The mistake: "Halfway out, the energy is split half and half."

Why it is wrong: Us goes with x². At x = A/2, Us = ¼E and K = ¾E. Equal sharing happens at x ≈ 0.71A.

How to spot it: square the fraction of A to get the fraction of E in the spring.

The mistake: forgetting to square A or x, or dropping the ½.

Why it is wrong: Us = ½kx². With k = 100 N/m and x = 0.20 m that is 2.0 J, not 10 J or 20 J.

How to spot it: energies of classroom springs are usually well under 10 J.

The mistake: thinking bigger energy means a longer period.

Why it is wrong: more energy means a bigger amplitude and higher speeds; the period depends only on m and k.

How to spot it: energy questions change A and vmax; period questions need m, k, L, g.

The mistake: saying "energy is conserved" without naming the system or the reason.

Why it is wrong: on AP free response you must say which system (block + spring, or bob + Earth) and why (no friction, no outside work; tension does no work on a pendulum).

How to spot it: every energy argument should start with "the system is ..."