Energy of Simple Harmonic Oscillators: Study Guide (Physics 1 7.4)

This guide is for the algebra-based Physics 1 course. It uses the energy ideas from Unit 3 and needs no calculus. Topic 7.3 described where an oscillator is fastest; energy explains why, and lets you calculate how fast.

The system and its energy

Choose the system first. For a mass on a spring, take the spring–object system. For a pendulum, take the bob–Earth system. The total mechanical energy is:

E_total = K + U

• Kinetic energy: K = ½mv². It is never negative.
• Spring potential energy: U = ½kx², with x measured from the spring’s relaxed length. On a level surface that is also the equilibrium position.
• Gravitational potential energy: U = mgh, with h measured up from the lowest point of the swing.

If there is no friction or air resistance and nothing outside the system does work on it, the total energy stays constant. The energy moves back and forth between K and U, but the sum does not change.

Trading energy back and forth

Follow a spring–object system through one swing:

• At a turning point (x = ±A): the object is momentarily at rest, so K = 0. This is the smallest possible kinetic energy. The spring is stretched or compressed the most, so U is at its largest: U = ½kA².
• At equilibrium (x = 0): on a level surface the spring is relaxed there, so U = 0, its smallest value. All the energy is kinetic, so K is at its largest.
• In between: some of each. K + U is the same everywhere.

Because the total energy equals the potential energy at a turning point:

E_total = ½kA² (spring–object system)

At equilibrium all of that is kinetic energy:

½mv_max² = ½kA², so v_max = A√(k/m)

At any displacement x:

½mv² + ½kx² = ½kA²

This last line is the most useful one. It links speed and position without needing any time.

Energy bar charts

An energy bar chart shows each kind of energy as a bar at one instant. For an oscillator, the rules are:

• Every bar starts at the zero line. K can never go below it.
• The total height of the bars is the same at every instant (if no energy leaves the system).
• A zero quantity is shown as a line on the zero mark, not left blank.

For the puck in Worked example 1, the bars at three positions would be:

Position	K (J)	U (J)	Total (J)
x = +0.15 m (turning point)	0	0.36	0.36
x = +0.090 m	0.23	0.13	0.36
x = 0 (equilibrium)	0.36	0	0.36

Energy against time

K and U also change with time. Starting from x = +A, U begins at its largest value and K at zero. After a quarter period they have swapped; after half a period they have swapped back. So each energy reaches its maximum twice in every period of the motion. Their sum is a flat line.

Changing the amplitude

Since E = ½kA², the total energy depends on the square of the amplitude:

Change	Total energy	Maximum speed	Period
A doubled	× 4	× 2	unchanged
A tripled	× 9	× 3	unchanged
A halved	× ¼	× ½	unchanged

The maximum speed scales with A because v_max = A√(k/m). The period does not change, as Topic 7.3 showed. Changing the amplitude changes the largest potential energy, and so the total energy.

Worked example 1: speed from energy

Question. A 0.50 kg puck on a frictionless air table is attached to a spring with k = 32 N/m. Take +x along the spring, from equilibrium. The puck is pulled to x = 0.15 m and released. Find (a) the total energy, (b) the maximum speed and (c) the speed at x = 0.090 m.

1. (a) At release, K = 0 and all the energy is spring energy: E = ½kA² = ½ × 32 × 0.15² = 0.36 J.
2. (b) At equilibrium U = 0, so ½mv_max² = 0.36 J. v_max = √(2 × 0.36 ÷ 0.50) = √1.44 = 1.2 m/s.
3. (c) At x = 0.090 m: U = ½ × 32 × 0.090² = 0.1296 J ≈ 0.13 J. So K = 0.36 − 0.1296 = 0.2304 J ≈ 0.23 J.
4. v = √(2K/m) = √(2 × 0.2304 ÷ 0.50) = √0.9216 = 0.96 m/s.

Check. v_max = A√(k/m) = 0.15 × √64 = 0.15 × 8.0 = 1.2 m/s, which agrees. At 60% of the amplitude the puck still has 80% of its top speed: it slows down mostly near the turning points.

Worked example 2: a pendulum

Question. A 0.15 kg bob hangs on a light string 0.60 m long. It is pulled aside until it is 0.015 m higher than its lowest point and released from rest. Take the bob–Earth system and h = 0 at the lowest point. Find (a) the total energy, (b) the speed at the bottom, and (c) the speed when the bob is 0.0075 m above the bottom.

1. (a) At release K = 0: E = mgh = 0.15 × 9.8 × 0.015 = 0.022 J (0.02205 J).
2. (b) At the bottom U = 0, so ½mv² = mgh and v_max = √(2gh) = √(2 × 9.8 × 0.015) = 0.54 m/s.
3. (c) At h = 0.0075 m, half the energy is still potential, so K = 0.02205 − 0.011025 = 0.011025 J. v = √(2 × 0.011025 ÷ 0.15) = 0.38 m/s.

Interpretation. The mass cancels in step 2. A 0.30 kg bob released from the same height has twice the energy (0.044 J) but the same speed at the bottom. The angle here is about 13°, small enough for the motion to be close to SHM. Energy conservation itself works at any angle.

Worked example 3: finding the amplitude from a push

Question. A 0.50 kg cart on a level, frictionless track sits at rest at the equilibrium position of a spring with k = 18 N/m. It is given a quick push so that it starts moving at 0.30 m/s. (a) Find the amplitude. (b) The experiment is repeated with a starting speed of 0.60 m/s. Find the new amplitude, and compare the energies and periods.

1. (a) The cart starts at x = 0, so all the energy is kinetic: E = ½ × 0.50 × 0.30² = 0.0225 J.
2. At the turning point all of it is spring energy: ½kA² = 0.0225 J, so A = √(2 × 0.0225 ÷ 18) = √0.0025 = 0.050 m.
3. (b) E = ½ × 0.50 × 0.60² = 0.090 J, which is 4 times as much. A = √(2 × 0.090 ÷ 18) = 0.10 m, twice as far.
4. Period in both cases: T = 2π√(0.50 ÷ 18) = 1.0 s (1.05 s unrounded). It does not depend on the push.

Check. v_max = A√(k/m) gives 0.050 × 6.0 = 0.30 m/s and 0.10 × 6.0 = 0.60 m/s, matching the starting speeds.

Quick checks with extreme cases

• x = 0: ½mv² + 0 = ½kA² gives v = v_max, as it should.
• x = ±A: ½mv² + ½kA² = ½kA² gives v = 0, the turning point.
• A → 0: no energy, no motion. The object just sits at equilibrium.
• A stiffer spring, same amplitude: E = ½kA² grows with k, and so does v_max. The period gets shorter, because T = 2π√(m/k).

Background: real oscillators lose energy

Real springs and pendulums slowly stop. Friction and air resistance do negative work and turn mechanical energy into thermal energy, so E falls and the amplitude shrinks. The SHM model in this course assumes these effects are negligible over the time you study.

Common misconceptions

• “The kinetic energy is largest at the turning points, where the force is largest.” The force is largest there, but the object is at rest, so K = 0.
• “Energy is lost when the object stops.” It is stored as potential energy and returned.
• “At half the amplitude, the energy is half kinetic and half potential.” At x = A/2, U = ¼E and K = ¾E. K equals U at x = A/√2 ≈ 0.71A.
• “Doubling the amplitude doubles the energy.” It multiplies the energy by 4, because E = ½kA².
• “A heavier pendulum bob swings faster at the bottom.” v_max = √(2gh) does not depend on mass (Worked example 2).
• “More energy means a longer period.” The period depends on m and k (or ℓ and g), not on the energy.

Where this leads

This completes Unit 7. Next, Unit 8 turns to fluids, starting with Topic 8.1, Internal Structure and Density. Before moving on, try the practice questions, then use the revision notes and the checklist. To review graphs of SHM, go back to Topic 7.3, or return to the course roadmap.