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

How this differs from the Physics 1 version. Physics C: Mechanics and Physics 1 are separate courses that both have a Topic 7.4 called Energy of Simple Harmonic Oscillators. This guide is the calculus-based one. It proves that the total energy is constant, writes K and U as functions of time, and links energy to the amplitude and phase from Topic 7.3. The algebra-based treatment is in the Physics 1 study guide; do not mix the two when you revise.

The system and its energy

Choose the system first. For a block on a horizontal spring, take the block–spring system. Its total mechanical energy is the sum of two parts:

E = K + U, with K = ½mv_x² and U = ½kx²

Here x is measured from equilibrium, where the spring is at its natural length. The potential energy comes from Topic 3.3: U = −∫F_x dx = −∫(−kx) dx = ½kx², taking U = 0 at x = 0.

If the system were only the block, the spring would be outside it. Then the spring would do work on the block and the block’s energy would not be constant. Including the spring in the system turns that work into a change of internal potential energy.

Why the total energy is constant

From Newton’s second law. For the block, m dv_x/dt = −kx. Differentiate E with respect to time, using the chain rule:

dE/dt = mv_x (dv_x/dt) + kx (dx/dt) = v_x (m dv_x/dt + kx)

The bracket is zero by Newton’s second law. So dE/dt = 0: the total energy does not change. This is conservation of energy for a system with no friction and no external work.

From x(t). Use x = A cos(ωt + φ₀) and v_x = −Aω sin(ωt + φ₀) from Topic 7.3, with k = mω²:

• U = ½kA² cos²(ωt + φ₀)
• K = ½mA²ω² sin²(ωt + φ₀) = ½kA² sin²(ωt + φ₀)
• K + U = ½kA² (sin² + cos²) = ½kA²

So E = ½kA². The amplitude fixes the energy, and the energy fixes the amplitude.

Trading energy back and forth

The total stays the same, but the split changes all the time:

Position	K	U
Turning point, x = ±A	0 (minimum)	½kA² = E (maximum)
Equilibrium, x = 0	½mv_max² = E (maximum)	0 (minimum)
General x	½k(A² − x²)	½kx²

Kinetic energy is greatest where potential energy is least, and the other way round. The smallest kinetic energy is zero, at each turning point.

Setting ½mv_max² = ½kA² gives v_max = A√(k/m) = Aω, the same result Topic 7.3 found by differentiating.

Where are K and U equal? Set ½kx² = ½k(A² − x²): x² = A²/2, so x = ±A/√2 ≈ ±0.71A. At x = A/2 the potential energy is only a quarter of the total, because U depends on x².

Energy against position

On an energy–position graph (Figure 1) U = ½kx² is a parabola and E is a horizontal line. At any position, the gap between the line and the parabola is the kinetic energy. The object can only be where K ≥ 0, so it is trapped between the two points where the line meets the parabola. Those points are the turning points, x = ±A.

Energy against time

Using the double-angle identities, with φ₀ = 0:

U(t) = ½E[1 + cos(2ωt)] and K(t) = ½E[1 − cos(2ωt)]

Each one oscillates at angular frequency 2ω, so it repeats every T/2. That makes sense: U reaches its maximum at x = +A and again at x = −A, twice per cycle. Both average to E/2 over a full cycle. The sum is a flat line at E (Figure 2).

Changing the amplitude, mass or spring

• Amplitude. E = ½kA², so doubling A multiplies E by 4. v_max = Aω doubles. The period does not change (Topic 7.3).
• Mass, at the same amplitude. E is unchanged, because it contains only k and A. A heavier object has a smaller v_max = A√(k/m) and a longer period.
• Spring constant, at the same amplitude. A stiffer spring stores more energy, E ∝ k, and gives a shorter period.

Vertical springs and pendulums

Vertical spring. Hang a mass on a spring. At equilibrium the spring is already stretched by d = mg/k. Measure y from this new equilibrium, +y upward. The spring potential energy is ½k(d − y)² and the gravitational potential energy is mgy. Their sum is ½kd² − kdy + ½ky² + mgy. Because kd = mg, the two middle terms cancel:

U_total = ½ky² + constant

So the energy pattern is the same as for a horizontal spring, as long as you measure from the hanging equilibrium. The minimum of U_total is not zero, but that does not matter: only changes in U appear in the energy equation.

Pendulum. For a bob of mass m on a string of length L, U = mgL(1 − cos θ). For small angles this is very close to ½(mg/L)s², where s = Lθ is the arc length. So a pendulum behaves like a spring with “k” = mg/L, and all the energy ideas above apply. Topic 7.5 develops this.

Worked example 1: speed and position from energy

Question. Take +x to the right, origin at equilibrium. A 0.60 kg block on a frictionless surface is attached to a spring with k = 150 N/m. It oscillates with amplitude 0.080 m. Find (a) the total energy, (b) the maximum speed, (c) the speed at x = +0.040 m, and (d) where, and how soon after passing equilibrium, the kinetic and potential energies are equal.

1. (a) E = ½kA² = ½ × 150 × 0.080² = 0.48 J.
2. (b) ½mv_max² = E, so v_max = √(2 × 0.48/0.60) = 1.26 m/s. Check: ω = √(150/0.60) = 15.8 rad/s and Aω = 0.080 × 15.8 = 1.26 m/s.
3. (c) U = ½ × 150 × 0.040² = 0.12 J, so K = 0.48 − 0.12 = 0.36 J and v = √(2 × 0.36/0.60) = 1.10 m/s. Halfway out, the block still has 75% of its kinetic energy.
4. (d) K = U when x = ±A/√2 = ±0.057 m. Starting at equilibrium, x = A sin(ωt), so sin(ωt) = 1/√2 and ωt = π/4. t = π/(4 × 15.8) = 0.050 s, which is T/8.

Interpretation. In the first T/8 after equilibrium the block covers 0.057 m. In the next T/8 it covers only 0.080 − 0.057 = 0.023 m, because it is slowing down as energy moves into the spring.

Worked example 2: amplitude from a starting position and velocity

Question. Take +x to the right, origin at equilibrium. A 0.40 kg glider on a frictionless track is attached to a spring with k = 40 N/m. At t = 0 it is at x = +0.050 m moving at +1.2 m/s. Find the total energy, the amplitude, v_max and the period. What would the amplitude be if the glider had been given twice as much energy?

1. Energy at the start: U₀ = ½ × 40 × 0.050² = 0.050 J; K₀ = ½ × 0.40 × 1.2² = 0.288 J. E = 0.338 J.
2. Amplitude: at a turning point all of E is potential: ½kA² = 0.338 J, so A = √(2 × 0.338/40) = 0.13 m.
3. Maximum speed: v_max = √(2E/m) = √(2 × 0.338/0.40) = 1.3 m/s.
4. Period: ω = √(40/0.40) = 10 rad/s, so T = 2π/10 = 0.63 s. It does not depend on the energy.
5. Twice the energy: A ∝ √E, so A′ = √2 × 0.13 = 0.18 m. The period is still 0.63 s.

Check. Topic 7.3’s formula gives the same amplitude: A = √(x₀² + (v_x0/ω)²) = √(0.050² + 0.12²) = 0.13 m. The energy method avoids the phase and is often quicker.

Note. The glider has the same speed, 1.2 m/s, at x = −0.050 m, because U depends on x², not x.

Common misconceptions

• “Doubling the amplitude doubles the energy.” E ∝ A², so it quadruples.
• “K = U at half the amplitude.” At A/2, U is only E/4. They are equal at A/√2.
• “K and U oscillate at the same frequency as x.” They repeat twice per cycle, every T/2.
• “A heavier block on the same spring and amplitude has more energy.” E = ½kA² has no m in it.
• “The energy is zero at equilibrium.” Only U is zero there; K is at its maximum.
• Leaving the spring out of the system. Then the block’s energy alone is not conserved.
• Using x measured from the natural length for a vertical spring. Measure from the hanging equilibrium, or include gravitational energy separately.

Where this leads

Topic 7.5 (Simple and Physical Pendulums) applies Newton’s second law in rotational form to swinging bodies and uses the energy pattern you met here: read the Topic 7.5 study guide. Now try the practice questions, then use the revision notes and the checklist. You can also return to the course roadmap.