Defining Simple Harmonic Motion (SHM): Study Guide (Physics C: Mechanics 7.1)

How this differs from the Physics 1 version. Physics C: Mechanics and Physics 1 are separate courses that both have a Topic 7.1 with this title. This guide is the calculus-based one. It writes Newton’s second law as a differential equation and uses derivatives of potential energy to decide whether motion is simple harmonic. The algebra-based treatment is in the Physics 1 study guide; do not mix the two when you revise.

Periodic motion and SHM

Motion that repeats itself in equal time intervals is periodic. A ball bouncing on a floor, a planet in orbit and a block on a spring are all periodic.

Simple harmonic motion (SHM) is a special case of periodic motion. It is not defined by the shape of the path or by how often the motion repeats. It is defined by the force that drives it. To state that force, you need two ideas.

Equilibrium position and restoring force

The equilibrium position is a place where the net force on the object or system is zero. If the object sits there at rest, it stays at rest.

• A glider on a level air track attached to a horizontal spring: equilibrium is where the spring is relaxed.
• A block hanging from a vertical spring: equilibrium is where the spring’s upward pull balances the weight (Worked example 2).

Take the axis along the motion and put x = 0 at equilibrium. Then x is the displacement from equilibrium.

A restoring force points opposite to the displacement: back toward equilibrium. When x > 0 it points in −x, and when x < 0 it points in +x.

The condition for SHM

An object moves in SHM when the net force on it is a restoring force whose size is proportional to the displacement from equilibrium. Put that force into Newton’s second law along x, and you get the equation for this topic:

m a_x = −kΔx

The left side is the net force, F_net,x = m a_x. Δx is the displacement from the equilibrium position, and k is a positive constant (in N/m). The minus sign makes the force restoring. The proportionality means doubling the displacement doubles the force. For an ideal spring on a level, smooth surface, k is the spring constant from Topic 2.8. For other systems, k is an effective spring constant.

If you put the origin at equilibrium, Δx is simply x, so F_net,x = −kx. Now write a_x = d²x/dt²:

m d²x/dt² = −kx, so d²x/dt² = −(k/m)x

This is a second-order differential equation. It says the acceleration is proportional to the displacement and opposite to it. It is the signature of SHM: any system whose equation of motion can be put in this form, for any variable (a position, an arc length, an angle), is in SHM. Topic 7.2 names the constant √(k/m) the angular frequency ω and uses it for the period. Topic 7.3 gives the solution x(t); you do not need to prove that it solves the equation.

Three equivalent tests for SHM:

1. Force law. m a_x = F_net,x = −kΔx about some equilibrium position, with k constant.
2. Graph. A plot of net force (or acceleration) against displacement is a straight line through the equilibrium point with a negative slope.
3. Potential energy. U(x) is a parabola, U = ½kx² + constant, about its minimum, since F_x = −dU/dx = −kx.

Constant forces shift the equilibrium

Suppose F_net,x = F₀ − kx, where F₀ is constant. Set F_net = 0: equilibrium is at x_e = F₀/k. Let u = x − x_e be the displacement from that point. Then F_net = F₀ − k(u + x_e) = −ku. The motion is still SHM, with the same k, centred on x_e. Gravity on a vertical spring is the commonest case.

Small oscillations about any stable equilibrium

This section is an extension: it combines Topic 7.1 with the stability ideas of Topic 3.3, and uses a second derivative that the Topic 7.1 statements do not ask for directly. Many systems have a force that is not exactly linear. Near a stable equilibrium at x₀ (a minimum of U, Topic 3.3), expand U to second order:

U(x) ≈ U(x₀) + ½U″(x₀)(x − x₀)²

The first-derivative term is missing because U′(x₀) = 0 at equilibrium. So for small displacements the force is F_x ≈ −U″(x₀)(x − x₀): approximately SHM, with k_eff = d²U/dx² at the minimum. The approximation gets worse as the displacement grows (Worked example 3). At a maximum of U, U″ < 0, the force pushes the object away, and there is no oscillation at all.

The pendulum: SHM in an angle

A simple pendulum (bob of mass m on a light string of length ℓ) has a restoring torque about the pivot: τ = −mgℓ sin θ. With I = mℓ² and τ = I d²θ/dt² (Topic 5.6):

d²θ/dt² = −(g/ℓ) sin θ

This is not exactly the SHM equation, because sin θ is not proportional to θ. For small angles in radians, sin θ ≈ θ, and the equation becomes d²θ/dt² = −(g/ℓ)θ: SHM in the angle. At θ = 0.20 rad (about 11°), sin θ = 0.199, only 0.67% smaller than θ. Topic 7.2 uses this result for the period.

Worked example 1: is it SHM?

Question. Take +x to the right. Each of three 0.50 kg objects moves along x, and the net force on it is given (x in m, F in N). For each, find any equilibrium position and decide whether the motion is SHM.

(a) F_a = 6.0 − 40x (b) F_b = −250x³ (c) F_c = +30x

1. (a) Set F_a = 0: x_e = 6.0 ÷ 40 = 0.15 m. With u = x − 0.15, F_a = −40u. This is a restoring force proportional to displacement from 0.15 m, so the motion is SHM with k = 40 N/m. The equation of motion is d²u/dt² = −(40 ÷ 0.50)u = −(80 s⁻²)u. At x = 0.25 m (u = +0.10 m), a_x = −8.0 m/s².
2. (b) F_b = 0 only at x = 0. The force is restoring (negative for x > 0, positive for x < 0), so the object oscillates. But F_b/x = −250x² is not constant: at x = 0.10 m the force is −0.25 N, at 0.20 m it is −2.0 N. Doubling x multiplies the force by 8, not 2. The motion is periodic but not SHM.
3. (c) F_c = 0 at x = 0, but at x = +0.10 m the force is +3.0 N, pointing away from equilibrium. Any small push grows. This equilibrium is unstable, and there is no oscillation.

Check. Only (a) passes the straight-line, negative-slope test on an F–x graph. (b) is curved; (c) is straight but has a positive slope.

Worked example 2: a vertical spring

Question. Take +y upward. A 0.80 kg block hangs from an ideal spring with k = 240 N/m. (a) How far is the spring stretched at equilibrium? Let y be the block’s displacement from equilibrium. (b) Show that the net force is −ky. (c) Find the net force and acceleration at y = −5.0 cm and at y = +5.0 cm, and describe the spring force in each case.

1. (a) At equilibrium the spring force balances the weight: kd = mg, so d = (0.80 × 9.8) ÷ 240 = 0.0327 m ≈ 3.3 cm.
2. (b) When the block is at displacement y, the stretch is d − y. Spring force (up) = +k(d − y). Net force: F_net,y = k(d − y) − mg = (kd − mg) − ky = −ky, since kd = mg. So d²y/dt² = −(k/m)y = −(300 s⁻²)y: SHM about the equilibrium position.
3. (c) At y = −0.050 m: stretch = 0.0827 m, spring force = 19.8 N up, weight = 7.84 N down. F_net = +12 N, a_y = +15 m/s².
4. At y = +0.050 m: stretch = 0.0327 − 0.050 = −0.0173 m. The spring is compressed by 1.7 cm and pushes down with 4.16 N. F_net = −4.16 − 7.84 = −12 N, a_y = −15 m/s².

Interpretation. The net force has the same size at equal distances above and below equilibrium, as −ky requires. The spring force alone does not: it is 19.8 N in one case and 4.16 N in the other. The restoring force in SHM is the net force, not the spring force.

Worked example 3: small oscillations in a double well

Question. Take +x to the right. A 0.50 kg particle moves along x in a system with U(x) = (4.0 J/m⁴)x⁴ − (8.0 J/m²)x² (Figure 2). (a) Find the equilibrium positions and their stability. (b) Find k_eff for small oscillations about x = 1.0 m and write the approximate equation of motion. (c) Compare the exact force with the SHM model at x = 1.01 m and x = 1.05 m.

1. (a) F_x = −dU/dx = −16x³ + 16x = 16x(1 − x²). F_x = 0 at x = 0 and x = ±1.0 m. U″ = 48x² − 16. At x = ±1.0 m, U″ = +32 J/m² > 0: stable (minima, U = −4.0 J). At x = 0, U″ = −16 J/m² < 0: unstable (a maximum).
2. (b) k_eff = U″(1.0) = 32 N/m. For small u = x − 1.0, d²u/dt² ≈ −(32 ÷ 0.50)u = −(64 s⁻²)u.
3. (c) At x = 1.01 m: exact F = −0.325 N, model −32 × 0.01 = −0.320 N. They differ by about 1.5%.
4. At x = 1.05 m: exact F = −1.72 N, model −1.60 N. They now differ by about 7%. On the other side, at x = 0.95 m, exact F = +1.48 N against the model’s +1.60 N.

Interpretation. The SHM model is excellent for small displacements and gets worse for larger ones. The errors also differ in sign on the two sides, because the well is not symmetric about x = 1.0 m.

Common misconceptions

• “Any repeating motion is SHM.” Periodic is not enough; the net force must be −kΔx (Worked example 1(b)).
• “Restoring force means spring force.” In a vertical spring, the restoring force is the net of spring force and weight (Worked example 2).
• “Gravity spoils SHM.” A constant force only shifts the equilibrium.
• Measuring x from the wrong point. x must be measured from equilibrium, not from the relaxed spring length, when a constant force acts.
• “A positive force constant means SHM.” F = +kx pushes away from equilibrium: no oscillation.
• Forgetting the small-oscillation condition. k_eff = U″ is only an approximation near the minimum; pendulums need small angles.
• Using degrees in sin θ ≈ θ. The approximation needs radians.

Where this leads

Topic 7.2 turns d²x/dt² = −(k/m)x into the period and frequency of the motion, for springs and small-angle pendulums. Topic 7.3 gives x(t) and the velocity and acceleration graphs, Topic 7.4 adds energy, and Topic 7.5 extends the pendulum to rigid bodies. Try the practice questions now, then use the revision notes and the checklist. When you are ready, move on to Topic 7.2, Frequency and Period of SHM, or return to the course roadmap.