Waves & Oscillations · Chapter 9

Simple Harmonic Motion

Any system near a stable equilibrium oscillates the same way — from atomic vibrations to suspension bridges.

Prerequisites

Newton's Laws \cdot Energy & Work \cdot Basic calculus (derivatives

Learning Goals

Recognize simple harmonic motion from a restoring force or equation of motion.
Derive the mass-spring frequency and period from Newton's second law.
Explain why a pendulum is only approximately SHM, and identify when the approximation fails.
Use energy conservation to connect amplitude, speed, and total oscillator energy.
Interpret damping, resonance, and phase-space plots qualitatively.

9.1 What is Simple Harmonic Motion?

A system undergoes simple harmonic motion (SHM) when the restoring force is proportional to — and directed opposite to — the displacement from equilibrium. This is Hooke's Law in its most general form, and it applies to an enormous range of physical systems.

Definition 9.1 — Condition for SHM

A system undergoes SHM whenever its equation of motion has the form:

d^{2}x/dt^{2} = -\omega^{2}x

The general solution is x(t) = A cos(\omegat + \phi), where A is amplitude, \omega is angular frequency

and \phi is the initial phase. The period T = 2\pi/\omega depends only on the system — not on A

This last point — isochronism — is profound. A pendulum swinging in a large arc takes the same time as one swinging in a small arc (for small angles). Galileo allegedly discovered this by timing swinging lamps in Pisa Cathedral with his pulse.

The practical test is simple: if the force points back toward equilibrium and grows linearly with displacement, the motion is sinusoidal. If the restoring force is only approximately linear, the system behaves like SHM only near equilibrium.

9.2 The Mass-Spring System

A mass m on a spring of stiffness k is the canonical SHM system. The restoring force is F = −kx (Hooke's Law), giving Newton's second law:

m d^{2}x/dt^{2} = -kx \qquad \to \qquad d^{2}x/dt^{2} = -(k/m)x

(9.1)

Comparing with Definition 9.1, we see ω² = k/m, so:

\omega = \sqrt(k/m) \qquad T = 2\pi\sqrt(m/k) \qquad f = (1/2\pi)\sqrt(k/m)

(9.2)

A stiffer spring (larger k) oscillates faster; a heavier mass oscillates slower. The amplitude has no effect on the period.

Theorem 9.1 — Energy in SHM

The total mechanical energy of a mass-spring system is constant:

E = \frac{1}{2}mv^{2} + \frac{1}{2}kx^{2} = \frac{1}{2}kA^{2} = constant

At the equilibrium position (x = 0), all energy is kinetic: v_{max} = A\sqrt(k/m) = A\omega. At maxim

um displacement (x = \pmA), all energy is potential, and v = 0. Energy sloshes back and fort

h between KE and PE at twice the oscillation frequency.

Example 9.1 — Spring-Mass Period

A 0.5 kg mass hangs on a spring. When pulled 8 cm and released, it oscillates with period 1.2 s. Find the spring constant k.

Use T = 2π√(m/k):

T^{2} = 4\pi^{2}m/k \to k = 4\pi^{2}m/T^{2} = 4\pi^{2}(0.5)/(1.2)^{2} =

13.7 N/m

Max speed:

v_{max} = A\omega = A(2\pi/T) = (0.08)(2\pi/1.2) =

0.419 m/s

9.3 The Simple Pendulum

A pendulum of length L, displaced by a small angle θ₀, experiences a restoring torque τ = −mgL sin θ ≈ −mgLθ for small θ. This gives SHM with:

\omega = \sqrt(g/L) \qquad T = 2\pi\sqrt(L/g)

(9.3)

Note that T is independent of both mass and amplitude (for small angles). A pendulum of length L = 1 m on Earth has T ≈ 2.006 s — this is the basis of the seconds pendulum used in early clocks. The approximation breaks down above about 15°; at 90° the true period is about 18% longer.

Figure 9.1. Pendulum simulation with phase-space portrai

t. Try a small angle first: the phase plot is nearly an ellipse and the period follows T =

nonlinear distor

tion, increase damping to watch the orbit spiral inward, and change length to verify the \sqrt

Theorem 9.2 — Small-Angle Approximation

The pendulum equation is exactly \theta¨ = -(g/L) sin \theta. It becomes SHM only after the approxim

ation sin \theta \approx \theta

\theta¨ = -(g/L)\theta \qquad only when |\theta| is small

Mass cancels from the torque equation, so the period does not depend on bob mass. Amplitude cancels only in the small-angle limit; large-amplitude pendulums run slow.

Example 9.2 — Pendulum on the Moon

$A pendulum has period T = 2.0 s on Earth. What is its period on the Moon (g_{Moon} = 1.62 m/$ $s^{2}$ ?

Find length:

T_{E} = 2\pi\sqrt(L/g_{E}) \to L = g_{E}(T_{E}/2\pi)^{2} = 9.81(2/2\pi)^{2} =

0.994 m

Moon period:

T_{M} = 2\pi\sqrt(L/g_{M}) = 2\pi\sqrt(0.994/1.62) = 2\pi \times 0.783 =

4.92 s

Ratio:

T_{M}/T_{E} = \sqrt(g_{E}/g_{M}) = \sqrt(9.81/1.62) = 2.46 — the pendulum runs 2.46\times slower

9.4 Damped and Driven Oscillations

Real oscillators lose energy to friction and air resistance. The equation of motion with a damping force −bẋ (proportional to velocity) is:

m ẍ + b ẋ + kx = 0 \qquad \to \qquad x(t) = Ae^{-\gammat} cos(\omega′t + \phi)

(9.4)

where γ = b/2m is the damping coefficient and ω′ = √(ω₀² − γ²) is the damped frequency. When a periodic driving force F₀ cos(ωt) is added, the system reaches a steady state with amplitude:

A(\omega) = F_{0}/m / \sqrt((\omega_{0}^{2} - \omega^{2})^{2} + (b\omega/m)^{2})

(9.5)

The amplitude peaks near ω = ω₀ — this is resonance. At resonance, even a small driving force can build up a very large amplitude if damping is small. This destroyed the Tacoma Narrows Bridge in 1940 and must be engineered around in every building, bridge, and engine.

Definition 9.2 — Common Traps

Amplitude independence is not universal: it is exact for ideal springs, approximate for pendulums.
Mass does not affect a simple pendulum's period: heavier bobs have larger weight and larger inertia in the same proportion.
Damping removes energy: the phase-space orbit spirals inward instead of closing on itself.
$Resonance is not always exactly at \omega_{0}$ : damping shifts the maximum response slightly lower.

Exercises — 9.1–9.4 Simple Harmonic Motion

A 0.5 kg block is attached to a spring with k = 10 N/m. What is the period of oscillation

Straightforward

What length pendulum has a period of exactly 2.0 s on Earth? (g = 9.81 m/s^{2}

Straightforward

A 0.5 kg block on a spring (k = 18 N/m) is displaced 10 cm from equilibrium and released f

rom rest. What is the maximum speed it reaches?

m/s

Intermediate

Three pendulums all have length L = 1 m: (a) mass 100 g, amplitude 5°; (b) mass 500 g, amp

litude 5°; (c) mass 100 g, amplitude 30°. Which has the longest period, and why?

Intermediate

A driven oscillator has resonance amplitude 5 cm with damping coefficient b = 0.2 N\cdots/m. H

ow would you reduce the resonance amplitude to 2.5 cm? Give two independent methods.

Challenging

Key Takeaways

$SHM occurs whenever restoring force \propto -displacement: d^{2}x/dt^{2} = -\omega^{2}x$ .
$Mass-spring: \omega = \sqrt(k/m) \cdot period T = 2\pi\sqrt(m/k) — independent of amplitude$ .
$Simple pendulum: T = 2\pi\sqrt(L/g) — independent of mass and amplitude (small angles$ .
$Energy alternates between KE and PE; total E = \frac{1}{2}kA^{2} = constant$ .
$Resonance: driven at \omega_{0}, amplitude grows dramatically — limited only by damping$ .