Difficulty: Intermediate | Prerequisites: PHYS 121 mechanics (forces, energy conservation), basic calculus
Simple harmonic motion is the backbone of PHYS 123. Nearly every wave phenomenon you will meet this quarter, from standing waves on strings to sound in air, rests on the maths of oscillation. If you can write down the position, velocity, and acceleration of a mass on a spring, and if you understand how energy flows between kinetic and potential forms, you already have the toolkit for everything that follows. You should be comfortable with trigonometric functions, basic calculus (derivatives and integrals of sine and cosine), and conservation of energy from PHYS 121.
A mass-spring system oscillates sinusoidally. Its displacement, velocity, and acceleration are all determined by the amplitude, angular frequency, and phase. Energy swaps between kinetic and potential, but the total stays constant. Standing waves on strings are the spatial cousin of this idea, with boundary conditions setting which frequencies are allowed.
Simple harmonic motion (SHM)
Motion in which the restoring force is directly proportional to the displacement from equilibrium and directed toward it. Think of it as: the further you pull, the harder the system pulls back, and it does so in exact proportion.
Angular frequency (omega, ω)
The rate of oscillation measured in radians per second. Related to the period by ω = 2π/T. In simple terms, this tells you how quickly the system cycles through its motion.
Amplitude (A)
The maximum displacement from the equilibrium position. This sets the scale of the motion but does not affect the frequency or period.
Period (T)
The time for one complete oscillation. T = 2π/ω = 1/f. In simple terms, this is the time it takes the mass to go out and come back.
Phase constant (φ)
The angle (in radians) that sets where in its cycle the oscillation starts at t = 0. Think of it as the "starting position" on the oscillation's clock.
Equilibrium position
The point where the net force on the object is zero. For a spring, this is its natural (unstretched) length. Stable equilibrium means a small nudge produces a restoring force back toward the point; unstable equilibrium means a small nudge drives the object further away.
Kinetic energy in SHM
KE = ½mv². At equilibrium the speed is maximum and all energy is kinetic. At the turning points the speed is zero and all energy is potential.
Potential energy in SHM
PE = ½kx². Maximum at the turning points (x = ±A), zero at equilibrium.
Position: x(t) = A cos(ωt + φ)
Velocity: v(t) = -Aω sin(ωt + φ)
Acceleration: a(t) = -Aω² cos(ωt + φ) = -ω²x
Maximum speed occurs at equilibrium (x = 0): v_max = Aω
Maximum acceleration occurs at the turning points (x = ±A): a_max = Aω²
When you know x₀ and v₀ at t = 0, the amplitude comes from energy conservation:
Total energy E = ½kA² = ½kx₀² + ½mv₀²
Rearranging: A² = x₀² + (v₀/ω)²
The speed at equilibrium equals Aω (all energy is kinetic there)
a_max = Aω². If you double both A and ω, the new maximum acceleration is (2A)(2ω)² = 8Aω², which is eight times the original. This is a common exam trap: students multiply by 4 instead of 8 because they forget the ω is squared.
Total mechanical energy: E = ½kA² (constant throughout the motion)
At any position x: KE = ½k(A² - x²), PE = ½kx²
At equilibrium: KE = E, PE = 0
At turning points: KE = 0, PE = E
If KE = ½ E, then ½k(A² - x²) = ½(½kA²), giving x² = A²/2, so x = A/√2
If a mass starts at x = 0 moving in the +x direction and at t = 1.0 s has KE = ½ of its initial KE (while still moving in the +x direction), then x = A/√2 at t = 1.0 s. That is one-eighth of a full cycle (from 0 to A/√2 is the first eighth). The period is therefore T = 8.0 s.
Given a force F(x), equilibrium points are where F(x) = 0. To test stability:
Compute dF/dx at each equilibrium point
If dF/dx < 0, the equilibrium is stable (a restoring force)
If dF/dx > 0, the equilibrium is unstable
For F(x) = -x(x² - 1) = -x³ + x:
F(x) = 0 at x = 0, x = 1, x = -1 (three equilibrium points)
dF/dx = -3x² + 1
At x = 0: dF/dx = +1 > 0, so unstable
At x = 1: dF/dx = -2 < 0, so stable
At x = -1: dF/dx = -2 < 0, so stable
Result: three equilibrium points, two of them stable
x(t) = A\cos(\omega t + \phi)v(t) = -A\omega\sin(\omega t + \phi)a(t) = -A\omega^2\cos(\omega t + \phi) = -\omega^2 xv_{\max} = A\omega \qquad a_{\max} = A\omega^2A = \sqrt{x_0^2 + \left(\frac{v_0}{\omega}\right)^2}E = \tfrac{1}{2}kA^2 = \tfrac{1}{2}kx^2 + \tfrac{1}{2}mv^2\omega = \sqrt{\frac{k}{m}} = \frac{2\pi}{T}A fixed end (tied to a wall) is always a node.
A free end (driven up and down) is always an antinode.
The pattern of nodes and antinodes determines which wavelengths fit on a string of length L.
Both ends are nodes.
Allowed wavelengths: λ_n = 2L/n, where n = 1, 2, 3, ...
Allowed frequencies: f_n = nv/(2L)
One end is a node, the other is an antinode.
The fundamental mode fits one quarter-wavelength on the string: λ₁ = 4L
Fundamental frequency: f₁ = c/(4L)
Only odd harmonics exist: f₁, 3f₁, 5f₁, ...
The next mode has three quarter-wavelengths fitting in the string: λ = 4L/3, which gives two antinodes (including the one at the free end).
Wave speed on a string: c = √(T/μ), where T is tension and μ is mass per unit length.
If you keep the driving frequency fixed and reduce the tension, the wave speed drops. Since c = fλ and f is fixed, the wavelength must shrink.
A shorter wavelength means more antinodes fit on the same string.
A fixed-free string of length L driven at f₁ = c/(4L) initially has one antinode (λ = 4L). To get two antinodes at the same frequency, you need λ = 4L/3. Since f is the same:
c_new = f₁ × (4L/3) = [c/(4L)] × (4L/3) = c/3
Since c = √(T/μ), the new tension is T_new = μ(c/3)² = T/9
The tension must decrease by a factor of 9.
The average power transmitted by a sinusoidal wave on a string is:
P = \tfrac{1}{2}\mu\omega^2 A^2 cwhere μ is the linear mass density, ω is the angular frequency, A is the amplitude, and c is the wave speed.
If μ, ω, and A all stay the same but c changes, the power scales directly with c. In the standing-wave problem above, c drops to c/3, so the power in the new configuration is one-third of the original. Equivalently, the original delivers 3 times as much power as the reduced-tension case.
Musical instruments rely on standing waves with specific boundary conditions. A guitar string is fixed-fixed; a clarinet (closed at one end) is effectively fixed-free, which is why it only produces odd harmonics and sounds different from a flute (open-open) at the same pitch. Tuning a guitar string by tightening it increases the tension, raising the wave speed and therefore the pitch, exactly as the c = √(T/μ) formula predicts.
Students often think that doubling the amplitude doubles the maximum acceleration. It does not on its own: a_max = Aω², so doubling A alone only doubles a_max. You need to track what happens to ω as well.
Students confuse nodes and antinodes. A node is where displacement is always zero; an antinode is where displacement is maximum. A fixed end is a node, not an antinode.
Students assume that "half the kinetic energy" means "half the speed." It does not. KE = ½mv², so half the KE means v = v₀/√2, not v₀/2.
Students forget that a fixed-free string supports only odd harmonics. If the fundamental is f₁, the next allowed frequency is 3f₁, not 2f₁.
⚠️ Finding amplitude from initial conditions (x₀ and v₀) is a favourite exam question. Use A² = x₀² + (v₀/ω)², then v_max = Aω.
⚠️ Scaling problems ("double A and ω, find new a_max") appear frequently. Write out a_max = Aω² and substitute carefully.
⚠️ Standing wave problems on fixed-free strings require you to know that λ₁ = 4L and only odd harmonics are allowed.
⚠️ Tension-change problems test whether you can hold frequency fixed, work out the new wavelength from boundary conditions, then find the new wave speed and back out the tension ratio from c = √(T/μ).
True or false: the period of SHM depends on the amplitude. (False. Period depends on k and m, not A.)
Fill in the blank: the maximum speed in SHM is ______. (Aω)
True or false: at the equilibrium position, the acceleration in SHM is at its maximum. (False. Acceleration is zero at equilibrium.)
Fill in the blank: for a fixed-free string, the fundamental wavelength is ______. (4L)
True or false: doubling both amplitude and angular frequency increases the maximum acceleration by a factor of 4. (False. It increases by a factor of 8.)
Q: A mass on a spring has ω = 4 rad/s. At t = 0 it is at x = 1.5 m with v = 10.4 m/s. What is its speed at x = 0?
A: First find the amplitude: A² = (1.5)² + (10.4/4)² = 2.25 + 6.76 = 9.0, so A = 3.0 m. At equilibrium, v_max = Aω = 3.0 × 4 = 12.0 m/s.
Q: A mass passes equilibrium at t = 0 moving in the +x direction and keeps moving that way until t = 1.0 s, at which point its KE is half its initial KE. What is the period?
A: At equilibrium all energy is kinetic. KE drops to half when x = A/√2. Since the mass started at x = 0 and has only reached x = A/√2 (still heading in the +x direction), it has covered 1/8 of a full cycle. So T = 8 × 1.0 s = 8.0 s.
Q: F(x) = -x(x² - 1). How many equilibrium points and how many stable?
A: Set F = 0: x = 0, x = 1, x = -1. Compute dF/dx = -3x² + 1. At x = 0, dF/dx = +1 (unstable). At x = ±1, dF/dx = -2 (stable). Three equilibrium points, two stable.
Q: a_max is 30 m/s² for a certain A and ω. You double both. New a_max?
A: a_max = Aω². Doubling both gives (2A)(2ω)² = 8Aω² = 8 × 30 = 240 m/s².
Q: A fixed-free string of length L has wave speed c. What is the fundamental frequency?
A: The fundamental fits one quarter-wavelength on the string, so λ = 4L and f₁ = c/(4L).
Q: If you reduce the tension by a factor of 9, what happens to the standing wave pattern at the same driving frequency?
A: The wave speed drops by a factor of 3 (since c = √(T/μ)). Wavelength drops by the same factor (λ = c/f). The string now fits three quarter-wavelengths instead of one, giving two antinodes.
SHM connects directly to wave motion: a travelling wave is just many coupled oscillators, each slightly out of phase with its neighbour. The energy relationships here (KE + PE = constant) reappear in wave energy and intensity calculations. The standing-wave material leads into interference and superposition, which are central to the next portion of the course.
Simple harmonic motion, SHM, mass-spring system, angular frequency, omega, amplitude, period, frequency, phase constant, equilibrium, stable equilibrium, unstable equilibrium, restoring force, kinetic energy in oscillation, potential energy in oscillation, energy conservation in SHM, standing waves, nodes, antinodes, boundary conditions, fixed end, free end, harmonics, odd harmonics, fundamental frequency, wave speed on a string, tension and wave speed, power in a wave, PHYS 123, UW physics