The Timpson Lyceum · University Physics I (2025)

Mechanical Waves and the Wave Equation - Lecture 15 (University Physics I)

Overview

This lecture introduces mechanical waves — disturbances that transport energy through a medium without permanently transporting mass. It distinguishes transverse from longitudinal waves, builds the sinusoidal wave function $y(x,t)$ that describes the displacement of every point on a wave at every instant, and shows that this function obeys the wave equation. It then derives the speed of a wave on a string from the physical properties tension and mass density, computes the power and intensity carried by a wave, and closes with superposition, standing waves, and the harmonic frequencies of a fixed string. The central challenge is conceptual: $y$ is a function of two variables, position $x$ and time $t$ , so you must keep straight whether you are watching the whole string frozen in time (a snapshot) or one point of the string over time.

Key Concepts

Waves transport energy, not mass

A wave moves energy through a medium while each particle of the medium only oscillates about its equilibrium position and returns there.
Mechanical waves (ropes, strings, sound) require a medium; there is no sound in a vacuum because there is nothing to vibrate.

Transverse vs. longitudinal

Transverse wave: particle motion is perpendicular to the direction the wave travels (e.g., a wave on a string). This lecture focuses on these.
Longitudinal wave: particle motion is parallel to the wave's travel direction (e.g., sound, which is alternating regions of high and low air pressure — compressions and rarefactions).

Describing a wave — two complementary "slices"

A snapshot in time (fix $t$ , vary $x$ ): the distance between successive crests is the wavelength $\lambda$ (a length).
One point over time (fix $x$ , vary $t$ ): the time between successive crests is the period $T$ (a time).
Frequency $f = 1/T$ is cycles per second (Hz). Period and frequency are reciprocals, not the same quantity.
These are linked by the fundamental wave relation $v = \lambda f = \lambda / T$ , where $v$ is how fast the waveform (not the medium) travels.

The sinusoidal wave function

Taking $t=0$ at a crest and $x=0$ at the driven end (driven in simple harmonic motion), the wave is $y(x,t) = A\cos\!\left[\omega\left(t - \dfrac{x}{v}\right)\right].$
The term $-x/v$ is a rightward time-shift: a point at position $x$ repeats what the source did, delayed by the travel time $x/v$ .
Angular frequency $\omega = 2\pi f$ . (Note: this $\omega$ is a fixed constant, unlike the angular velocity $d\theta/dt$ from circular motion, which can vary.)
Wave number $k = \dfrac{2\pi}{\lambda}$ — the number of radians of phase per unit length (how many wavelengths fit in $2\pi$ ).
The same wave can be written four equivalent ways; the most-used form is $y(x,t) = A\cos(kx - \omega t), \qquad v = \dfrac{\omega}{k}.$

The wave equation

Any $y(x,t)$ satisfying $\dfrac{\partial^2 y}{\partial x^2} = \dfrac{1}{v^2}\,\dfrac{\partial^2 y}{\partial t^2}$ is a wave traveling at speed $v$ . Differentiating the sinusoidal form gives $\partial^2 y/\partial x^2 = -k^2 y$ and $\partial^2 y/\partial t^2 = -\omega^2 y$ ; their ratio yields $v^2 = \omega^2/k^2$ , confirming the form is a solution.

Wave speed on a string (a property of the medium)

From momentum impulse on the lifted segment, the speed depends only on the restoring force (tension $F$ ) and the inertia (linear mass density $\mu$ , in kg/m): $v = \sqrt{\dfrac{F}{\mu}}.$
Higher tension → faster; denser/heavier string → slower. Speed does NOT depend on amplitude or on how fast you wiggle the source. The general intuition: $v \propto \sqrt{\text{restoring force} / \text{inertia}}$ .

Power and intensity

Power delivered: $P(x,t) = \sqrt{\mu F}\,\omega^2 A^2 \sin^2(kx-\omega t)$ .
Maximum power $P_{\max} = \sqrt{\mu F}\,\omega^2 A^2$ ; since the average of $\sin^2$ over a cycle is $\tfrac12$ , the average power is $P_{\text{avg}} = \tfrac12 P_{\max}$ .
Intensity $I = P/A_{\text{area}}$ . For a source radiating uniformly in 3-D, power spreads over a sphere of area $4\pi r^2$ , giving the inverse-square law: $\dfrac{I_1}{I_2} = \dfrac{r_2^{\,2}}{r_1^{\,2}}$ . Double the distance, one-quarter the intensity.

Superposition and standing waves

Superposition principle: where two waves overlap, the total displacement is the algebraic sum of the individual displacements (the basis of noise-cancellation via destructive interference).
A wave reflecting off a fixed end returns inverted; adding $y_1 = A\cos(kx-\omega t)$ and $y_2 = -A\cos(kx+\omega t)$ gives the standing wave $y_{\text{SW}}(x,t) = 2A\,\sin(kx)\,\sin(\omega t).$
Nodes (points that never move) occur where $\sin(kx)=0$ , i.e. $x = n\dfrac{\lambda}{2}$ .

Harmonics on a fixed string of length $L$

Both ends must be nodes, so allowed wavelengths are $\lambda_n = \dfrac{2L}{n}$ and allowed frequencies are $f_n = n\dfrac{v}{2L} = n f_1$ — integer multiples of the fundamental $f_1 = \dfrac{v}{2L}$ . With $v=\sqrt{F/\mu}$ , pitch is set by length, tension, and density (why fretting, tuning, and string gauge change the note).

Worked Example

A clothesline is wiggled in a sinusoidal transverse wave with frequency $f = 2.00\ \text{Hz}$ , amplitude $A = 0.075\ \text{m}$ , and wave speed $v = 12.0\ \text{m/s}$ , with the held end at maximum displacement at $t=0$ . Find $\omega$ , $T$ , $\lambda$ , $k$ , and write $y(x,t)$ .

Angular frequency:

\begin{aligned} \omega &= 2\pi f \\ \omega &= 2\pi(2.00) = 4\pi \approx 12.6\ \text{rad/s} \end{aligned}

Period:

\begin{aligned} T &= \frac{1}{f} \\ T &= \frac{1}{2.00} = 0.500\ \text{s} \end{aligned}

Wavelength and wave number:

\begin{aligned} \lambda &= \frac{v}{f}, \qquad k = \frac{2\pi}{\lambda} \\ \lambda &= \frac{12.0}{2.00} = 6.00\ \text{m}, \qquad k = \frac{2\pi}{6.00} \approx 1.05\ \text{rad/m} \end{aligned}

Wave function (and the displacement of the held end at $x=0$ ):

\begin{aligned} y(x,t) &= A\cos(kx - \omega t) \\ y(x,t) &= 0.075\cos\big(1.05\,x - 12.6\,t\big), \qquad y(0,t) = 0.075\cos(12.6\,t) \end{aligned}

As a second short example, a 2.00 kg rope of length 80.0 m hangs in a mine shaft holding a 20.0 kg box, so $\mu = 2.00/80.0 = 0.0250\ \text{kg/m}$ and the tension at the bottom is $F = (20.0)(9.8)$ N. The wave speed near the box is

\begin{aligned} v &= \sqrt{\frac{F}{\mu}} \\ v &= \sqrt{\frac{(20.0)(9.8)}{0.0250}} \approx 8.85\ \text{m/s} \end{aligned}

Summary

A wave carries energy through a medium while the medium's particles only oscillate about equilibrium; transverse motion is perpendicular to travel, longitudinal motion is parallel.
Wavelength is a snapshot-in-time distance, period is a single-point time interval, and they connect through $v = \lambda f$ ; frequency $f = 1/T$ and $\omega = 2\pi f$ .
The sinusoidal wave function $y(x,t) = A\cos(kx-\omega t)$ (with $k = 2\pi/\lambda$ , $v = \omega/k$ ) gives the displacement of every point at every time and satisfies the wave equation $\partial^2 y/\partial x^2 = (1/v^2)\,\partial^2 y/\partial t^2$ .
Wave speed on a string is a property of the medium, $v = \sqrt{F/\mu}$ — set by tension and mass density, independent of amplitude and drive frequency.
Power scales as $P \propto \sqrt{\mu F}\,\omega^2 A^2$ (average $= \tfrac12$ the max), and for 3-D radiation intensity obeys the inverse-square law $I \propto 1/r^2$ .
Overlapping waves add by superposition; a wave reflected from a fixed end interferes with itself to form a standing wave $y_{\text{SW}} = 2A\sin(kx)\sin(\omega t)$ with nodes at $x = n\lambda/2$ , giving a fixed string its harmonic series $f_n = n\,v/2L$ .