29 May 2025

Progressive Waves and Superposition

Plane and spherical waves, longitudinal and transverse motion, the wave equation, particle and wave velocities, beats, and Lissajous figures.

waves-and-optics wave-equation progressive-waves superposition beats lissajous-figures

A wave is a disturbance that transports energy without a net transport of the medium. In a transverse wave the displacement is perpendicular to the direction of propagation; in a longitudinal wave it is parallel.

Wave equation and travelling waves

For a stretched string of tension $T$ and linear mass density $\mu$, take an element $dx$ whose slope is small. The transverse resultant of the tensions is

\[T\left[\frac{\partial y}{\partial x}(x+dx,t) -\frac{\partial y}{\partial x}(x,t)\right] =T\frac{\partial^2y}{\partial x^2}\,dx.\]

Newton’s second law therefore gives

\[\mu\,dx\frac{\partial^2y}{\partial t^2} =T\frac{\partial^2y}{\partial x^2}\,dx,\]

or

\[\boxed{\frac{\partial^2y}{\partial x^2} =\frac1{v^2}\frac{\partial^2y}{\partial t^2}}, \qquad \boxed{v=\sqrt{\frac{T}{\mu}}}.\]

The dimensions are $[T/\mu]={\rm m^2\,s^{-2}}$, so $v$ has units ${\rm m\,s^{-1}}$. For a right-moving profile $y=f(x-vt)$, put $\xi=x-vt$. Then

\[y_{xx}=f''(\xi), \qquad y_{tt}=v^2f''(\xi),\]

so the profile satisfies the wave equation. Similarly, $g(x+vt)$ travels in the $-x$ direction. The general one-dimensional solution is

\[\boxed{y(x,t)=f(x-vt)+g(x+vt)}.\]

For a harmonic progressive wave,

\[y=A\cos(kx-\omega t+\phi), \qquad k=\frac{2\pi}{\lambda}, \qquad \omega=2\pi f,\]

a point of constant phase obeys $kx-\omega t={\rm constant}$; hence

\[\boxed{v=\frac{dx}{dt}=\frac{\omega}{k}=f\lambda}.\]

This wave velocity is distinct from the transverse velocity of a string particle:

\[\boxed{u_y(x,t)=\frac{\partial y}{\partial t} =\omega A\sin(kx-\omega t+\phi)}.\]

The particle velocity oscillates and can be positive, negative, or zero while the profile continues to move in $+x$.

In three dimensions the scalar wave equation is $\nabla^2\psi=v^{-2}\psi_{tt}$. A plane harmonic wave is

\[\psi(\mathbf r,t)=A\cos(\mathbf k\cdot\mathbf r-\omega t+\phi),\]

whose constant-phase surfaces are planes normal to $\mathbf k$. For spherical symmetry, setting $U=r\psi$ reduces the radial equation (for $r>0$) to $U_{rr}=v^{-2}U_{tt}$. Thus an outgoing spherical wave is

\[\boxed{\psi(r,t)=\frac{1}{r}F(r-vt)}\]

and, for a harmonic source, $\psi=(A/r)\cos(kr-\omega t+\phi)$. Since intensity is proportional to amplitude squared,

\[I(r)\propto\frac1{r^2},\]

so $4\pi r^2I(r)$ is constant when there is no absorption.

Linearity and equal-frequency superposition

If $L[y]=y_{xx}-v^{-2}y_{tt}$, then

\[L[ay_1+by_2]=aL[y_1]+bL[y_2].\]

Therefore any linear combination of solutions is also a solution. For two collinear oscillations of the same angular frequency,

\[y_1=A_1\cos\omega t, \qquad y_2=A_2\cos(\omega t+\delta),\]

their sum is

\[y=(A_1+A_2\cos\delta)\cos\omega t -A_2\sin\delta\sin\omega t.\]

Write this as $y=R\cos(\omega t+\alpha)$. Comparing the cosine and sine coefficients,

\[R\cos\alpha=A_1+A_2\cos\delta, \qquad R\sin\alpha=A_2\sin\delta.\]

Squaring and adding gives

\[\boxed{R=\sqrt{A_1^2+A_2^2+2A_1A_2\cos\delta}}, \qquad \boxed{\tan\alpha= \frac{A_2\sin\delta}{A_1+A_2\cos\delta}}.\]

The same result is obtained graphically by adding phasors of lengths $A_1$ and $A_2$ separated by $\delta$. In particular, $R=A_1+A_2$ for $\delta=0$, while $R=\lvert A_1-A_2\rvert$ for $\delta=\pi$.

Different frequencies and beats

Take two equal-amplitude waves:

\[y_1=A\cos(k_1x-\omega_1t), \qquad y_2=A\cos(k_2x-\omega_2t).\]

Using $\cos p+\cos q=2\cos[(p-q)/2]\cos[(p+q)/2]$,

\[\boxed{y=2A\cos\left(\frac{\Delta k\,x-\Delta\omega\,t}{2}\right) \cos(\bar kx-\bar\omega t)},\]

where $\Delta k=k_1-k_2$, $\Delta\omega=\omega_1-\omega_2$, $\bar k=(k_1+k_2)/2$, and $\bar\omega=(\omega_1+\omega_2)/2$. At a fixed position the rapidly oscillating factor has frequency close to $\bar f$, while the magnitude of the envelope is

\[A_{\rm env}=2A\left\lvert\cos\left(\frac{\Delta k\,x-\Delta\omega\,t}{2}\right)\right\rvert.\]

Successive envelope maxima are separated by $T_b=2\pi/\lvert\Delta\omega\rvert$. Hence

\[\boxed{f_b=\frac1{T_b}=\frac{\lvert\Delta\omega\rvert}{2\pi} =\lvert f_1-f_2\rvert}.\]

Because intensity is proportional to $A_{\rm env}^2$, the sound or light intensity rises and falls at this beat frequency.

Lissajous figures and their uses

Two perpendicular simple harmonic motions,

\[x=A\sin\omega_xt, \qquad y=B\sin(\omega_yt+\delta),\]

trace a Lissajous figure. For equal frequencies, put $X=x/A=\sin\omega t$ and expand

\[\frac yB=\sin\omega t\cos\delta+\cos\omega t\sin\delta.\]

Thus $y/B-X\cos\delta=\cos\omega t\sin\delta$. Squaring and using $\cos^2\omega t=1-X^2$ gives

\[\boxed{\frac{x^2}{A^2}+\frac{y^2}{B^2} -2\frac{x}{A}\frac{y}{B}\cos\delta=\sin^2\delta}.\]

It is generally an ellipse. It becomes a straight line for $\delta=0$ or $\pi$, and a circle for $A=B$ with $\delta=\pi/2$ or $3\pi/2$. If the ellipse cuts the $y$-axis at $y_0$, then

\[\boxed{\lvert\sin\delta\rvert=\frac{\lvert y_0\rvert}{B}},\]

while the slope/orientation identifies the appropriate phase quadrant.

If $\omega_x/\omega_y=p/q$ with coprime integers $p,q$, the curve closes after a common period. Counting non-degenerate vertical and horizontal tangencies gives

\[\boxed{\frac{f_x}{f_y}=\frac{N_{\rm vertical}}{N_{\rm horizontal}}}.\]

On an oscilloscope in $X$-$Y$ mode, these facts allow measurement of an unknown frequency against a standard and measurement of phase difference for equal frequencies.

Every curve is plotted from the equations in the note with its parameter values printed beside the panel.

The wave-equation, resultant-amplitude, beat, and Lissajous identities are checked in the Unit I Maxima worksheet.