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

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

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.

Solved Problems

1. Resultant of two equal-frequency oscillations

Two collinear oscillations have amplitudes $A_1=3\,\mathrm{mm}$ and $A_2=4\,\mathrm{mm}$, with the second leading the first by $90^\circ$. Find the resultant amplitude and phase lead.

Step 1: Use the phasor components. With $\delta=\pi/2$,

\[R\cos\alpha=A_1+A_2\cos\delta=3\,\mathrm{mm},\] \[R\sin\alpha=A_2\sin\delta=4\,\mathrm{mm}.\]

Step 2: Determine the amplitude.

\[R=\sqrt{3^2+4^2}\,\mathrm{mm}=5\,\mathrm{mm}.\]

Step 3: Determine the phase.

\[\tan\alpha=\frac{4}{3} \quad\Longrightarrow\quad \alpha=\tan^{-1}\!\left(\frac43\right)\simeq53.13^\circ.\]

The component check is $5\cos\alpha=3$ and $5\sin\alpha=4$, so both the amplitude and quadrant are consistent.

2. Phase from an equal-frequency Lissajous figure

An equal-frequency Lissajous ellipse has vertical semiaxis $B=6\,\mathrm{cm}$ and crosses the positive $y$-axis at $y_0=3\,\mathrm{cm}$. Its major axis has positive slope. Find the phase difference.

Step 1: Use the intercept relation.

\[\left\lvert\sin\delta\right\rvert =\frac{\lvert y_0\rvert}{B}=\frac{3}{6}=\frac12.\]

Thus the possible phases in $0\leq\delta\leq\pi$ are $30^\circ$ and $150^\circ$.

Step 2: Use the orientation. A positive-slope ellipse has $\cos\delta>0$, which selects the first quadrant:

\[\boxed{\delta=30^\circ}.\]

Substitution gives $\lvert\sin30^\circ\rvert=1/2$, reproducing the measured intercept.

Descriptive Questions

1. How does particle velocity differ from wave velocity in a transverse progressive wave?
2. Why does the amplitude of an ideal spherical wave vary as $1/r$ while its intensity varies as $1/r^2$?
3. How does linearity of the wave equation lead to the superposition principle?
4. How can a Lissajous figure determine an unknown frequency and an equal-frequency phase difference?

Numerical Problems

1. A string has tension $64\,\mathrm{N}$ and linear density $1.00\times10^{-2}\,\mathrm{kg\,m^{-1}}$. A travelling wave on it has wavelength $0.400\,\mathrm{m}$. Find its speed and frequency.

   Answer: $v=80.0\,\mathrm{m\,s^{-1}}$ and $f=200\,\mathrm{Hz}$.

2. The displacement amplitude of an outgoing spherical wave is $3.00\,\mathrm{mm}$ at $r=2.00\,\mathrm{m}$. Find its amplitude at $r=5.00\,\mathrm{m}$ and the corresponding intensity ratio $I(5)/I(2)$.

   Answer: $A(5)=1.20\,\mathrm{mm}$ and $I(5)/I(2)=4/25=0.160$.

3. Two tuning forks of frequencies $256\,\mathrm{Hz}$ and $260\,\mathrm{Hz}$ sound together. Find the beat frequency and the time between successive intensity maxima.

   Answer: $f_b=4.00\,\mathrm{Hz}$ and $T_b=0.250\,\mathrm{s}$.

4. A closed Lissajous figure has three vertical tangencies and two horizontal tangencies. If the standard signal applied along $y$ has frequency $f_y=200\,\mathrm{Hz}$, find the unknown frequency $f_x$ using the tangency-count relation stated above.

   Answer: $f_x=(3/2)f_y=300\,\mathrm{Hz}$.

The solved results and all numerical answers are verified by exact residuals in the Unit I Maxima worksheet.

References

1. Wave equation - Wikipedia
2. F. S. Crawford Jr., Waves, Berkeley Physics Course, Vol. 3, McGraw-Hill, sections on progressive waves and superposition.
3. H. J. Pain, The Physics of Vibrations and Waves, Wiley, chapters on travelling waves and beats.
4. A. P. French, Vibrations and Waves, MIT Introductory Physics Series, W. W. Norton, chapters on coupled harmonic motions and Lissajous figures.