Superposition and interference of waves

Why superposition holds at all. The 1D wave equation $\partial^2 y/\partial t^2 = v^2\,\partial^2 y/\partial x^2$ is linear: every term involves $y$ or its derivatives to the first power only, with no $y^2$, $y\,\partial y/\partial x$, or similar terms. A defining property of linear equations is that if $y_1$ and $y_2$ are each solutions, so is $y_1+y_2$. Physically this means overlapping waves add algebraically at every point and instant, and once they separate each continues exactly as before — neither wave leaves a permanent mark on the other. This holds for small-amplitude waves in an ordinary medium; it breaks down in nonlinear media (shock waves, very loud sound, intense light in some materials) where the governing equation picks up higher powers of $y$.

Two equal-frequency waves: spatial interference. Take two harmonic waves of the same $\omega$, $k$, and amplitude $A$, offset by a fixed phase $\delta$:

where: $A$ is the amplitude of each individual wave, $k=2\pi/\lambda$ the wavenumber, $\omega=2\pi f$ the angular frequency, and $\delta$ the constant phase difference between the two waves at a given point. Adding them (phasor or direct trigonometric addition) gives

The result is still a wave of the same frequency, but with amplitude $2A\cos(\delta/2)$ that depends only on $\delta$ — fixed in time at any given point in space. This is why interference produces a stationary pattern: constructive when $\delta=2m\pi$ (amplitude $2A$, a bright fringe or loud spot), destructive when $\delta=(2m+1)\pi$ (amplitude $0$, a dark fringe or silent spot), for any integer $m$.

Two close but unequal frequencies: beats. Now take two waves of equal amplitude but slightly different angular frequencies $\omega_1$ and $\omega_2$, observed at a fixed point (the spatial part is dropped since we only care about the time variation there):

The same sum-to-product identity now groups the time-varying parts instead of a fixed phase:

where: the second cosine, at the mean angular frequency $\bar\omega=(\omega_1+\omega_2)/2$, is the rapidly oscillating carrier; the first cosine, at the small angular frequency $\Delta\omega/2=(\omega_1-\omega_2)/2$, is the slowly varying envelope amplitude $2A\cos(\Delta\omega\,t/2)$ that rises and falls in time. A detector (ear, photodiode, meter) responds to intensity, which goes as the envelope squared, so it registers loudness swelling and fading — a beat — every time the envelope passes through zero. This happens twice per envelope period, because $\cos^2\theta$ completes two full cycles for every one cycle of $\cos\theta$.

Two tuning forks of frequency $256\ \text{Hz}$ and $260\ \text{Hz}$ are struck together, each producing a sound wave of amplitude $A=1.0\times10^{-3}\ \text{m}$ at a listener's ear. A third fork of unknown frequency $f_3$ is now struck alongside only the $256\ \text{Hz}$ fork, and the listener counts $6$ beats every $2$ seconds. Find (a) the beat frequency and mean carrier angular frequency for the original pair, (b) the time between successive silences for that pair, and (c) the two possible values of $f_3$.

Step 1. For the original pair, the beat frequency is $f_{\text{beat}}=|f_1-f_2|=|260-256|=4\ \text{Hz}$, and the mean (carrier) frequency is $\bar f=(256+260)/2=258\ \text{Hz}$, so $\bar\omega=2\pi\bar f=2\pi(258)\approx1621\ \text{rad/s}$.

Step 2. The envelope is $2A\cos(\Delta\omega\,t/2)$ with $\Delta\omega=2\pi(4)=8\pi\ \text{rad/s}$. Successive silences occur at consecutive zeros of this envelope, separated by $\Delta t = \dfrac{2\pi}{\Delta\omega}=\dfrac{1}{\Delta f}=\dfrac{1}{4}\ \text{s}=0.25\ \text{s}$, matching the beat period $1/f_{\text{beat}}$ as expected.

Step 3. For the new pair (fork 1 at $256\ \text{Hz}$ and fork 3 at $f_3$), $6$ beats in $2$ seconds is a beat frequency of $6/2=3\ \text{Hz}$.

Step 4. Since $f_{\text{beat}}=|f_1-f_3|=3\ \text{Hz}$ is consistent with $f_3$ lying either above or below $256\ \text{Hz}$, there are two candidates: $f_3 = 256+3 = \mathbf{259\ Hz}$ or $f_3 = 256-3 = \mathbf{253\ Hz}$. Beat counting alone cannot distinguish the two; a standard trick (loading fork 3 with wax to lower its frequency slightly and seeing whether the beat rate rises or falls) is needed to pick between them.

Competitive stretch

If instead the two original forks were combined with a coherent light analogue — two sources of wavelength $600\ \text{nm}$ meeting at a point with path difference $\Delta x=1.5\ \mu\text{m}$ — the phase difference is $\delta=\dfrac{2\pi}{\lambda}\Delta x=\dfrac{2\pi}{600\times10^{-9}}(1.5\times10^{-6})=5\pi\ \text{rad}$, an odd multiple of $\pi$, so the resultant amplitude $y_{\text{res}}=2A|\cos(\delta/2)|=2A|\cos(5\pi/2)|=0$: a dark fringe, the spatial twin of a beat-envelope silence, but fixed in space rather than repeating in time.

• Reporting the beat frequency as $\Delta\omega/2$ (the amplitude envelope's own angular frequency) instead of $|f_1-f_2|$; the intensity envelope beats at twice the rate of the amplitude envelope because $\cos^2\theta$ has half the period of $\cos\theta$.
• Confusing interference (same-frequency waves producing a fixed spatial pattern of bright/dark or loud/quiet points) with beats (different-frequency waves producing a temporal throb at one fixed point) — same identity, different physical setup.
• Assuming superposition holds for any amplitude in any medium; it strictly requires the governing equation to stay linear, which fails for large-amplitude or nonlinear-medium waves.
• Equating "path difference equal to a whole number of wavelengths" with constructive interference without checking for an extra half-wavelength shift from reflection, which flips bright and dark conditions.

Ask the tutor: if the sum-to-product step from $\cos(\omega_1 t)+\cos(\omega_2 t)$ to the envelope-times-carrier form ever feels like a trick you can't reproduce from scratch, ask the AI tutor to rebuild it line by line with a fresh pair of frequencies.