Interpreting wave evidence and model limits

A wave is a model that describes energy transfer without matter transfer; its key parameters are linked by $v = f\lambda$. Direct measurements give only the positions of crests, nodes, or maxima angles, while wavelength, source coherence, and point-source behavior are already model inferences. Interference and diffraction provide quantitative tests: for two coherent sources, for example, $d\sin\theta = m\lambda$. The model works only within its range of validity: if the obstacle size $a$ is much larger than $\lambda$, diffraction is small and the wave pattern is barely visible. Therefore, when interpreting wave evidence, you must name the assumptions explicitly (coherence, far field, point source) and check them against deviations of the data from the prediction.

In a ripple tank, 11 crests pass a marker in $6.0$ s. The distance between the first and the eleventh crest is $90$ cm. Treating the wave as steady and periodic, find $\lambda$, $f$, and $v$.

Two coherent dippers in a ripple tank are separated by $d = 5.0$ cm. The third antinodal line is observed at an angle $\theta = 20^\circ$ from the central line. Find $\lambda$ using $d\sin\theta = m\lambda$, state one assumption, and say how to test it.

A student claims: “Sound bends around a doorway $1.0$ m wide, so a green laser must bend around it the same way — otherwise light is not a wave.” For sound, $\lambda \approx 0.34$ m; for light, $\lambda \approx 5.5\times 10^{-7}$ m. Evaluate the ratio $\lambda/a$ for both cases and check the claim.

A safe self-study plan using a simulation (for example, PhET “Wave Interference”).

1. Turn on two coherent point sources and fix the frequency. Record the positions of antinodal lines for $m = 1, 2, 3$.
2. Plot $\sin\theta$ against $m$. The slope of the line gives $\lambda/d$.
3. Vary the source separation $d$ and check whether the slope follows the prediction.
4. Model limits: increase the slit width or switch off coherence. Note where the pattern no longer matches the simple formula $d\sin\theta = m\lambda$.
5. Write three conclusions: what was measured, what the model predicted, and where the model failed.

Perform all actions only in the simulation; do not use lasers, sharp edges, or high-volume sound sources.

Separate measured data from the model curve and see at which point the model stops describing the observations.

A sketch of a ripple tank with two coherent sources: labeled crests, nodal and antinodal lines. Beside it, a graph of $\sin\theta$ versus $m$: data points, the best-fit straight line, and a visible deviation of the points at large $m$, where the far-field approximation breaks down.

You stand in a room with a Wi-Fi router emitting at $f = 2.4\ \text{GHz}$. The signal travels at the speed of light, $c = 3.0 \times 10^8\ \text{m/s}$, so its wavelength is $\lambda = c/f \approx 0.125\ \text{m}$. The signal reflects off walls, floor, and metal objects, and the reflected waves interfere with the direct wave. Walk slowly across the room: the signal strength on your phone drops from about $-40\ \text{dBm}$ to $-80\ \text{dBm}$ every $\lambda/2 \approx 6.25\ \text{cm}$. These dead spots are real evidence of wave interference, but the exact pattern is never a simple set of parallel lines.

This scenario exercises the relation $v = f\lambda$ from the prerequisite: you convert a known frequency into a wavelength, then predict the spacing of interference nodes as $\lambda/2$. The limit appears when the simple plane-wave model (one direct wave + one reflected wave) is used: in a real room there are many reflections (multipath), so the model predicts typical spacing but not the exact dead-spot map. Interpreting the phone's signal bars requires knowing both what the wave model gives and where it stops being quantitative.

A microwave oven uses electromagnetic waves at $f = 2.45\ \text{GHz}$. Inside the metal cavity, the waves reflect and form a standing wave pattern. The wavelength in air is $\lambda = c/f \approx 0.122\ \text{m}$ (about $12.2\ \text{cm}$). The electric field has nodes every $\lambda/2 \approx 6.1\ \text{cm}$. If you put a bar of chocolate or a slice of cheese on a plate and heat it for $20\ \text{s}$ without the turntable, you will find melted spots separated by roughly $6\ \text{cm}$. The turntable rotates at about $6\ \text{rpm}$ to average these hot spots.

This scenario tests the standing-wave model built on the prerequisite: you use $v = f\lambda$ to find $\lambda$, then predict node spacing $\lambda/2$. The evidence (melted spots) directly shows where the electric field is strong. The model's limit is that it assumes a perfectly rectangular cavity and a single frequency; real ovens have mode stirrers, imperfect walls, and food that absorbs and re-emits energy, so the actual pattern is more complex and the turntable is needed.

An earthquake off Japan can launch a tsunami with wavelength $\lambda = 200\ \text{km}$ in the open ocean where depth is $h = 4000\ \text{m}$. For such long waves, the speed depends only on depth: $v = \sqrt{gh}$, with $g = 9.8\ \text{m/s}^2$, giving $v \approx 198\ \text{m/s}$ (about $713\ \text{km/h}$). The wave period is $T = \lambda/v \approx 1010\ \text{s}$ (about $17\ \text{min}$). A tide gauge 5000 km away records the wave about $7\ \text{h}$ later. As the wave enters shallow water, $h$ decreases, $v$ drops, and the height grows from $0.5\ \text{m}$ to $10\ \text{m}$ or more.

This scenario uses the prerequisite's relation $v = \lambda/T$ (equivalent to $v = f\lambda$) to connect wavelength, speed, and period, and then applies a specific long-wave dispersion relation. The limit is that the simple shallow-water model assumes constant depth and small amplitude. Near the coast, the model fails: nonlinear steepening, wave breaking, and complex coastal geometry make the simple prediction of run-up unreliable. Interpreting tsunami warnings means knowing which parts of the model are robust (arrival time) and which are not (exact run-up height).