
Physics III course sequence · Problem studio
Wave-phenomena data studio
You have the recording, not the answer. This studio runs the section backwards: from a waveform or a spectrogram to the wavelength, phase, beat rate or source speed that produced it, with the record's own sampling and window limits deciding what can be known, and an error bar on every number you infer.
UPIII-03.09 · Lesson 9 of 9 in this unit. Position is not a completion score.Build the model
Run the section backwards: from the recording to the parameter, with error bars.
The key idea
A wave-phenomena data studio is the inverse problem of the section: starting from a sampled signal or its spectrogram, recover the wavelength, phase difference, beat rate or source speed that produced it, and state how uncertain each recovered number is.
Example. A horn passes a microphone: plateaus at 512.4 Hz approaching and 471.6 Hz receding, each ±1.2 Hz from 0.25 s windows, give vₛ = 343 × 40.8/984.0 = 14.2 ± 0.6 m s⁻¹ without ever knowing the horn's own pitch.
The complete physical model and assumptions
Every result in this section is a forward model: a path difference goes in and a phase and an amplitude come out; two frequencies go in and a beat envelope comes out. A data studio runs those models backwards, and inversion has its own rules. The record sets two limits before any physics: the sampling rate caps the highest frequency it can hold at fₛ/2, and the duration T of the record or window fixes how finely frequencies can be told apart, at about 1/T. Inside those limits the inversion is algebra — δ = (m + ½)λ solved for λ, Δφ = 2πfΔt for Δφ, vₛ = v(fₐ − fᵣ)/(fₐ + fᵣ) — but each inverted formula carries a trap: a delay is a phase only modulo 2π, a beat count gives |f₁ − f₂| but not which tone is higher, and an unresolved pair of tones is the same signal as one amplitude-modulated tone. Because beats, a swept interference pattern and a moving source can produce the same frequency structure, a recovered parameter means nothing until a second measurement has chosen the model. Every inferred number inherits an uncertainty: the bin width or timing resolution of the record, propagated through the derivative of the inversion. The model stops working where its assumptions do: a loud source adds harmonics that alias, a signal that changes inside the window smears its own spectrum, and independent sources with drifting phase never form the stable pattern the formulas assume.
1Two limits the record imposes before any physics
A digital record is a list of samples taken fₛ times per second for a duration T, and those two numbers bound everything you can infer from it. Sampling at fₛ can represent frequencies only up to the Nyquist limit fₛ/2; a component above it is not lost but folded to a false frequency |f − n fₛ|. At fₛ = 8.00 kHz a 5.20 kHz overtone appears at 2.80 kHz, indistinguishable from a real 2.80 kHz tone, which is why an anti-alias filter must act before sampling, never after. The duration sets the other limit: a spectrum computed from a window of length T has bins 1/T apart, so a 0.250 s window gives 4.00 Hz bins and cannot separate two tones 3 Hz apart. A peak read from its tallest bin is known to about Δf/(2√3) = 1.15 Hz, the a/√3 rule applied to a half-width of half a bin. The two limits are independent: sampling faster does nothing for resolution, and recording longer does nothing for bandwidth.
2Wavelength: fit every minimum, and use the exact geometry
Two in-phase sources d apart produce minima wherever the path difference δ is a half-integer number of wavelengths, δ = (m + ½)λ. A microphone slid along a line at distance L records the minima positions yₘ, and the honest inversion converts each into δₘ = √(L² + (yₘ + d/2)²) − √(L² + (yₘ − d/2)²) before anything is fitted. Plot δₘ against m + ½: the slope is λ, and the intercept tests the model, because a phase offset φ₀ between the sources shifts every minimum by −φ₀λ/2π. A non-zero intercept is the evidence that the sources were not in phase. The first-year shortcut yₘ = (m + ½)λL/d is the small-angle limit, and a survey reaching y/L = 0.8 is nowhere near it: for d = 0.800 m and L = 2.000 m the m = 2 minimum of a 0.200 m wave sits at y = 1.62 m, where the shortcut returns λ = 0.260 m, 30% high. Uncertainty flows through the derivative dδ/dy = sin θ₁ − sin θ₂, which shrinks with order, so the outermost minima carry the most weight in the fit.
3Phase: a delay is a phase only inside half a period
Two microphones recording one 1250 Hz tone give two channels, and the cross-correlation peak between them locates the lag Δt of one behind the other. The phase difference the section's formulas want is Δφ = 2π f Δt, and it inherits u(Δφ) = 2π f u(Δt). A lag of 0.160 ± 0.010 ms gives Δφ = 1.257 ± 0.079 rad, one fifth of a cycle. But a pure tone's correlation repeats every period, 0.800 ms here, so the lag is measured only modulo the period and the phase only modulo 2π. Geometry removes the ambiguity: with the microphones 0.120 m apart the true lag cannot exceed d/v = 0.350 ms, less than half a period, so the 0.160 ms peak is the only candidate. The general condition is d < λ/2, and λ = 0.274 m passes. The delay then gives the direction of arrival: sin α = vΔt/d = 0.457, so the source lies 27.2° off the perpendicular, with u(α) = (v/d) u(Δt)/cos α = 0.032 rad = 1.8°.
4Beats or two tones: the window decides, the count resolves
Two equal tones 2.67 Hz apart add to 2 cos(πΔf t) cos(2πf̄ t): a carrier at the mean frequency whose amplitude passes through zero 2.67 times a second. Nothing in the signal says whether it is beats or two lines; the analysis window does. A 0.20 s window holds 0.53 of a beat, and its spectrum shows one line at f̄ whose height changes from window to window. A 1.00 s window holds 2.67 beats, and the same signal now shows two lines 2.67 bins apart. Both pictures are correct; the second has resolved what the first only modulated. The count in time is the more precise route: N beats spanning Tcount give |f₁ − f₂| = N/Tcount, with an uncertainty set by two edge timings, not by any bin. What the count cannot give is the sign: 440.0 Hz beating at 2.67 Hz means a partner at 442.67 or 437.33 Hz. Lengthen the window until the lines split, or detune one source and watch whether the beat slows or quickens: either measurement decides.
5Source speed: read both plateaus, and mind the glide
A source moving past a fixed microphone leaves a spectrogram track that starts on the approach plateau fₐ = f₀v/(v − vₛ), glides through f₀ at closest approach and settles on fᵣ = f₀v/(v + vₛ). Dividing the plateaus cancels f₀: vₛ = v(fₐ − fᵣ)/(fₐ + fᵣ), and f₀ = 2fₐ fᵣ/(fₐ + fᵣ) comes free. With equal uncertainties uf on both plateaus, propagation gives u(vₛ) ≈ v uf/(√2 f₀), so 1.2 Hz on a 491 Hz horn costs 0.6 m s⁻¹. The glide is where the window must be chosen with care. Writing f′ = f₀v/(v − vₛ cos θ) with θ the angle between the velocity and the line to the microphone, and using dθ/dt = vₛ/b at closest approach, the received frequency there changes at df′/dt = −f₀vₛ²/(vb). For 14.2 m s⁻¹ at b = 6.0 m that is 48 Hz s⁻¹, so a 0.25 s window sees the line move 12 Hz, three bins, and smears it. The blur stays under a bin only when Tw² |df′/dt| < 1, here Tw < 0.14 s. The plateaus, 3 s out, glide at 0.1 Hz s⁻¹ and tolerate the long window.
6One spectrum, three stories: name the measurement that decides
A recorded 500 Hz line whose strength rises and falls 2.9 times a second has three candidate causes, each a different model. Beats: a second source at 500 ± 2.9 Hz, with pulsing at the same rate at every microphone position. Interference: two coherent sources set up a stationary pattern in space, and a microphone on a cart at 1.00 m s⁻¹ crossing nodes λ/2 = 0.343 m apart reports 2.92 Hz pulses that stop the instant the cart stops. Doppler: relative motion shifts the line's frequency, not its strength; a source rocking toward and away from the microphone shows a track wobbling in frequency under a constant envelope. The discriminating measurements are cheap: move the microphone and see whether the amplitude changes (interference); lengthen the window until the line splits (beats); look for a frequency glide that reverses at closest approach (Doppler). Only once the model is chosen does the recovered parameter carry meaning, and it is reported with the uncertainty the record allows, 0.29/T on a bin-read peak or the timing resolution on a count, propagated through the inversion.
Key equations
Read the units, assumptions and limits before substituting.
Sampling limit and aliasing
fₛₑₑₙ = |f − n fₛ|, with n the integer nearest f/fₛ
Fixes the highest frequency a record can hold. A tone above the Nyquist limit fₛ/2 folds down to a false lower one, and no later processing can unfold it.
f the true frequency and fₛ the sampling rate, both in Hz; fₛₑₑₙ equals f only when f < fₛ/2Frequency resolution of a record
Δf = 1/T
Two lines closer than 1/T merge. A peak read from the tallest bin is uncertain by Δf/(2√3) ≈ 0.29/T, the resolution rule applied to one bin.
T the record or window length in s; Δf the spacing of the spectrum's bins in HzWavelength from a survey of minima
√(L² + (yₘ + d/2)²) − √(L² + (yₘ − d/2)²) = (m + ½)λ
Exact path difference, so a fit of δₘ against m + ½ has slope λ; only for yₘ ≪ L does it collapse to yₘ = (m + ½)λL/d.
in-phase sources d apart; yₘ the m-th minimum along a line L away; all lengths in mPhase from a measured delay
Δφ = 2π f Δt
Turns a cross-correlation lag into the phase difference the interference formulas take, with u(Δφ) = 2π f u(Δt).
Δt the lag between two channels in s, f in Hz, Δφ in rad; unique only if |Δt| < 1/(2f) is guaranteedBeat rate from a count
|f₁ − f₂| = N/Tcount
Recovers a separation far finer than the spectrum's bins, but not which tone is higher; that needs a second measurement.
N complete beats, envelope minimum to minimum, in a stretch Tcount in s; f₁, f₂ in HzSource speed from a pass-by
vₛ = v (fₐ − fᵣ)/(fₐ + fᵣ)
The source's own frequency cancels, so an unknown horn still yields its speed; for vₛ ≪ v this is vₛ ≈ v (fₐ − fᵣ)/(2f₀).
fₐ, fᵣ the approach and recede plateaus in Hz; v the speed of sound and vₛ the source speed in m s⁻¹Interactive lab
Predict first. Explore the model. Compare the result.
Explore the model
Try this. Set Δf = 4 Hz and drag Tw from 0.05 s to 1 s: the single pulsing line grows a dip once the window holds one full beat, at Tw = 0.25 s, and splits cleanly by 0.5 s. Halve Δf to 2 Hz and each stage takes twice as long: only the product Δf Tw matters.
Live results
- BEAT PERIOD 1/Δf
- 0.250 s
- BIN WIDTH 1/Tw
- 5.00 Hz
- BEATS IN WINDOW = BINS APART
- 0.80
- PEAK READ TO ±
- 1.44 Hz
Results follow this lesson’s model and assumptions. Displayed values are rounded.
Predict → change one variable → check
Plan a change before running it. Every other input stays at the saved baseline.
Saved baseline. Tone separation Δf: 4.0 Hz; Window length Tw: 0.20 s
This experiment is kept only while this page is open. It does not change saved course progress.
Catch the common trap
Sampling faster does not resolve closer tones. Asked to separate two lines 2 Hz apart, students reach for a higher sampling rate, because more samples must mean more detail. The sampling rate sets only the bandwidth, the highest frequency the record can hold, fₛ/2; the spacing of the spectrum's bins is 1/T and depends only on how long the window is. A 96 kHz recording 0.2 s long has 5 Hz bins and shows one merged line; an 8 kHz recording 1 s long has 1 Hz bins and shows two. The converse trap is as common: concluding that a separation finer than the bin width is unmeasurable. It is measurable in the time domain, by counting beats over the whole record, which is why the window is a choice about display and the record length is the only real limit.
Problem-solving studio
3 authored examples. Make an attempt, then reveal the reasoning one step at a time.
EasyExample 1
A clip of two tuning forks, one known to be 440.0 Hz, shows envelope minima at t = 0.212, 0.588, 0.961, 1.338 and 1.712 s, each located to ±0.008 s. Find the beat frequency with its uncertainty, list the possible frequencies of the second fork, and state the shortest window whose spectrum would show two lines.
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 5 steps revealed.
- Successive minima are one beat period apart. Four intervals span t₄ − t₀ = 1.712 − 0.212 = 1.500 s, so Tb = 1.500/4 = 0.3750 s and fb = 1/Tb = 2.667 Hz.
- Averaging the four intervals would use the same two numbers: (t₁ − t₀) + (t₂ − t₁) + (t₃ − t₂) + (t₄ − t₃) telescopes to t₄ − t₀, so the inner minima cancel and only the end timings carry uncertainty: u(t₄ − t₀) = √2 × 0.008 = 0.0113 s, giving u(Tb) = 0.0113/4 = 0.0028 s.
- fb = 1/Tb is a power of Tb, so the relative uncertainty carries over: u(fb) = fb × 0.0028/0.3750 = 0.020 Hz. Hence fb = 2.67 ± 0.02 Hz.
- The beat rate is |f₁ − f₂| with no sign, so the second fork is at 440.00 + 2.67 = 442.67 Hz or 440.00 − 2.67 = 437.33 Hz. Loading the unknown fork with a little wax lowers its pitch: if the beats slow, it was the higher one.
- Two lines separate once the window holds about one beat, T ≥ 1/fb = 0.375 s, and a comfortable 2/fb = 0.75 s puts them 2 bins apart. The clip holds 1.5 s of data, so one 1.5 s spectrum would show the pair 4 bins apart and settle the sign directly.
Answer. fb = 2.67 ± 0.02 Hz; the second fork is at 442.67 Hz or 437.33 Hz; a window of at least 0.375 s, comfortably 0.75 s, splits the pair.
MediumExample 2
A car horn passes a roadside microphone. A spectrogram built from 0.250 s windows shows plateaus at 512.4 Hz while the car approaches and 471.6 Hz after it has passed. Air is at 20 °C, v = 343 m s⁻¹. Find the car's speed and the horn's frequency, each with uncertainty, and check the result against the low-speed approximation.
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 5 steps revealed.
- Each plateau is read from a 0.250 s window, whose bins are 1/T = 4.00 Hz apart, so a bin-read peak carries uf = 4.00/(2√3) = 1.15 Hz ≈ 1.2 Hz on both fₐ and fᵣ.
- Moving source: fₐ = f₀v/(v − vₛ) and fᵣ = f₀v/(v + vₛ). Dividing, fₐ/fᵣ = (v + vₛ)/(v − vₛ), which rearranges to vₛ = v(fₐ − fᵣ)/(fₐ + fᵣ) = 343 × 40.8/984.0 = 343 × 0.04146 = 14.22 m s⁻¹, about 51 km h⁻¹.
- Propagate: ∂vₛ/∂fₐ = 2v fᵣ/(fₐ + fᵣ)² and ∂vₛ/∂fᵣ = −2v fₐ/(fₐ + fᵣ)², so u(vₛ) = 2v uf √(fᵣ² + fₐ²)/(fₐ + fᵣ)² = 686 × 1.2 × 696.4/968 256 = 0.59 m s⁻¹. Hence vₛ = 14.2 ± 0.6 m s⁻¹.
- Eliminating vₛ instead gives f₀ = 2fₐ fᵣ/(fₐ + fᵣ) = 2 × 512.4 × 471.6/984.0 = 491.2 Hz; the same kind of derivatives give u(f₀) = 0.85 Hz, so the horn sounds at 491.2 ± 0.9 Hz.
- Low-speed check: fₐ − fᵣ ≈ 2f₀vₛ/v = 2 × 491.2 × 14.22/343 = 40.7 Hz against the measured 40.8 Hz; the 0.1 Hz gap is the (vₛ/v)² = 0.17% term the approximation drops, well inside the uncertainty. As vₛ → 0 both plateaus collapse onto f₀, as they must.
Answer. vₛ = 14.2 ± 0.6 m s⁻¹ (51 ± 2 km h⁻¹); f₀ = 491.2 ± 0.9 Hz; the low-speed form reproduces the 40.8 Hz split to 0.1 Hz.
HardExample 3
Two loudspeakers 0.800 m apart are driven from one oscillator whose frequency was not logged. Along a line 2.000 m from the speakers, a microphone finds the first three intensity minima on one side of centre at y = 0.258, 0.821 and 1.623 m, each ±0.003 m. Recover the wavelength and the frequency with uncertainties (v = 343.0 m s⁻¹), test whether the speakers were in phase, and show why the small-angle formula must not be used.
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 6 steps revealed.
- In-phase sources give minima where the path difference δ = r₁ − r₂ = (m + ½)λ, with r₁ = √(L² + (y + d/2)²) and r₂ = √(L² + (y − d/2)²). The outermost point has y/L = 0.81, an angle of 39°, so the exact geometry is compulsory.
- Path differences: m = 0: r₁ = 2.1055, r₂ = 2.0050, δ₀ = 0.1004 m. m = 1: r₁ = 2.3433, r₂ = 2.0438, δ₁ = 0.2994 m. m = 2: r₁ = 2.8447, r₂ = 2.3443, δ₂ = 0.5004 m.
- Fit δₘ = (m + ½)λ through the origin by least squares: λ = Σ(m + ½)δₘ / Σ(m + ½)² = (0.5 × 0.1004 + 1.5 × 0.2994 + 2.5 × 0.5004)/(0.25 + 2.25 + 6.25) = 1.7505/8.75 = 0.20005 m.
- Uncertainty: dδ/dy = (y + d/2)/r₁ − (y − d/2)/r₂ = 0.383, 0.315 and 0.189 for the three orders, so u(δₘ) = 0.00115, 0.00095 and 0.00057 m. The slope inherits u(λ) = √Σ[(m + ½)u(δₘ)]² / 8.75 = 0.0021/8.75 = 0.00024 m; the ±1 mm on d and L, propagated through ∂δ/∂d and ∂δ/∂L, lift it to 0.0003 m.
- Residuals δₘ − (m + ½)λ = +0.0004, −0.0007, +0.0003 m all sit inside u(δₘ), so no intercept is needed: the speakers were in phase, since an offset φ₀ would shift every δₘ by −φ₀λ/2π. The small-angle estimate yₘ d/((m + ½)L) instead returns 0.206, 0.219 and 0.260 m, high by 3%, 9% and 30%, an error a hundred times the measurement uncertainty.
- f = v/λ = 343.0/0.20005 = 1714.6 Hz with u(f) = f × 0.0003/0.2000 = 2.6 Hz. Accessible orders: (m + ½)λ ≤ d gives m ≤ 3.5, so an m = 3 minimum exists at δ = 0.700 m, which the geometry puts at y = 3.64 m, beyond the survey; m = 4 would need δ = 0.900 m > d and does not exist.
Answer. λ = 0.2000 ± 0.0003 m and f = 1715 ± 3 Hz, with the speakers in phase; the small-angle formula would have been wrong by up to 30%. A 1 K uncertainty in air temperature would add another ±3 Hz through v.
Quick check
Select an answer and explain it before checking.
Review & continue
A wave-phenomena data studio is the inverse problem of the section: starting from a sampled signal or its spectrogram, recover the wavelength, phase difference, beat rate or source speed that produced it, and state how uncertain each recovered number is.
- Explain the relationship without looking at the equation.
- Check a sign, a unit and a model assumption in your working.
- Return to the lab or worked example if a step is still unclear.