Superposition · Stationary Waves

Stationary Waves Lab

y₁ + y₂ = 2A sin kx · cos ωt

Station 1

Formation by the graphical method

y₁ = A sin(kx − ωt) · y₂ = A sin(kx + ωt)

Two identical waves, opposite directions — and their sum

t / T

0.000

cos ωt

1.000

resultant amplitude

0.300 m

× each wave's A

2.00 ×

Try first: Hit t = T/4 — the whole string goes flat.

Time scrubber

Instant t/T0.000

y₁ and y₂ are in phase everywhere — full constructive, double amplitude.

ⓘ more

The snap sequence 0 → T/8 → T/4 → 3T/8 → T/2 is the graphical construction the syllabus asks for. Only the two component waves move; the resultant just breathes between +2A and −2A on a fixed frame of nodes.

Controls

Wavelength of each wave λ1.00 m

Wave speed v2.00 m s⁻¹

f = v/λ

2.00 Hz

T = λ/v

0.500 s

node spacing λ/2

0.500 m

Each wave has A = 0.15 m fixed, so the resultant maxes at 2A = 0.30 m.

Do the nodes ever move?

predicted nodes m·λ/2

9 in view

zeros measured now

all x

worst node drift

0.0000 m

watched for

0.000 s

measured x: — every point is a zero at the flat instant

ⓘ more

Each frame the resultant is bisected to find its actual zeros, then each is compared with the predicted m·λ/2. The worst mismatch ever seen is the drift figure — it stays at the bisection tolerance, so the nodes are pinned in place while the loops flip.

Window 4.0m wide. Both component waves have identical A, λ and v — that is what "identical waves" means in the syllabus statement.

Station 2

Vibrating systems and harmonics

f = v/λ

String · 3f₁λ = 0.667 m · f = 510 Hzslow motion ×1/283

drag — orbit · scroll — zoom

Measurement overlay

ⓘ more

Nodes and antinodes are computed from ψ(x) = 0 and |ψ(x)| = 1 for the selected mode, then the brackets simply subtract their positions. λ/2 and λ/4 come out as results, not as inputs. Where the particles bunch up sits a displacement node — that is a pressure antinode, which is what a microphone actually hears.

System

String, fixed both ends — node at both ends.

Every whole-number harmonic fits, so the series is f₁, 2f₁, 3f₁, …

Controls

Length L1.00 m

Wave speed v340 m s⁻¹

λ = 2L/h

0.667 m

f = v/λ

510 Hz

f / f₁

3.00 ×

ⓘ more

Real f₁ here is 170 Hz, far too fast to watch, so time is scaled by 1/283. The shape, the phase and every number quoted are the true ones; only the playback rate is slowed. Higher harmonics still vibrate visibly faster (on-screen 1.80 Hz).

Spacing measured off the markers

nodes in mode

antinodes in mode

N → N

0.333 m

A → A

0.333 m

N → nearest A

0.167 m

λ/4

0.167 m

λ/2 = 0.333 m · N→N ÷ (λ/2) = 1.000 · N→A ÷ (λ/4) = 1.000

Phase check inside the stationary wave

Marker M₁ x₁0.100 m

Marker M₂ x₂0.450 m

ANTIPHASE — Δφ = π

1 node between → odd → ψ(x₁)·ψ(x₂) < 0

ψ(x₁) · s₁/a₀ now

0.809 · 0.809

ψ(x₂) · s₂/a₀ now

-0.891 · -0.891

ⓘ more

Drag both markers inside one loop: the two live displacements always share a sign, so Δφ = 0. Drag one across a node: they always carry opposite signs, so Δφ = π. Every point in a stationary wave is therefore either exactly in phase or exactly antiphase with every other — never anything between.

Station 3

Progressive vs stationary

P = −T (∂y/∂x)(∂y/∂t)

Energy transfer — measured, not asserted

⟨P⟩ of y₁ alone

0.000 W

⟨P⟩ of y₂ alone

0.000 W

⟨P⟩ of y₁ + y₂

0.0000 W

averaged over whole cycles

— s

P of y₁ right now

0.000 W

P of y₂ right now

0.000 W

P of resultant now

0.000 W

|⟨P⟩ net| ÷ |⟨P⟩ of one wave|

0.0e+0

Each travelling wave carries power one way (0.000 W measured against the analytic ½TA²kω = 0.000W). Their superposition carries none: the resultant's instantaneous power swings hard both ways and integrates to zero.

ⓘ more

Power crossing one cross-section of a string is P = −T (∂y/∂x)(∂y/∂t), integrated here by midpoint substeps inside the animation loop with tension T = 10.0 N at x = 0.000m — a point λ/8 from a node, chosen so the resultant's instantaneous power is at full swing rather than trivially zero. Each frame is split at exact period boundaries so the average always covers complete cycles. Energy is not destroyed: it sloshes between kinetic and potential inside each loop, but no loop ever hands energy to the next.

The other four rows of the comparison

Feature	Progressive wave	Stationary wave
Energy	Transfers energy along the direction of travel.	No net energy transfer — see the measured ⟨P⟩ above.
Amplitude	Every particle has the same amplitude A.	Amplitude depends on position: 0 at a node, 2A at an antinode.
Phase	Neighbouring particles are out of step: Δφ = 2πΔx/λ takes any value.	Only two answers exist: in phase within a loop, antiphase across a node.
Waveform	The profile advances one λ every period.	The profile stays put and only changes size; nodes never move.
Wavelength	Distance between adjacent particles in phase.	Twice the node-to-node distance (λ = 2 × N→N).
Frequency	Every particle vibrates at the source frequency.	Every particle vibrates at the same frequency, except the nodes, which never move.

Station 4

Resonance tube and the end correction

L + c = (2n − 1)λ/4

Fixed frequency, adjustable air column

quiet

The air above the rim still moves, so the antinode sits c beyond the mouth, not at it.

ⓘ more

The column can only hold s(x) = B sin(kx), because the water surface is rigid and forces a displacement node there. Its antinodes therefore sit at fixed heights (2m − 1)λ/4 above the water, set by λ alone. Sliding the water level moves the rim relative to those fixed antinodes; the column roars when the rim plus c lands exactly on one. Off resonance the driver is fighting the column, B collapses, and the sideways plot shrinks — the pattern has not moved, it has just gone quiet.

Loudness against air-column length

ⓘ more

Plotted amplitude is γ/√(cos²(k·L_eff) + γ²) with a loss factor γ = 0.05, which is what stops a driven column going infinite. γ sets only how sharp each peak is; the peak POSITIONS come from cos(k(L + c)) = 0 alone, so they are exactly the quarter-wave lengths and nothing about them was put in by hand.

Try first: drag L until it roars, Mark it, then find the next one.

Driver and tube

Driving frequency f512 Hz

Speed of sound v340 m s⁻¹

Tube radius r20 mm

λ = v/f

0.6641 m

λ/4

0.1660 m

c = 0.6 r

12.0 mm

ⓘ more

c ≈ 0.6 r is an approximation, not a law: the exact value for an unflanged pipe is 0.61 a, a flanged one is nearer 0.82 a, and it drifts slightly with frequency. That is precisely why no measurement below is allowed to depend on it.

Air column

Length of air column L0.100 m

amplitude ÷ peak

10.2 %

level re peak

-19.9 dB

L + c

0.1120 m

(L + c) ÷ (λ/4)

0.675

nearest resonance n = 1 at L = 0.154 m · you are -54.0 mm short of it

ⓘ more

Marking is blocked below 90 % of peak amplitude, the same way a real experiment refuses to give you a length until you can hear the maximum. Whatever length you mark is recorded exactly as it stands, so a sloppy peak-hunt shows up honestly in the wavelength below. Tune to loudest lands on the exact quarter-wave length if you want the clean numbers.

Marks → λ, v and c

No marks yet. Find the first loud length and mark it.

ⓘ more

Both resonance conditions carry the same unknown c: L₁ + c = λ/4 and L₂ + c = 3λ/4. Subtracting kills it — L₂ − L₁ = λ/2 whatever c happens to be — which is why the experiment is always written as a difference of two lengths. Change the radius slider and watch the two lengths both shift while their difference does not move at all. Once λ is known, c comes back for free from c = λ/4 − L₁ = (L₂ − 3L₁)/2, and it agrees with the 0.6 r that was fed in. The naive route never recovers: 4L₁ is short by exactly 4c, so it reports a speed of sound that is too low by 4c/λ.

What this confirms

Formation by superposition — LO 12(a), 12(d). Station 1 draws two identical waves travelling in opposite directions and adds them point by point. The bold curve is the literal sum, never the collapsed 2A sin kx cos ωt identity, so the standing pattern is a result of the addition.
The canonical T/8 sequence — LO 12(d) "graphical method". t = 0: the two waves coincide, resultant 2A. t = T/4: they are antiphase at every point and the string is completely flat. t = T/2: coincident again but inverted, −2A. The amplitude readout shows the ×2A factor collapsing to 0 and back.
Nodes are pinned — LO 12(d) "identify nodes and antinodes". The zeros of the resultant are found by bisection every frame and compared with the predicted m·λ/2. The worst drift stays at the numerical tolerance for the whole run, while the loops flip up and down between them.
λ/2 and λ/4 spacings — LO 12(d). The brackets subtract computed marker positions: node to node and antinode to antinode both equal λ/2, node to nearest antinode equals λ/4. The ratio readouts sit at 1.000 for every mode and every system.
Harmonics of a string — LO 12(c) "stretched strings". Fixed ends force a node at each end, so only λ = 2L/n fits and f_n = n·f₁. Every whole harmonic exists; the L and v sliders move f = v/λ live.
Closed pipe: odd harmonics only — LO 12(c) "air columns". A displacement node at the closed end and an antinode at the open end forces λ = 4L/(2n−1). The selector shows the even harmonics struck out because they cannot satisfy both boundary conditions, and f₁ = v/4L is an octave below the open pipe of the same length.
Open pipe: antinodes at both ends — LO 12(c). λ = 2L/n and the full harmonic series returns, so the same L gives double the fundamental of the closed pipe.
Longitudinal reality behind the transverse picture — LO 12(c). The air particles oscillate along the pipe and bunch up at displacement nodes, which is why those points are pressure antinodes. The green curve is the same displacement plotted sideways so its zeros are visible.
No net energy transfer — Station 3. P = −T (∂y/∂x)(∂y/∂t) is integrated numerically. One travelling wave gives a one-signed average matching ½TA²kω; the superposition averages to zero at a point where its instantaneous power swings at full amplitude.
Phase inside a stationary wave — LO 12(b). Two placeable markers report the sign of ψ(x₁)·ψ(x₂) alongside the count of nodes between them. Same loop gives Δφ = 0, across one node gives Δφ = π, and no intermediate value is ever produced — unlike a progressive wave, where Δφ = 2πΔx/λ takes any value.
End correction — LO 12(c), and the resonance-tube experiment. Station 4 drives a closed tube at fixed f and varies the air column. Resonance is detected where the driven amplitude peaks, and those peaks land at L + c = (2n − 1)λ/4 with the antinode drawn c beyond the rim, where the air is still moving. Marking two of them gives λ = 2(L₂ − L₁), which matches v/f because c subtracts out, while λ = 4L₁ misses by 4c — the percentage error is printed alongside. c itself is then recovered from (L₂ − 3L₁)/2 and agrees with the 0.6 r that was used.