IB Physics HL Topic 3 — Oscillations & Waves Paper 1 & 2 Fixed = node, free = antinode ~15 min read

Standing Wave Boundary Conditions

A standing wave is a fussy guest. It cannot settle just anywhere — the ends of the string or pipe lay down the law. Tie an end down and the wave is forced to be still there. Let an end move freely and the wave insists on swinging hardest there. Those two demands are the boundary conditions, and they decide which standing waves are even allowed to exist.

📘 What you need to know

Why an end forces the wave’s hand

Everything on this page comes from one question: what happens to the wave when it hits the end?

What the end does to the reflectionFIXED END FREE ENDincoming reflected flipped over → a NODE here not flipped → an ANTINODE here
The ring on the rod is the classic picture of a free end: the string can still slide up and down there, so nothing forces it to be still.
Fixed / closed end
cannot move
reflection
inverted
Waves cancel
destructive
so
NODE
Free / open end
free to move
reflection
upright
Waves add
constructive
so
ANTINODE
Two words, and you never have to think again: tied means still. If an end is tied down, closed off, clamped, it is a node. If it is loose, open, free to wobble, it is an antinode. Every standing wave question begins by asking yourself this about both ends.

Boundary conditions on a string

A string has two ends, and each one can be fixed or free. That gives three combinations. Here is the simplest pattern each one allows.

The three string boundary conditionsfixed — fixed N Nfree — free A Afixed — free N AN = node    A = antinode    dashed = half a period later
Notice the bottom string: only a quarter of a loop fits. A fixed–free string is a different animal from a fixed–fixed one, and its allowed frequencies are different too.
The guitar connection: a guitar string is fixed at both ends, so both ends are nodes. How fast it vibrates depends on the tension (set by the tuning pegs) and the mass per unit length (which is why the low strings are the thick ones).

Boundary conditions in a pipe

Blow across the top of a bottle and the air column inside vibrates. These are longitudinal standing waves, but the rules are identical: closed end, node. Open end, antinode.

The three pipe boundary conditionsopen — open A Aclosed — closed N Nopen — closed A Nthe curve shows how far the air moves, not the shape of the pipe
Careful: the wiggly line is a graph of air displacement along the pipe. The air itself sloshes back and forth along the tube, not up and down.
End of the mediumReflected waveWhat forms there
Fixed end of a stringInverted — in anti-phaseNode
Closed end of a pipeInverted — in anti-phaseNode
Free end of a stringUpright — in phaseAntinode
Open end of a pipeUpright — in phaseAntinode

Only certain waves are allowed

Now put the two ideas together. The ends demand nodes or antinodes in particular places, and nodes are spaced λ/2 apart. Between them, they leave only a short list of wavelengths that fit.

🧭 Working out what fits

  1. Label both ends first. Fixed or closed → N. Free or open → A.
  2. Sketch the simplest curve that obeys both labels. Don’t add loops you don’t need.
  3. Measure it in quarters. N to N is λ/2. N to A is λ/4. A to A is λ/2.
  4. Set that equal to L and solve for λ. Only then use v = .
WE 1

A pipe of length 0.85 m is open at one end and closed at the other. State the boundary condition at each end and determine the longest wavelength that can form a standing wave in the pipe.

Step 1 — label the ends Open end → antinode. Closed end → node. Step 2 — the simplest pattern that fits Just a node at one end and an antinode at the other, with nothing in between. That distance is a quarter of a wavelength: L = λ/4 Step 3 — rearrange λ = 4L = 4 × 0.85 λ = 3.4 m Longest wavelength means simplest pattern. Add loops and the wavelength only gets shorter.
WE 2

A string of length 1.5 m is fixed at both ends. A standing wave on it has a total of 3 nodes. The speed of waves on the string is 60 m s⁻¹. Determine the wavelength and the frequency of the vibration.

Step 1 — turn nodes into loops 3 nodes (one at each end, one in the middle) → 2 loops Step 2 — each loop is half a wavelength L = 2 × λ/2 = λ λ = 1.5 m Step 3 — use the wave equation f = v/λ = 60 / 1.5 f = 40 Hz Both ends fixed, so both ends are nodes. That is what made “3 nodes” mean “2 loops”.
WE 3

A pipe of length 0.34 m is open at both ends. Taking the speed of sound in air as 343 m s⁻¹, calculate the lowest frequency at which a standing wave can form in the pipe.

Step 1 — label the ends Both open → an antinode at each end, so one node sits in the middle. Step 2 — antinode to antinode is half a wavelength L = λ/2 → λ = 2 × 0.34 = 0.68 m Step 3 — use the wave equation f = v/λ = 343 / 0.68 f = 504 Hz Lowest frequency goes with the longest wavelength, which is the simplest pattern. That is why we drew only one node.

💡 Top tips

⚠ Common mistakes

Quick recap: The ends decide everything. A fixed string end or closed pipe end reflects the wave inverted, so it must be a node. A free string end or open pipe end reflects it upright, so it must be an antinode. Match those demands against the λ/2 and λ/4 spacings and only a short list of wavelengths survives — the natural frequencies of the string or pipe.
You have just found the simplest pattern for each set of ends. But the ends will happily accept more loops: two, three, ten. Each one is a new allowed wavelength and a new allowed frequency, and they come in a beautiful sequence. Those are the harmonics, and they are the next page.

Boundary conditions boxing you in?

Book a free meeting and we’ll practise labelling ends, sketching patterns and turning them into wavelengths and frequencies.

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