IB Physics HLTopic 3 — Oscillations & WavesPaper 1 & 2Nodes every λ/2~14 min read
Nodes & Antinodes
Last page you met the two points on a standing wave that behave strangely: the ones that never budge, and the ones that swing hardest of all. Time to name them. Nodes stand perfectly still. Antinodes go wild. And here is the useful part — they are spaced out along the wave with a ruler-like regularity, so once you can spot them you can read a wavelength straight off a picture.
📘 What you need to know
A node is a point of zero amplitude — it never moves
An antinode is a point of maximum amplitude — it swings the furthest
Adjacent nodes are λ/2 apart. So are adjacent antinodes
A node and the next antinode are λ/4 apart
Nodes form by destructive interference; antinodes by constructive interference
Nodes and antinodes do not move along the wave. Antinodes only oscillate up and down
Two points on a standing wave are either in phase (0) or in anti-phase (π) — nothing in between
The anatomy of a standing wave
Every standing wave is a chain of loops. Where two loops meet, the string is nailed down. In the belly of each loop, the string is free to swing.
Nodes — zero amplitude. Take a photograph at any instant and the string is on the axis there
Antinodes — maximum amplitude. This is where the loop is fattest
Between a node and an antinode, every point has its own amplitude, somewhere in between
The dashed curve is the same wave half a period later. Notice it passes through the same nodes — those points are pinned for good.
Spacings you must knownode → next node = λ / 2antinode → next antinode = λ / 2node → next antinode = λ / 4
Here’s the trick students always miss: one loop is only half a wavelength, not a whole one. Your eye sees a fat bump and thinks “that’s a wave”. It isn’t. You need two loops, one up and one down, before you have paid for a full λ. Count loops, then halve.
Why nodes and antinodes exist
Remember where the standing wave came from: two waves running through each other in opposite directions. At some places along the string the two waves are always in step. At others they are always exactly opposed.
Crest meets trough always in anti-phase
gives
Total cancellation destructive
so you get a
NODE
Crest meets crest always in phase
gives
Displacements add constructive
so you get an
ANTINODE
At a node the two waves arrive in anti-phase at every single moment, so they cancel at every single moment. At an antinode they arrive in phase, so the displacements add and that point reaches 2A, twice the amplitude of either wave on its own.
Node
Antinode
Amplitude
Zero — never moves
Maximum, equal to 2A
Interference there
Destructive
Constructive
The two waves arrive
In anti-phase
In phase
Spacing to the next one
λ/2
λ/2
Does it move along the string?
No
No — it only oscillates up and down
Careful with “moves”: an antinode is a place, and that place stays put. What moves is the piece of string sitting at that place, which oscillates vertically with the largest amplitude on the wave.
Phase on a standing wave
On a progressive wave, phase difference slides smoothly with distance — two points can be 37° out of phase if you like. A standing wave refuses to do that. Every point is either moving exactly with another point, or exactly against it.
All the points in the same loop are in phase — they go up together and down together
Points in next-door loops are in anti-phase — one goes up as the other goes down
So: odd number of nodes between two points → anti-phase (π rad)
Even number of nodes between them (including zero) → in phase (0 rad)
A, B and D all ride the loops that bulge upwards at this instant; C is in the loop that bulges down. Half a period later they all swap — but A, B and D still agree with each other, and still disagree with C.
Don’t try to measure phase with a ruler here. Just count the nodes in between. Odd number, the points fight each other. Even number, they cooperate. That is the whole rule, and it earns easy marks in Paper 1.
Reading a wavelength off a picture
Exam questions love handing you a diagram and a length. Count the loops, and you’re done.
Three loops fit into 1.2 m, so each loop is 0.40 m — and each loop is λ/2.
📏 Getting λ out of a standing wave diagram
Count the loops between the two ends of the pattern.
Each loop is λ/2. So the length shown = (number of loops) × λ/2.
Rearrange for λ. Nothing else is needed.
Need a speed? Only now bring in v = fλ.
WE 1
On a vibrating string, two adjacent nodes are measured to be 12 cm apart. The string oscillates at 40 Hz. Calculate the wavelength, the speed of the waves on the string, and the distance from a node to the nearest antinode.
Step 1 — adjacent nodes are half a wavelength apartλ/2 = 0.12 m → λ = 0.24 mStep 2 — use the wave equationv = fλ = 40 × 0.24v = 9.6 m s⁻¹Step 3 — node to nearest antinode is a quarter wavelengthλ/4 = 0.24 / 4 = 0.060 m6.0 cmThat speed belongs to the two travelling waves on the string. The standing wave pattern itself still goes nowhere.
WE 2
A standing wave on a string of length 1.2 m shows three loops, as in the diagram above. Determine the wavelength, and state the number of nodes and antinodes.
Step 1 — each loop is half a wavelengthL = 3 × λ/2Step 2 — rearrange and substituteλ = 2L / 3 = 2 × 1.2 / 3λ = 0.80 mStep 3 — count them off the diagram
Nodes: one at each end plus two inside → 4 nodes
Antinodes: one per loop → 3 antinodesAlways one more node than antinode when both ends are nodes. Handy check.
WE 3
Using the phase diagram above, state the phase difference between points A and C, and between points A and D. Explain your reasoning.
Step 1 — count the nodes between A and C
One node lies between them → odd numberA and C: π rad (anti-phase)Step 2 — count the nodes between A and D
Two nodes lie between them → even numberA and D: 0 rad (in phase)Step 3 — justify it
Points in the same loop move together; each node you cross flips the motion.
Notice A and D have different amplitudes but the same phase. Amplitude and phase are separate questions — don’t mix them up.
💡 Top tips
One loop = half a wavelength. Burn this in.
Node to node λ/2; node to nearest antinode λ/4. Sketch it if you blank.
Phase on a standing wave is only ever 0 or π. Count nodes: odd = anti-phase, even = in phase.
If both ends are nodes, nodes = antinodes + 1. Use it to check your counting.
An antinode reaches 2A, twice the amplitude of either travelling wave.
⚠ Common mistakes
Calling one loop a full wavelength — it is λ/2
Saying nodes are λ apart instead of λ/2
Thinking nodes and antinodes travel along the string — their positions are fixed
Quoting a phase difference like 90° between two points on a standing wave — impossible
Assuming two points with different amplitudes must be out of phase
Forgetting that a node has zero amplitude, not just a small one
Quick recap:Nodes are points of zero amplitude formed by destructive interference; antinodes are points of maximum amplitude (2A) formed by constructive interference. Adjacent nodes, and adjacent antinodes, sit λ/2 apart; a node and its neighbouring antinode are λ/4 apart. Neither travels along the wave. Two points are in phase if an even number of nodes separates them, and in anti-phase if an odd number does.
Now a question that decides everything: what is happening at the ends? Tie the string down and the end must be a node. Leave it free, or open a pipe to the air, and the end becomes an antinode. Those choices are called boundary conditions, and they decide which standing waves a string or pipe is even allowed to have. That is the next page.
Nodes and antinodes tying you in knots?
Book a free meeting and we’ll practise reading wavelengths off diagrams and nailing the phase rule together.
