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A sound wave question shows a graph of air displacement and asks where the pressure is highest. Many students pick the crest. That is wrong, and the reason is one of the cleanest ideas in the waves chapter. This post explains why pressure and displacement in a sound wave are 90° out of phase, works through a full numerical example, and shows how the idea settles closed-pipe and open-pipe questions.
Where is this in the syllabus?
The JEE (Main) 2026 syllabus lists longitudinal waves, the displacement relation for a progressive wave, and standing waves in organ pipes. The NEET (UG) 2026 syllabus uses the same wording. Neither names "pressure waves" as a separate line. The NCERT Class 11 chapter on waves (Reprint 2026-27) does use the idea: it describes sound as compressions and rarefactions, links pressure change to the bulk modulus, says that at the closed end of a pipe "the pressure changes are the largest, while the displacement is minimum", and asks in Exercise 14.19 why "a displacement node is a pressure antinode". Check the syllabus for your own exam year.
Why is pressure a quarter wavelength away from displacement?
Think of thin layers of air. A layer that is displaced a long way is not squeezed if its neighbours are displaced by almost the same amount. What matters for pressure is how the displacement changes from one layer to the next. OpenStax University Physics, section 17.3, writes this as
Δp = −B ∂s/∂x
where B is the bulk modulus and s the displacement. If s = s₀ cos(kx − ωt), as in section 17.1, then
Δp = Bks₀ sin(kx − ωt)
A cosine turned into a sine: the two curves are shifted by a quarter wavelength, which is a phase difference of 90°. Where s is at its maximum, Δp is zero. Where s passes through zero, Δp is at its largest.
Is that point a compression or a rarefaction? Look at the neighbours. If the layer just to the left is pushed right and the layer just to the right is pushed left, they crowd together: compression. In the equation, that is where ∂s/∂x is negative, so Δp is positive.
Worked example: from pressure amplitude to displacement
Sample values for a classroom problem: a 500 Hz tone in air has a pressure amplitude of 0.60 Pa. Take the speed of sound as 340 m/s and the density of air as 1.2 kg/m³. Find the wavelength, the displacement amplitude, the largest speed of an air layer and the intensity.
Bulk modulus. NCERT gives v = √(B/ρ), so B = ρv² = 1.2 × 340² = 1.39 × 10⁵ Pa.
Wavelength and k. λ = v/f = 340/500 = 0.68 m, and k = 2π/λ = 9.24 rad/m.
Displacement amplitude. s₀ = Δp₀/(Bk) = 0.60/(1.39 × 10⁵ × 9.24) = 4.7 × 10⁻⁷ m. The air moves less than half a micrometre.
Particle speed. The largest speed of a layer is ωs₀ = 2π × 500 × 4.7 × 10⁻⁷ = 1.5 × 10⁻³ m/s. Check: ρv(ωs₀) = 1.2 × 340 × 1.47 × 10⁻³ = 0.60 Pa, which is the pressure amplitude we started with.
Intensity. OpenStax gives I = Δp₀²/(2ρv) = 0.36/(2 × 1.2 × 340) = 4.4 × 10⁻⁴ W/m².
Where are the peaks? A point of largest displacement and the nearest point of largest pressure change are λ/4 = 0.17 m apart.
Notice how small the pressure change is: 0.60 Pa against normal air pressure of about 1.0 × 10⁵ Pa, roughly six parts in a million.
What does this mean for closed and open pipes?
In a standing wave the 90° shift becomes a rule about ends. At a closed end the air cannot move, so it is a displacement node, and NCERT says the pressure changes there are the largest: a pressure antinode. At an open end the air meets the atmosphere, so the pressure stays close to normal: a pressure node and a displacement antinode. In a closed pipe in its fundamental mode, L = λ/4, so the pressure swings most at the closed end and hardly at all at the open end. Our guide to standing waves in strings and pipes works through the harmonics.
Common mistakes and how to avoid them
Putting the pressure maximum at the displacement crest. It sits a quarter wavelength away, where the displacement is zero.
Using Bs₀ instead of Bks₀. Check units: B is in pascals and s₀ in metres, so you need k (per metre) to get pascals.
Confusing particle speed with wave speed. Here ωs₀ is about 1.5 mm/s while the wave moves at 340 m/s.
Using B = P for air. The bulk modulus for sound in a gas is the adiabatic one, γP. Our post on Laplace's correction explains why.
If you find it hard to picture layers bunching up while the wave moves on, the Fermi Concept Library covers every concept in physics, so you can read up on longitudinal waves before you try the numbers again.
A practice routine for this week
Draw s = s₀ cos kx for one wavelength. Under it, draw Δp by following the slope, and label each compression and rarefaction.
Redo the worked example for a 1000 Hz tone with the same pressure amplitude. Predict first whether s₀ goes up or down.
Answer NCERT Exercise 14.19(a) in four lines.
For a closed pipe in its third harmonic, mark every displacement node and pressure antinode.
Start now: take any displacement graph from your textbook and mark the three points of greatest compression.
