Sound is a longitudinal wave: air molecules are pushed back and forth along the direction the wave travels. You can describe it two ways. The displacement $s(x, t)$ says how far each thin layer of air has moved from its usual place. The pressure $\Delta p(x, t)$ says how much the pressure there is above or below normal. The two describe the same wave, but their peaks are in different places, as pipes already hinted.
🔑 Key idea
Sound moves at $$v = \sqrt{\frac{B}{\rho}},$$ where $B$ is the bulk modulus (stiffness) and $\rho$ the density of the medium. For a displacement wave $s = s_{\max}\cos(kx - \omega t)$, the pressure wave is $$\Delta p = \Delta p_{\max}\sin(kx - \omega t),\qquad \Delta p_{\max} = Bks_{\max} = \rho v\omega s_{\max}.$$ Pressure is greatest where the displacement is zero, a quarter wavelength from where the air swings most.
💡 Why it works
Pressure changes when a layer of air is squeezed or stretched. A layer between $x$ and $x + \Delta x$ changes its volume by $A_{\text{c}}[s(x + \Delta x) - s(x)] \approx A_{\text{c}}\,\frac{\partial s}{\partial x}\,\Delta x$, so its fractional volume change is $\frac{\partial s}{\partial x}$. The bulk modulus is defined by $\Delta p = -B\,\frac{\Delta V}{V}$, so $$\Delta p = -B\,\frac{\partial s}{\partial x}.$$ For $s = s_{\max}\cos(kx - \omega t)$ this gives $Bks_{\max}\sin(kx - \omega t)$. The pressure depends on the slope of the displacement, and a cosine's slope is largest where the cosine itself is zero. Newton's second law on the same layer gives the wave equation with $v^2 = B/\rho$, just as tension and mass gave $v^2 = F_T/\mu$ for a string.
Since $B = \rho v^2$, the pressure amplitude can also be written $\rho v\omega s_{\max}$. Stiffness speeds sound up and density slows it down. Water is about 800 times as dense as air but about 15,000 times as stiff, so sound moves faster in water.
Worked examples
✏️ Worked example: Speed in water
Water has bulk modulus $B = 2.2 \times 10^9$ Pa and density 1000 kg/m³. Find the speed of sound in it.
$= 0.026$ Pa, about $2.6 \times 10^{-7}$ of atmospheric pressure.
Answer: $\Delta p_{\max} \approx 0.026$ Pa
The air layers in that tone move only 10 nanometers, far less than the wavelength of visible light, yet you hear it clearly. Your ear is a very sensitive pressure sensor.
⚠️ Common mistake
Assuming the pressure and the displacement peak at the same places. Where the air swings the farthest (a displacement maximum), the layers on both sides move together, so the air there is neither squeezed nor stretched: $\Delta p = 0$. The pressure peaks where neighboring layers move toward or away from each other, which is where $s = 0$. In a pipe, an open end is a displacement antinode and a pressure node; a closed end is a displacement node and a pressure antinode.
A pipe closed at one end. Switch the view between air displacement and pressure: the nodes of one are the antinodes of the other.
Practice
Pass with 3 right answers, the last 2 in a row. Each right answer earns 2 XP.