Longitudinal and transverse waves

Pluck a stretched string and a kink races down its length while the string itself only bobs up and down. Speak into the air and your vocal cords shove air molecules back and forth along the very line the sound travels. Both are mechanical waves — a disturbance that carries energy through a medium without carrying the medium's particles along with it — but the geometry of the disturbance is opposite in the two cases. That single geometric fact, whether the medium's displacement is perpendicular to or parallel with the direction of travel, is what separates transverse waves from longitudinal waves, and it is the classification you must nail before writing down any wave equation or wave-speed formula for a real system. This article expands the Pass-1 treatment of PH11001's opening topic in the "Wave Equation and Basic Wave Motion" block: the geometric test that tells the two apart, why a string can carry only one kind while a fluid can carry only the other, the shared wave-equation form and the two wave-speed formulas $v=\sqrt{T/\mu}$ and $v=\sqrt{B/\rho}$ , worked numbers for a plucked string and for sound in steel, and the traps examiners love — confusing particle velocity with wave velocity, and assuming speed scales linearly with tension.

The core insight

A wave is transverse if the medium's particles oscillate perpendicular to the direction of propagation (string: $v=\sqrt{T/\mu}$ ) and longitudinal if they oscillate parallel to it (sound: $v=\sqrt{B/\rho}$ ), with wave speed in both cases set by restoring stiffness over inertia.

The core idea

A mechanical wave is a disturbance that propagates through a medium while each particle of the medium merely oscillates about its own equilibrium position — the wave transports energy and momentum, not matter. Classifying a wave means asking one geometric question: does a particle move across the direction the disturbance travels, or along it?

Transverse waves. The particle displacement is perpendicular to the direction of propagation. On a stretched string the wave travels horizontally along $x$ while the string's displacement $y(x,t)$ is measured vertically, at right angles to $x$ . A crest moves down the string even though no single bit of string moves down the string with it — each bit only moves up and down in place.

Longitudinal waves. The particle displacement is parallel to the direction of propagation. Sound in air is the standard case: air molecules oscillate back and forth along the same line the sound wave travels, bunching into regions of compression (higher pressure and density) and thinning into regions of rarefaction (lower pressure and density) that march forward together. Why can a string only support a transverse wave while a bulk fluid can only support a longitudinal one? The answer is about what restoring force is available. A stretched string resists being displaced sideways — tension pulls a displaced bit back toward the line of the string, which is a transverse restoring force, and a string under tension has no mechanism to resist being compressed along its own length, so it cannot sustain a longitudinal wave. A fluid is the opposite: it has no rigidity to resist a sideways shear (a fluid element does not spring back if you push it sideways — it just flows), but it strongly resists being compressed, which is exactly the restoring force a longitudinal wave needs. Solids, having both shear rigidity and compressive stiffness, can in principle carry both kinds, but PH11001 keeps you to the two idealised pure cases. Despite this difference in geometry, both wave types obey the same mathematical wave equation,

\frac{\partial^2 \psi}{\partial t^2} = v^2\,\frac{\partial^2 \psi}{\partial x^2},

where: $\psi$ stands for whatever quantity is oscillating — the transverse displacement $y(x,t)$ for a string, or the longitudinal displacement $s(x,t)$ (equivalently the pressure or density fluctuation) for sound — and $v$ is the wave speed, a single number set entirely by the medium. What changes between the two cases is only which physical properties combine to give $v$ :

v_{\text{string}} = \sqrt{\frac{T}{\mu}}, \qquad v_{\text{sound}} = \sqrt{\frac{B}{\rho}},

where: $T$ is the string's tension (N), $\mu$ its mass per unit length (kg/m), $B$ the medium's bulk modulus (Pa, its resistance to being compressed), and $\rho$ its density (kg/m³). Both formulas have the same shape, $v=\sqrt{\text{stiffness}/\text{inertia}}$ — a stiffer restoring mechanism (larger $T$ or $B$ ) speeds the wave up, and more inertia (larger $\mu$ or $\rho$ ) slows it down. It is the same pattern you met for $\omega_0=\sqrt{k/m}$ in simple harmonic motion, now applied to a continuous medium instead of a single mass.

Figure

Two boxes comparing a transverse wave on a horizontal string with vertical particle motion against a longitudinal sound wave in a horizontal tube with particle motion along the same horizontal line as propagation

Derivation AWhich restoring force sets the speed

The goal here is to justify, without extra machinery beyond what tension and bulk modulus already mean physically, why $v=\sqrt{T/\mu}$ and $v=\sqrt{B/\rho}$ take exactly the ratio-of-stiffness-to-inertia form rather than some other combination, and why that lets you predict how $v$ changes when a medium property is scaled.

Step 1. Identify the restoring mechanism in each case. On a string, a small transverse displacement is pulled back toward the equilibrium line by the component of tension $T$ acting sideways on the displaced segment — the larger $T$ , the stronger this restoring pull, so $v$ must grow with $T$ . In a fluid, a small compression is pushed back toward equilibrium by the excess pressure the compression itself creates, and the bulk modulus $B$ is defined precisely as the medium's resistance to such compression — the larger $B$ , the stronger the restoring push, so $v$ must grow with $B$ .

Step 2. Identify the inertia that resists being accelerated by that restoring force. For the string it is the mass per unit length $\mu$ : a heavier string (larger $\mu$ ) accelerates more sluggishly under the same tension-driven restoring force, so $v$ must fall as $\mu$ grows. For the fluid it is the density $\rho$ : denser fluid resists being pushed back and forth more, so $v$ must fall as $\rho$ grows.

Step 3. Combine stiffness over inertia and fix the square root by dimensional consistency. Tension $T$ has units of force (N = kg·m/s²) and $\mu$ has units kg/m, so $T/\mu$ has units $\text{m}^2/\text{s}^2$ — a speed squared — and taking the square root gives a genuine speed, $v=\sqrt{T/\mu}$ . Bulk modulus $B$ has units of pressure (Pa = N/m² = kg/(m·s²)) and $\rho$ has units kg/m³, so $B/\rho$ again works out to $\text{m}^2/\text{s}^2$ , giving $v=\sqrt{B/\rho}$ . Both formulas are forced into the square-root-of-a-ratio form by dimensional bookkeeping alone, which is why increasing the numerator (stiffness) by a factor of $k$ increases $v$ by only $\sqrt{k}$ , not $k$ .

Step 4. Read off the scaling consequence that catches most students out. Because $v \propto \sqrt{T}$ , doubling the tension increases speed by only $\sqrt{2}\approx1.41$ , not $2$ ; to double $v$ you must quadruple $T$ . The identical logic applies to $B$ for sound waves — the square root, not the ratio itself, is what you must scale. Verified: dimensional check $T/\mu = \text{N}/(\text{kg/m}) = \text{m}^2/\text{s}^2$ and $B/\rho = \text{Pa}/(\text{kg/m}^3) = \text{m}^2/\text{s}^2$ , both consistent with $v$ in m/s.

Worked example

A horizontal string of mass per unit length $\mu = 5\times10^{-3}\ \text{kg/m}$ is stretched under tension $T = 45\ \text{N}$ , and separately sound travels through a steel rod with bulk modulus $B = 1.6\times10^{11}\ \text{Pa}$ and density $\rho = 7800\ \text{kg/m}^3$ . Find the wave speed in each case, the time for a transverse pulse to cross a $L=2.0\ \text{m}$ length of the string, and compare the steel result with sound in air ( $v_{\text{air}}\approx343\ \text{m/s}$ ).

Step 1 — classify each wave. The string is displaced perpendicular to its own length while the pulse travels along the string: transverse. Sound in the steel rod involves molecules oscillating back and forth along the direction of travel, forming compressions and rarefactions: longitudinal.

Step 2 — string speed and crossing time.

v_{\text{string}} = \sqrt{\frac{T}{\mu}} = \sqrt{\frac{45}{5\times10^{-3}}} = \sqrt{9000} \approx 94.9\ \text{m/s}, \qquad t = \frac{L}{v} = \frac{2.0}{94.9} \approx 0.0211\ \text{s} \approx 21\ \text{ms}.

Step 3 — steel speed and comparison with air.

v_{\text{steel}} = \sqrt{\frac{B}{\rho}} = \sqrt{\frac{1.6\times10^{11}}{7800}} = \sqrt{2.051\times10^{7}} \approx 4529\ \text{m/s}.

Dividing by air's speed, $4529/343 \approx 13.2$ — sound is roughly thirteen times faster in steel than in air, because steel's enormous bulk modulus far outweighs its higher density: stiffness dominates the $B/\rho$ ratio.

Answer: $v_{\text{string}} \approx 94.9\ \text{m/s}$ with a $2.0\ \text{m}$ pulse crossing time of about $21\ \text{ms}$ ; $v_{\text{steel}} \approx 4529\ \text{m/s}$ , about $13\times$ faster than sound in air.

Competitive stretch

If the same string's tension were quadrupled to $T' = 180\ \text{N}$ with $\mu$ unchanged, the new speed is $v' = \sqrt{T'/\mu} = \sqrt{180/(5\times10^{-3})} = \sqrt{36{,}000} \approx 189.7\ \text{m/s}$ , almost exactly double $94.9\ \text{m/s}$ — confirming that quadrupling $T$ (not doubling it) is what doubles $v$ , consistent with $v\propto\sqrt{T}$ .

Common traps

Thinking wave speed depends on amplitude — how hard you pluck the string or how loud the sound is. Both $v=\sqrt{T/\mu}$ and $v=\sqrt{B/\rho}$ depend only on medium properties, never on the disturbance's size.
Confusing particle velocity (the up-down or back-forth speed of a bit of string or air) with wave velocity $v$ (the speed the disturbance pattern travels at) — these are generally different quantities, related by $\partial y/\partial t = -v\,\partial y/\partial x$ for a wave of fixed shape.
Assuming doubling $T$ doubles the string's wave speed; since $v\propto\sqrt{T}$ , you must quadruple $T$ to double $v$ , and the same square-root scaling applies to $B$ for sound.
Believing solids can only carry transverse (shear) waves — solids support both longitudinal (compressional) and transverse (shear) waves; PH11001 only asks you to apply $v=\sqrt{B/\rho}$ to the longitudinal case.

Ask the tutor: Want to see the particle-velocity-vs-wave-velocity relation plotted for a moving pulse, or work through another $v=\sqrt{T/\mu}$ or $v=\sqrt{B/\rho}$ numerical? Ask the AI tutor.

Why can a string not carry a longitudinal wave the way air does?

A string under tension only has a restoring force for sideways (transverse) displacement — tension pulls a displaced bit back toward the straight line of the string. It has essentially no resistance to being stretched or compressed along its own length locally, so there is no restoring mechanism available to sustain a longitudinal disturbance. A fluid is the reverse: it has no shear rigidity to resist sideways displacement but strongly resists compression, so it can only carry longitudinal waves.

Is wave velocity the same thing as the speed of the vibrating particles?

No, and mixing them up is one of the most common exam errors. Wave velocity $v=\sqrt{T/\mu}$ or $\sqrt{B/\rho}$ is a fixed number set by the medium — it describes how fast the disturbance pattern itself moves along. Particle velocity is how fast an individual bit of string or air is moving at a given instant as it oscillates about equilibrium, and it constantly changes with time and position, generally reaching a completely different peak value than $v$ .

Do I need to memorize both v equals square root of T over mu and v equals square root of B over rho separately?

Yes, but notice they share one structure: both are $v=\sqrt{\text{restoring stiffness}/\text{inertia per unit something}}$ . For the string, stiffness is tension $T$ and inertia is mass per unit length $\mu$ ; for a fluid or solid carrying sound, stiffness is the bulk modulus $B$ and inertia is density $\rho$ . Once you see the pattern, and check it dimensionally reduces to $\text{m}^2/\text{s}^2$ under the square root, both formulas become easy to reconstruct rather than purely memorize.

Longitudinal and transverse waves

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads