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Lesson 3 of 8

Waves · Lesson 3 of 8

What Determines Wave Speed

“Use restoring stiffness and inertia to calculate wave speeds, and explain Newton’s sound-speed estimate and Laplace’s correction.”

Learning Objectives

• Derive the dimensional dependence of transverse string-wave speed and explain what dimensional analysis cannot determine. • Apply v = √(F/μ), v = √(B/ρ) and v = √(Y/ρ) in their appropriate models. • Derive the isothermal and adiabatic bulk moduli of an ideal gas and explain why ordinary sound uses the latter. • Explain how temperature, pressure, humidity and changes of medium affect sound speed and wavelength.

Restoring action competes with inertia

Why does a pulse move faster on a taut rope than on a slack one? A greater tension produces a stronger restoring response when a string is bent. For a given restoring force, a more massive element accelerates less, so greater mass per unit length tends to slow transmission. These are properties of the prepared medium. The source sets how frequently it drives the motion; once speed and frequency are fixed, wavelength follows from λ = v/f. For a small wave on an ideal uniform string, increasing the driving frequency shortens the wavelength rather than making the wave travel faster.

Use F for the string tension in this lesson so that T can continue to mean oscillation period. The source sometimes uses T for tension. Linear mass density μ = m/L is mass per stretched length in kg m⁻¹, unlike the volume density ρ in kg m⁻³. The length and mass must refer to the same string section under the stated conditions.

Dimensional analysis for a stringLaTeX
The dimensional symbol 𝕋 here denotes time; C is dimensionless. Assume F and μ are the only relevant material parameters.

Matching mass, length and time powers with [v] = L𝕋⁻¹ gives α + β = 0, α − β = 1 and −2α = −1. The last equation gives α = 1/2, hence β = −1/2. Therefore v = C√(F/μ). Dimensional analysis cannot establish C = 1: it only fixes the powers. The detailed Newton-law string calculation, not developed in this source, supplies that constant for the ideal string model.

Transverse speed on a stretched stringLaTeX
Uniform flexible string, steady tension and sufficiently small transverse slopes; F in N and μ in kg m⁻¹ give v in m s⁻¹.

F/μ has units (kg m s⁻²)/(kg m⁻¹) = m² s⁻², so its square root has the correct speed unit. Quadrupling tension doubles speed; quadrupling μ halves it. Amplitude is absent because the ideal linear model assumes the disturbance does not appreciably alter the tension. Very large slopes, changing tension, bending stiffness or nonuniform string properties can invalidate that simplified model.

A steel wire under tension

Problem
A 0.72 m wire has mass 5.0 × 10⁻³ kg and tension 60 N. Calculate its transverse wave speed.

  1. 1.The relevant inertia is mass per length: μ = 0.0050/0.72 = 0.006944 kg m⁻¹.
  2. 2.Use the tension-supported transverse string model: v = √(60/0.006944) = √8640 = 92.95 m s⁻¹.
  3. 3.With the precision of the data, report 93 m s⁻¹. This is the transverse speed set by the external tension, not the speed of longitudinal elastic sound in bulk steel.
Travel time and the tension required for a chosen speed

Problem
(a) A 20.0 m string has mass 2.50 kg and tension 200 N. Find the one-way pulse time. (b) A 12.0 m steel wire of mass 2.10 kg must carry transverse waves at 343 m s⁻¹. Find its tension.

  1. 1.(a) μ = 2.50/20.0 = 0.125 kg m⁻¹, so v = √(200/0.125) = 40.0 m s⁻¹. The pulse time is length/speed = 20.0/40.0 = 0.500 s.
  2. 2.(b) μ = 2.10/12.0 = 0.175 kg m⁻¹. Rearrange v² = F/μ to F = μv².
  3. 3.The required tension is 0.175(343)² = 20 588.575 N ≈ 2.06 × 10⁴ N. Matching a sound speed in air does not change the wave’s transverse character on the wire.

Compression and the bulk modulus

For sound in a fluid, the restoring response comes from pressure changes when a small volume is compressed. Bulk modulus B measures the pressure increase required for a given fractional reduction in volume. If pressure rises, volume falls, so the minus sign makes the modulus positive for a stable material. Use sufficiently small changes, or the corresponding local derivative, to characterise the response near equilibrium.

Bulk modulus and fluid sound speedLaTeX
B is in Pa = kg m⁻¹ s⁻² and ρ in kg m⁻³. Use the modulus appropriate to the thermodynamic process of the sound disturbance.

The ratio B/ρ has dimensions (ML⁻¹𝕋⁻²)/(ML⁻³) = L²𝕋⁻². The same stiffness–inertia reasoning as for a string gives √(B/ρ), with the dimensionless factor supplied by the full dynamics. It is the ratio, not density alone, that determines speed. Liquids and solids are denser than gases, but their resistance to compression is generally much greater as well, so many transmit sound faster. This is a trend with exceptions, not a ranking rule based only on the material’s state.

For longitudinal extension and compression in a slender elastic rod, the relevant effective stiffness is Young’s modulus Y. In the elementary long-wavelength rod model, v = √(Y/ρ). This is not a universal formula for all longitudinal waves in an unbounded three-dimensional solid, where additional elastic moduli enter. A transverse wave on a tensioned steel wire, longitudinal sound in a rod, and bulk sound in steel are different mechanical models.

Medium / conditionRepresentative sound speed (m s⁻¹)What to notice
Air, 0 °C331Temperature matters even in the same gas
Air, 20 °C343Higher temperature raises the speed
Helium965Low molar mass gives a high gas sound speed
Hydrogen1284Still faster in this representative table
Water, 0 °C1402Liquid compressibility differs greatly from air
Water, 20 °C1482Use the value at the stated temperature
Seawater1522Composition also matters
Aluminium6420Large elastic stiffness can dominate greater density
Copper3560Elastic properties vary across solids
Steel5941Representative bulk value, not a tensioned-wire value
Granite6000Material and wave model matter
Vulcanised rubber54A solid need not have a high sound speed

Newton’s isothermal estimate

To estimate B for a gas we need a relation between pressure and volume during the disturbance. Newton assumed the compression and expansion were isothermal: temperature remains constant. For a fixed number N of ideal-gas molecules, PV = NkᴮT. At constant T, PV is constant. For small changes, (P + ΔP)(V + ΔV) − PV = 0. Ignoring the second-order product ΔPΔV gives VΔP + PΔV = 0. Rearranging yields −ΔP/(ΔV/V) = P, so the isothermal bulk modulus equals the equilibrium pressure.

Newton’s ideal-gas estimateLaTeX
This follows from an isothermal assumption. It is not the correct ordinary acoustic modulus for air.
Testing Newton’s prediction at STP

Problem
Take the mass of one mole of air as 29.0 × 10⁻³ kg, its volume as 22.4 L, and pressure as 1.01 × 10⁵ Pa. Estimate sound speed using Newton’s assumption, then compare with 331 m s⁻¹.

  1. 1.Convert litres to cubic metres: 22.4 L = 22.4 × 10⁻³ m³. Thus ρ = 0.0290/0.0224 = 1.29464 kg m⁻³.
  2. 2.The isothermal estimate is √(101000/1.29464) = 279.31 m s⁻¹, or about 280 m s⁻¹ with rounded input values.
  3. 3.The estimate is roughly 16% below 331 m s⁻¹ (often described as about 15%). The units and arithmetic are consistent, so examine the assumed physical process rather than forcing agreement by changing a number.

Laplace’s correction: compression is effectively adiabatic

The pressure changes in ordinary sound are rapid enough that neighbouring regions have little time to exchange the heat needed to maintain equal temperature during each cycle. A small compressed element warms and becomes stiffer than in an isothermal compression. Model the local reversible response as adiabatic, PVᵞ = constant, with γ = Cₚ/Cᵥ. For the same gas, γ is dimensionless and greater than one. This approximation describes ordinary small-amplitude sound; thermal conduction and losses can matter in other frequency or length regimes.

Differentiating the adiabatic relationLaTeX
The product rule differentiates both changing factors; a small volume change has Δ(Vᵞ) ≈ γVᵞ⁻¹ΔV.

Divide by Vᵞ to obtain ΔP + γP(ΔV/V) = 0. Move the volume term to the other side and divide by −ΔV/V: the adiabatic bulk modulus is γP. It is larger than the isothermal modulus by γ. Since speed depends on the square root of modulus, the sound-speed correction factor is √γ, rather than γ.

Laplace-corrected speedLaTeX
For ideal diatomic air near ordinary conditions, use γ ≈ 7/5 = 1.4.

Using the preceding unrounded density and pressure gives v ≈ 330.48 m s⁻¹. Using standard constants P = 1.01325 × 10⁵ Pa and molar volume 22.414 L gives approximately 331.1 m s⁻¹. Both agree at the precision of the stated textbook inputs with the quoted 331 m s⁻¹. Do not claim an extra decimal place that the rounded data do not support.

Pressure, temperature and humidity

Ideal-gas dependenceLaTeX
Tₐᵦₛ is absolute temperature in K; R = 8.314 J mol⁻¹ K⁻¹; M is molar mass in kg mol⁻¹.

At a fixed temperature and composition, increasing pressure also increases density in the same ratio, leaving P/ρ and therefore the ideal-gas sound speed unchanged. It would be wrong to change P alone while pretending ρ is independent of it. At fixed composition and approximately constant γ, speed grows as the square root of absolute temperature, so temperature ratios must use kelvin. Humid air contains water vapour of molar mass 18 g mol⁻¹, replacing some dry air of average molar mass about 29 g mol⁻¹. At equal temperature and pressure this reduces the mixture’s average molar mass and generally increases sound speed. Humidity is water vapour mixed into air, not suspended liquid droplets.

Changing the medium, not the source rhythm

At a stationary interface with steady media, the reflected and transmitted waves keep the driving frequency: adjacent regions must respond in step at the boundary. Their speeds can differ, so their wavelengths change according to λ = v/f. Increasing the frequency of the source does not by itself increase the speed in the ideal nondispersive medium. Similarly, moving the source does not alter the medium’s rest-frame elastic properties. Source motion can change the frequency detected elsewhere; a full Doppler treatment is outside the supplied chapter.

Ultrasound across an interface and in tissue

Problem
(a) A 1000 kHz ultrasonic wave in air meets stationary water. Use 340 m s⁻¹ in air and 1486 m s⁻¹ in water. (b) A tissue scanner works at 4.2 MHz where sound speed is 1.7 km s⁻¹. Find the relevant wavelengths.

  1. 1.(a) Convert 1000 kHz to 1.000 × 10⁶ Hz. The reflected wave remains in air, so λᵣ = 340/10⁶ = 3.40 × 10⁻⁴ m.
  2. 2.The transmitted frequency stays 10⁶ Hz but its speed becomes 1486 m s⁻¹. Thus λₜ = 1486/10⁶ = 1.486 × 10⁻³ m. The longer wavelength is caused by the larger speed, not a changed frequency.
  3. 3.(b) Convert 1.7 km s⁻¹ to 1700 m s⁻¹ and 4.2 MHz to 4.2 × 10⁶ Hz. Then λ = 1700/(4.2 × 10⁶) = 4.05 × 10⁻⁴ m, approximately 0.40 mm at two significant figures.

The frequency-independent speeds used above describe nondispersive models. A pulse is built from a range of sinusoidal components. In a dispersive medium the phase speed depends on frequency, so those components accumulate different relative phases as they propagate. Their superposition generally changes shape: the pulse broadens or distorts. A single pure sinusoid still has a defined wavelength and phase speed, but one universal speed need not describe an entire pulse. This explains why a pulse in a dispersive medium cannot generally be treated as one unchanged function F(x − vt).

Quick Quiz

Quiz

Quick check

Tension increases by a factor of four while μ stays fixed. The string-wave speed becomes

Explanation

Speed is proportional to √F.

Quick check

Dimensional analysis of F and μ alone determines

Explanation

Dynamics is needed to find the dimensionless factor.

Quick check

Which modulus applies to ordinary small-amplitude sound in ideal air?

Explanation

Rapid compression is approximately adiabatic.

Quick check

At fixed ideal-gas temperature and composition, pressure doubles. What happens to sound speed?

Explanation

P/ρ remains RT/M under these conditions.

Quick check

A wave crosses a stationary air–water interface. Which statement is correct?

Explanation

The boundary keeps the driving rhythm, while each medium sets its own speed.

Practice Problems

Practice Problems
  1. A 10.0 m rope of mass 0.50 kg has tension 80 N. Calculate μ, wave speed and the time for a pulse to reach the far end. Then predict the speed if the tension is halved.
  2. Derive v ∝ √(F/μ) by comparing dimensions. Explain why this calculation cannot determine the dimensionless factor or test the assumption that only F and μ matter.
  3. Starting from PV = constant and PVᵞ = constant separately, obtain the isothermal and adiabatic bulk moduli and explain physically which describes ordinary sound in air.
  4. At constant composition, compare the ideal-gas sound speeds at 27 °C and 127 °C. State why a ratio of the Celsius temperatures would be incorrect.
  5. Explain the independent effects of pressure at fixed temperature, temperature at fixed composition, and humidity at fixed temperature and pressure on sound speed in an ideal-gas model.
  6. A 1000 kHz wave reflects from air into water at a stationary surface. Calculate reflected and transmitted wavelengths using 340 and 1486 m s⁻¹, and explain the unchanged frequency.
  7. A scanner uses 4.2 MHz ultrasound in tissue with speed 1.7 km s⁻¹. Calculate its wavelength in metres and millimetres, showing all unit conversions.
  8. A slender rod has Young’s modulus 2.0 × 10¹¹ Pa and density 8.0 × 10³ kg m⁻³. Estimate its longitudinal rod-wave speed, and explain why the same formula does not describe transverse waves on a tensioned wire.

Key Takeaways

Key Takeaways

• Wave speed is set by the relevant restoring response and inertia in the chosen mechanical model. • For a string v = √(F/μ); for a fluid v = √(B/ρ); for a slender rod v = √(Y/ρ). • Ordinary ideal-gas sound uses B = γP, giving Laplace’s correction to Newton’s isothermal estimate. • At a stationary interface frequency is preserved, while wavelength follows the speed of the new medium.