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Speed of Transverse Waves

Level 1 - Physics topic page in Mechanical Waves.

Principle

The speed of a transverse wave on a string is set by the competition between tension and inertia.

Tension tries to straighten a curved segment, while the string's mass per unit length resists acceleration.

Notation

\(T\)
string tension
\(\mathrm{N}\)
\(\mu\)
mass per unit length
\(\mathrm{kg\,m^{-1}}\)
\(y(x,t)\)
transverse displacement
\(\mathrm{m}\)
\(\Delta x\)
small string element length
\(\mathrm{m}\)
\(v_w\)
wave speed
\(\mathrm{m\,s^{-1}}\)

Method

Derivation 1: Find the net transverse force on a short segment

Take a short string element of length \(\\Delta x\). The tension has nearly the same magnitude at both ends, but the directions differ because the string is curved.

Vertical force balance
\[F_y=T\sin\theta(x+\Delta x)-T\sin\theta(x)\]
Small-slope approximation
\[\sin\theta\approx\tan\theta\approx\frac{\partial y}{\partial x}\]
Slope difference
\[F_y\approx T\left[\frac{\partial y}{\partial x}(x+\Delta x)-\frac{\partial y}{\partial x}(x)\right]\]
Curvature form
\[F_y\approx T\frac{\partial^2 y}{\partial x^2}\Delta x\]

Derivation 2: Apply Newton's second law to the segment

The mass of the segment is \(\\mu\\Delta x\). Its transverse acceleration is \(\\partial^2 y/\\partial t^2\).

Segment mass
\[m=\mu\Delta x\]
Newton's law
\[\mu\Delta x\frac{\partial^2 y}{\partial t^2}=T\frac{\partial^2 y}{\partial x^2}\Delta x\]
Wave equation form
\[\frac{\partial^2 y}{\partial t^2}=\frac{T}{\mu}\frac{\partial^2 y}{\partial x^2}\]

Derivation 3: Read the wave speed from the equation

The standard one-dimensional wave equation is \(\\partial^2 y/\\partial t^2=v_w^2\\,\\partial^2 y/\\partial x^2\). Matching coefficients gives the string-wave speed.

Coefficient match
\[v_w^2=\frac{T}{\mu}\]
Transverse-wave speed
\[v_w=\sqrt{\frac{T}{\mu}}\]

Rules

These are the compact results from the derivation above.

String wave equation
\[\frac{\partial^2 y}{\partial t^2}=\frac{T}{\mu}\frac{\partial^2 y}{\partial x^2}\]
Transverse-wave speed
\[v_w=\sqrt{\frac{T}{\mu}}\]

Examples

Question
A string has tension
\[120\,\mathrm{N}\]
and linear density
\[0.030\,\mathrm{kg\,m^{-1}}\]
Find the wave speed.
Answer
Use
\[v_w=\sqrt{T/\mu}\]
\[v_w=\sqrt{\frac{120}{0.030}}=63.2\,\mathrm{m\,s^{-1}}\]

Checks

  • Increasing tension increases wave speed because the restoring force is larger.
  • Increasing mass per unit length decreases wave speed because the same force accelerates more mass.
  • The units of \(T/\\mu\) are meters squared per second squared, so the square root has units of speed.
  • This formula is for a stretched string with small transverse displacements.

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