From Beads to a String: The Continuum Limit
~7 min read · Tier toggle changes depth (T1 plain · T2 standard · T3 rigour)
You should be comfortable with the equations of motion for a discrete chain of coupled oscillators and the concept of normal modes. Familiarity with Taylor series expansions for small arguments is also assumed.
Why this matters
The transition from discrete beads to a continuous string is the conceptual bridge that turns mechanics into field theory.
So far, we have treated waves as collective motions of a finite number of distinct masses. However, a real guitar string or a sound wave in air does not consist of countable beads; it is a continuous medium. To describe such systems, we must take the limit where the number of oscillators goes to infinity and their spacing goes to zero. This process reveals that the underlying physics is governed by a partial differential equation rather than a set of coupled ordinary differential equations. Understanding this limit is crucial because it introduces the field concept: a quantity defined at every point in space, rather than just at discrete locations. This shift in perspective is fundamental to all of wave physics, from acoustics to electromagnetism.
The Idea in Plain Language
Imagine a long chain of beads connected by springs. If you look closely at one bead, its motion depends on the positions of its immediate neighbours. If the beads are far apart, the motion looks "jumpy" or discrete. But if you pack the beads very tightly together, the jumps become so small that the chain looks like a smooth, continuous rope. The "bead" becomes an infinitesimal element of the rope. The "spring" becomes the tension in the rope. The discrete difference between the positions of neighbouring beads becomes the slope of the rope. In this limit, we stop tracking individual beads and start tracking the shape of the entire string at any given moment. The physics changes from a system of many separate objects interacting to a single, continuous field evolving in space and time. This is the essence of the continuum limit: replacing a sum over particles with an integral over space, and replacing finite differences with derivatives.
From Discrete Differences to Derivatives
Let us return to the equation of motion for the \(n\)-th bead in a chain with mass \(m\), spacing \(a\), and tension \(T\). The force on the bead arises from the tension in the segments on either side. For small transverse displacements \(y_n(t)\), the net restoring force is proportional to the curvature of the chain. Mathematically, this is expressed using the second difference of the displacements:
To take the continuum limit, we identify the position of the \(n\)-th bead with the coordinate \(x = n a\). The displacement \(y_n(t)\) becomes a continuous function \(y(x,t)\). The term \(y_{n+1}\) corresponds to \(y(x+a, t)\), and \(y_{n-1}\) corresponds to \(y(x-a, t)\). We can expand these terms using a Taylor series around \(x\):
Substituting these expansions into the expression for the second difference \((y_{n+1}-2y_n+y_{n-1})\), the first-order terms cancel out, and we are left with:
As \(a \to 0\), the higher-order terms vanish, and the discrete second difference becomes \(a^2 \partial_x^2 y\). This is the key mathematical step: the finite difference operator converges to the second spatial derivative.
Defining the Linear Mass Density
In the discrete system, the mass \(m\) is a fixed property of each bead. In the continuum limit, we cannot keep \(m\) fixed while \(a \to 0\), because the total mass of the string would vanish. Instead, we must define a new quantity that remains finite: the linear mass density \(\mu\). This is the mass per unit length of the string. We define it as:
As the spacing \(a\) decreases, the mass of each individual element \(m\) must also decrease proportionally so that \(\mu\) stays constant. This ensures that we are describing a physical string with a definite mass per unit length, rather than a vanishingly light collection of points. We can rewrite the mass \(m\) in the equation of motion as \(m = \mu a\). Substituting this into our discrete equation of motion gives:
Now, we replace the second difference with its continuum limit \(a^2 \partial_x^2 y\). The equation becomes:
Notice that the factor of \(a\) on the left-hand side cancels with one factor of \(a\) on the right-hand side. This cancellation is critical; it shows that the resulting equation is independent of the spacing \(a\), which is exactly what we expect for a continuum theory.
The Wave Equation and Propagation Speed
After simplifying the equation, we arrive at the one-dimensional wave equation:
It is conventional to define the wave speed \(c\) as the square root of the ratio of tension to linear mass density:
Substituting this definition, we obtain the standard form of the wave equation:
This equation states that the acceleration of any element of the string (left-hand side) is proportional to the curvature of the string (right-hand side). The proportionality constant \(c^2\) determines how fast disturbances propagate. Notice that \(c\) depends only on the material properties of the string (tension and density) and not on the frequency or wavelength of the wave. This means the medium is non-dispersive: all frequencies travel at the same speed. This is a direct consequence of the linear, small-amplitude approximation and the uniformity of the string.
The derivation
The Wave Equation as a Continuum Limit
A one-dimensional chain of identical beads of mass \(m\), spaced by \(a\) on a light string of constant tension \(T\) and displaced transversely by small amounts \(y_n(t)\), obeys the discrete equations of motion \( m\,\ddot{y}_n = \dfrac{T}{a}\left(y_{n+1}-2y_n+y_{n-1}\right)\). In the limit \(a\to 0\) taken with the linear mass density \(\mu = m/a\) held fixed, the discrete second difference becomes a second spatial derivative and the chain obeys the one-dimensional wave equation \(\partial_t^2 y = c^2\,\partial_x^2 y\) with a single, dispersionless propagation speed \(c=\sqrt{T/\mu}\).
A common misconception is that the wave speed \(c\) depends on the amplitude of the wave. In the linear regime, it does not. However, if the displacements are large enough that the tension \(T\) varies significantly along the string, or if the string is not perfectly flexible, the relationship between \(T\) and \(\mu\) may become frequency-dependent, leading to dispersion. The derivation above assumes \(T\) is constant and the string is perfectly flexible, which is why we obtain a single, dispersionless speed. C
Worked example
Problem. A uniform string of length \(L = 1.2\ \mathrm{m}\) has a total mass \(M = 0.048\ \mathrm{kg}\). It is stretched with a tension \(T = 120\ \mathrm{N}\). Calculate the wave speed \(c\) on this string. Then, determine the time it takes for a pulse to travel from one end to the other.
- First, calculate the linear mass density \(\mu\). Since \(\mu = M/L\), we have \(\mu = \frac{0.048\ \mathrm{kg}}{1.2\ \mathrm{m}} = 0.04\ \mathrm{kg/m}\).
- Next, use the formula for wave speed \(c = \sqrt{T/\mu}\). Substituting the values, \(c = \sqrt{\frac{120\ \mathrm{N}}{0.04\ \mathrm{kg/m}}}\).
- Calculate the value inside the square root: \(\frac{120}{0.04} = 3000\ \mathrm{m^2/s^2}\).
- Take the square root: \(c = \sqrt{3000}\ \mathrm{m/s} \approx 54.77\ \mathrm{m/s}\).
- Finally, calculate the travel time \(t\) for a distance \(L\). Using \(t = L/c\), we get \(t = \frac{1.2\ \mathrm{m}}{54.77\ \mathrm{m/s}} \approx 0.0219\ \mathrm{s}\).
Units check. The units of \(T/\mu\) are \(\mathrm{N}/(\mathrm{kg/m}) = (\mathrm{kg \cdot m/s^2}) / (\mathrm{kg/m}) = \mathrm{m^2/s^2}\). The square root gives \(\mathrm{m/s}\), which is correct for speed. The time calculation uses \(\mathrm{m}/(\mathrm{m/s}) = \mathrm{s}\), which is correct.
Three quick tests
- What happens to the linear mass density \(\mu\) if you halve the spacing \(a\) between beads while keeping the total mass of the string constant?
Answer
\(\mu\) remains constant. Since \(\mu = m/a\) and the total mass \(M = N m\) is constant, if \(a\) is halved, \(N\) doubles, so \(m\) halves. Thus \(\mu = (m/2)/(a/2) = m/a\). - Why does the factor of \(a\) cancel out in the derivation of the wave equation?
Answer
It cancels because the mass of each element \(m = \mu a\) is proportional to \(a\), and the second difference is proportional to \(a^2\). The ratio \((\mu a) / (T/a \cdot a^2)\) simplifies to \(\mu/T\), removing the dependence on \(a\). - If the tension \(T\) is increased by a factor of 4, by what factor does the wave speed \(c\) increase?
Answer
By a factor of 2. Since \(c \propto \sqrt{T}\), increasing \(T\) by 4 gives \(c' = \sqrt{4T} = 2\sqrt{T} = 2c\).
Problem set for PU-103 →
Each derivation in this unit ends with five graded problems with full solutions.
Lecture notes drafted with a local model and checked line by line by the editor; every number in the worked example was recomputed.