Waves
Waves
Introduction
In the previous chapter, oscillatory systems were examined. At the microscopic level, matter is organized such that atoms can be modeled as oscillators when they are supplied with energy. In solids and liquids, atoms or molecules are constrained within crystalline or molecular structures and are effectively “bound” by electromagnetic forces, i.e., by the interparticle electromagnetic interactions that maintain the cohesion of the material.
In gases, by contrast, atoms or molecules are not fixed in place by such binding structures; however, their mutual interactions—primarily through collisions—can likewise be described in terms of electromagnetic forces between charged constituents.
A spatially and temporally ordered transfer of energy through a sequence of coupled oscillations, in the absence of any net transport of matter, is referred to as a wave.
Example 1: Transverse Waves on a Cable
Consider a tense cable connected to a wall at both ends, perfectly parallel to the floor (neglecting gravity).
When we slightly strike the cable, we observe that the original deflection travels with constant velocity along the cable as a wavefront. We say that the transverse deflection propagates as a wave. We notice that:
Every point of the cable oscillates perpendicular to the propagation of the wave.
A point on the cable remains at rest until the wave reaches it.
When the wave reaches the point, it is displaced from its rest position.
Ultimately, the point returns to its original equilibrium position.
It follows that the individual points on the cable are not transported by the wave; they only oscillate back and forth from their equilibrium state.
Example 2: Sound Waves in Gases
We now consider elastic waves originating from changes in gas pressure. Sound is the most important example of this kind of wave.
Such a wave can be generated by placing a loudspeaker in front of a glass tube. The sound propagates through the tube as a pressure wave of air at a specific sound velocity. Because gases are highly compressible, both the pressure and density of the gas will fluctuate.
Wave Types and Wave Propagation
Description of One-Dimensional Wave Propagation
We now want to describe the propagation of a one-dimensional wave quantitatively, considering transverse rope waves as an example.
At a time \(t = 0\), we can describe the shape of the wave through a function \(\xi(x)\), where \(\xi\) is the transverse displacement of the rope. The coordinate \(x\) corresponds to a point along the rope. After some time, the wave crest has moved, requiring a different function to describe the rope’s shape.
The displacement of the rope as a function of space and time can be expressed as:
\[\begin{equation} \xi = \xi(x,t) \end{equation}\]
Here, \(x\) is the spatial coordinate and \(t\) is the time. This is called the wave function, and it describes the propagation of the wave over time.
In general, the function \(\xi(x,t)\) describes the propagation of a one-dimensional wave:
Rope waves: \(\xi(x,t)\) describes the transverse displacement.
Spring waves: \(\xi(x,t)\) describes the longitudinal or transverse deformation.
Gas waves (sound): \(\xi(x,t)\) describes the pressure or density of the gas.
Simplification of the dispersion: Usually, the shape of a wave crest changes over time, an effect known as dispersion. We will initially neglect this effect and assume a stable shape for the wave crest.
Although \(\xi(x,t)\) appears to be a function of two independent variables (\(x\) and \(t\)), describing a wave restricts its possible forms:
If \(\xi(x,t)\) represents the temporal and spatial propagation of a wave, then \(x\) and \(t\) are not completely independent from one another.
The spatial dependence dictates the shape of the wave, while the time dependence dictates its propagation.
Assuming the shape of the wave at \(t = 0\) is given by \(f(x)\):
\[\begin{equation} \xi(x, 0) = f(x) \end{equation}\]
If we replace \(x\) with \(x - a\), we get:
\[\begin{equation} \xi = f(x - a) \end{equation}\]
This substitution translates the wave crest without altering its shape, shifting the wave to the right by the amount \(a\). Similarly:
\[\begin{equation} \xi = f(x + a) \end{equation}\]
describes a rigid shift to the left by the amount \(a\).
Substituting:
\[\begin{equation} a = vt, \quad \text{with} \quad v > 0 \end{equation}\]
where \(v\) is the wave velocity, yields the function for a wave propagating to the right or left, respectively:
\[\begin{equation} \xi = f(x \pm vt) \end{equation}\]
The velocity \(v\) in the above function is called the phase velocity (\(v_{Ph}\)). It represents the speed at which a given phase of the wave propagates.
The propagation velocity depends on at least two factors:
A restoring force that attempts to return the medium to equilibrium. A greater force results in faster propagation.
The mass (inertia) of the medium, which resists acceleration. A greater mass results in slower propagation.
Harmonic Waves
An important case arises when the wave takes the form of a periodic sine or cosine function. This is called a harmonic wave.
A harmonic wave can be described using a sine function: \[\begin{equation} \label{eq:harmonic_wave} \xi(x,t) = \xi_0 \sin [k(x \pm vt)] \end{equation}\] Here, \(k\) is the wave number (or wave vector), and \(\xi_0\) is the amplitude of the wave.
The spatial distance between two consecutive wave crests is the wavelength, \(\lambda\). The shape of a harmonic wave repeats periodically in both space and time.
The wave number is related to the wavelength:
\[\begin{equation} k(x + \lambda) = kx + 2\pi \implies k\lambda = 2\pi \implies k = \frac{2\pi}{\lambda} \end{equation}\]
Substituting this into Equation [eq:harmonic_wave] gives:
\[\begin{equation} \xi(x, t) = \xi_0 \sin \left( \underbrace{\frac{2\pi}{\lambda}}_{k} x \pm \underbrace{\frac{2\pi v}{\lambda}}_{\omega} t \right) \end{equation}\]
The temporal distance between two wave crests is the period, \(T\). In one period \(T\), the wave travels exactly one wavelength \(\lambda\). The phase velocity is therefore:
\[\begin{equation} v = \frac{\Delta x}{\Delta t} = \frac{\lambda}{T} = \frac{2\pi\lambda}{2\pi T} = \frac{\omega}{k} \end{equation}\]
where \(\omega\) is the angular frequency, defined as:
\[\begin{equation} \omega = \frac{2\pi}{T} = 2\pi f = vk \end{equation}\]
and \(f\) is the frequency of oscillation.
It follows that a harmonic wave can be written as:
\[\begin{equation} \xi(x,t) = \xi_0 \sin (\underbrace{kx \pm \omega t + \varphi_0}_{\text{Phase}}) \end{equation}\]
Using Euler’s identity for complex exponents, we can alternatively express the harmonic wave as:
\[\begin{equation} \xi(x,t) = \xi_0 e^{i(kx \pm \omega t)} \end{equation}\]
The wave equation in one dimension
We consider now the derivatives with respect to time and space of a harmonic wave:
\[\begin{equation} \xi (x,t) = \xi_0 \sin(kx-\omega t) = \xi_0 \sin[k(x-vt)] \end{equation}\]
Since \(\xi\) is a function of two variables we must use the partial derivatives:
\[\begin{equation} \pdv{\xi}{t} = \pdv{}{t}\xi_0\sin[k(x-vt)] = \xi_0(-kv)\cos[k(x-vt)] \end{equation}\] \[\begin{equation} \frac{\partial^2 \xi}{\partial t^2} = -\xi_0(kv)^2\sin[k(x-vt)] \end{equation}\]
and
\[\begin{equation} \pdv{\xi}{x} = \xi_0k\cos[k(x-vt)] \end{equation}\] \[\begin{equation} \frac{\partial^2 \xi}{\partial x^2} = -\xi_0k^2 \sin[k(x-vt)] \end{equation}\]
Confronting the two second derivatives, one notices that:
\[\begin{equation} \frac{\partial^2 \xi}{\partial t^2} = -\xi_0(kv)^2\sin[k(x-vt)] = v^2 \frac{\partial^2\xi}{\partial x^2} \end{equation}\]
The diligent reader can verify that the same result is obtained when the converse direction for the wave is applied:
\[\begin{equation} \xi (x,t) = \xi_0 \sin(kx+\omega t) = \xi_0 \sin[k(x+vt)] \end{equation}\]
In fact, we notice that the term \(vt\) has nothing to do with the second partial derivatives. To sum up everything that has been stated so far, we have found the following relation for a harmonic wave:
\[\begin{equation} \frac{\partial^2 \xi}{\partial t^2} = v^2 \frac{\partial^2 \xi}{\partial x^2} \implies \frac{\partial^2 \xi}{\partial t^2} - v^2 \frac{\partial^2 \xi}{\partial x^2} = 0 \end{equation}\]
We call Wave Equation, an equation of the form: \[\begin{equation} \frac{\partial^2 \xi}{\partial t^2} - v^2 \frac{\partial^2 \xi}{\partial x^2} = 0 \end{equation}\] where \(v\) is the velocity of the wave.
The general solution for this equation is of the form: \[\begin{equation} \xi(x,t) = f(x-vt) + g(x+vt) \end{equation}\] where \(f,g\) are two two-times differentiable functions.
Such a solution, satisfies always the wave equation, doesn’t matter what \(f,g\) are. The solution describes waves that move towards the right and the left.
Exercise
Prove that the given solution for the wave equation satisfies indeed the equation.
Hint: apply the following substitutions: \[\begin{equation} \alpha(x,t) = x-vt \quad \text{and} \quad \beta(x,t) = x+vt. \end{equation}\]
Transverse Waves
In a transverse wave, the deflection \(\xi\) is perpendicular to the direction of propagation. A representative example of a transverse wave is the propagation of a disturbance along a taut rope, in which the oscillations of the rope elements occur perpendicular to the direction of wave propagation.
We consider a wave \(\boldsymbol{\xi}\) with amplitude \(\mathbf{A} = A_x\boldsymbol{\hat{x}} + A_y\boldsymbol{\hat{y}}\), where the total displacement is equal to \(A = \abs{\mathbf{A}} = \sqrt{A_x^2 + A_y^2}\), which moves in the \(z\)-direction. The wave function is then of the form: \[\begin{equation} \boldsymbol{\xi}(x,t) = \mathbf{A}f(z-vt) \end{equation}\]
A harmonic wave as a special case is, for example, of the form: \[\begin{equation} \boldsymbol{\xi}(x,t) = A\cos(kx-\omega t) \boldsymbol{\hat{x}} \end{equation}\]
Example: Transverse Elastic Rope Waves
The relationship between the propagation velocity of the wave and the physical properties of the rope can be derived using Newton’s laws.
Imagine subdividing the rope into infinitesimal mass elements d\(m\). In this case, the rope can be seen as a mass-spring system, where the discrete sequence of masses is replaced by a continuous distribution of mass particles.
We can assume that the mass elements are allowed to move only vertically with respect to their rest position.
We now consider only one mass element d\(m\) with length d\(x\), where its starting point position is \(x\), and endpoint position is \(x+\text{d}x\) on the rope. The displacement is given by \(y = \xi(x)\). Par The force \(\mathbf{F}\) acting on the rope with section \(A\) generates a tensile stress \(\mathbf{S} = \mathbf{F}/A\). The resultant vertical component of the tension is
\[\begin{equation} \Delta S_y = S\sin(\alpha') -S\sin(\alpha) \end{equation}\]
where \(\alpha, \alpha'\) are the angles on both sides of the mass element with respect to the horizontal.
For small angles, the following approximation holds:
\[\begin{equation} \tan \alpha \approx \sin\alpha \end{equation}\]
it follows that
\[\begin{equation} \Delta S_y \approx S \tan (\alpha') - S \tan(\alpha). \end{equation}\]
The slope of the rope at the point \(x\) is equal to the derivative of \(\xi\) with respect to \(x\). Since \(\xi\) is a function of two variables, we must use the partial derivatives in the following way:
\[\begin{equation} \tan \alpha = \pdv{\xi(x,t)}{x} \quad \text{and} \quad \tan \alpha' = \pdv{\xi(x + \text{d}x,t)}{x} \end{equation}\]
As a consequence, we can write the resultant vertical tension as
\[\begin{equation} \Delta S_y = S\tan(\alpha') - S \tan (\alpha) = S \left [\pdv{\xi(x + \text{d}x,t)}{x}- \pdv{\xi(x,t)}{x} \right ]. \end{equation}\]
This equation reminds us of an incremental quotient, so for an infinitesimal d\(x\) we get
\[\begin{equation} \text{d}S_y = S \frac{\partial^2 \xi (x,t)}{\partial x^2} \text{d}x. \end{equation}\]
The velocity and acceleration of the differential segment can be obtained with the partial derivatives of our function \(\xi\):
\[\begin{equation} v_y(x,t) = \pdv{\xi}{t} \quad \text{and} \quad a_y (x,t) = \frac{\partial^2 \xi}{\partial t^2}. \end{equation}\]
Applying Newton’s second law:
\[\begin{equation} \text{d}S_y \cdot A = \text{d}m \cdot a_y \implies S \frac{\partial ^2 \xi}{\partial x^2} \underbrace{\text{d}x \cdot A}_{= \text{d}V} = \text{d}m \frac{\partial^2 \xi}{\partial t^2} \end{equation}\]
Dividing by \(V\):
\[\begin{equation} S\frac{\partial^2 \xi}{\partial x^2} = \rho \frac{\partial^2 \xi}{\partial t^2} \end{equation}\]
and dividing again by \(S\):
\[\begin{equation} \frac{\partial^2 \xi}{\partial x^2} = \frac{\rho}{S} \frac{\partial^2 \xi}{\partial t^2} \end{equation}\]
This is the differential equation of the wave propagation, and comparing this with the general form of the wave equation,
\[\begin{equation} \frac{\partial^2 \xi(x,t)}{\partial t^2}- v^2 \frac{\partial^2 \xi(x,t)}{\partial x^2} = 0 \end{equation}\]
we can deduce the value of the propagation velocity of the rope wave as
\[\begin{equation} v^2 = \frac{S}{\rho} \implies v = \pm \sqrt{\frac{S}{\rho}} \end{equation}\]
where \(S\) is the tension of the rope and \(\rho\) its density.
Longitudinal Waves
We refer to as longitudinal waves, waves whose deflection \(\boldsymbol{\xi}\) is parallel to the direction of propagation. If the wave propagates in the \(z\)-direction, the deflection is then given by:
\[\begin{equation} \boldsymbol{\xi}(z,t) = A f(z-vt) \hat{\boldsymbol{z}} \end{equation}\]
A typical example of longitudinal waves are sound waves. Due to the deflection of the atoms or molecules, we observe density fluctuations
\[\begin{equation} \rho (z,t) = \rho_0 [1 + Af(z-vt)] \quad \text{with} \abs{Af(z-vt)} \leq 1 \end{equation}\]
since a density cannot be negative.
The velocity of a sound wave in a rigid body is given by
\[\begin{equation} v = \sqrt{\frac{E}{\rho}} \end{equation}\]
where \(E\) is the module of elasticity and \(\rho\) the density.