5 - Waves at a boundary

Review of propagating waves in 1 dimension

Last time we developed expressions for travelling waves in a medium where the speed of propagation of the wave is \(v\), where the angular frequency of an oscillating particle at any point along the wave is \(\omega\), and the wavenumber, \(k\), which is the effectively the number of wavelengths per meter expressed in terms of radians. The relationship between \(\omega\), \(v\) and \(k\) is \(\omega =vk\), from which using \(k=2\pi/\lambda\) and \(\omega = 2\pi\nu\), we recover \(v=\lambda \nu\), the relationship between velocity, wavelength and frequency that you know from your A level work. For a travelling wave propagating in the direction of increasing \(x\), the wave disturbance can be represented by

(66)\[\begin{align} \psi_R(x,t)=\psi_{0R}\cos(kx-\omega t + \phi_0), \label{eq:waveright} \end{align}\]

where \(\psi_{0R}\) is the peak amplitude of the wave and \(\phi_0\) is a phase offset, the phase of the wave at position \(x=0\) and time \(t=0\). For a wave travelling to the left, we argued that the best representation is obtained by reversing the sign of \(k\), so that a left propagating wave can be represented by

(67)\[\begin{align} \psi_L(x,t)=\psi_{0L}\cos(-kx-\omega t + \phi_1), \label{eq:waveleft} \end{align}\]

where \(\psi_{0L}\) is the peak amplitude of the wave and \(\phi_1\) is a phase offset, the phase of the wave at position \(x=0\) and time \(t=0\).

From travelling waves to standing waves

In Lecture 3, we looked at standing waves – waves that don’t propagate, but yet still have a displacement that changes with both time and position. The key difference between a standing wave and a travelling wave is that the positional and temporal dependencies are separable, which ultimately means that all parts of the wave oscillate in-phase with each other.

When considering standing waves we came across an important concept - the boundary condition. Boundaries arise when a mathematical or physical system is constrained in one or more ways. As we saw in Lecture 3, these constraints can help us to “solve” the maths for the system, in a way that is analogous to the way that simultaneous equations allow us to determine unknown parameters.

In this lecture we’re going to see how, by again considering boundary conditions, we can generate a standing wave by combining two travelling waves travelling in opposite directions. As with Lecture 3, we’ll first consider waves travelling along a wire-under-tension of length \(L\) that is held in place at both ends: the familiar guitar string. We’ll define the leftmost boundary as being at position \(x=0\), and the rightmost boundary as being at \(x=L\).

We will start with the very most general solution for this guitar-string system, then use the boundaries to constrain various free parameters. Since this is a linear system, then we only need to think of waves in one dimension (i.e., along the \(x\)-dimension). However, waves could travel along the wire in two directions at velocity \(v\): from left-to-right and from right-to-left. We’ll call the left-to-right wave the incident wave, and the right-to-left wave the reflected wave. It is “reflected” at the boundary due to conservation of energy - there isn’t anywhere the left-to-right wave’s energy can go at the rightmost boundary, so it must be reflected.

From the last lecture, the displacements of the incident and reflected waves are given, respectively, by the following formulae:

(68)\[\begin{align} \psi_{I}(x,t) &= I\sin(kx - \omega t + \phi_I) \\ \psi_{R}(x,t) &= R\sin(-kx - \omega t + \phi_R), \end{align}\]

where \(I\) and \(R\) are the amplitudes of the incident and reflected waves, \(\phi_I\) and \(\phi_R\) their phase shifts, and \(k\) and \(\omega\) are the wavenumber and angular frequencies of the waves, which are related via \(\omega=vk\). Note that \(v\) is the same for both waves - it is a property of the system.

Since the incident and reflected waves are travelling along the same wire, the total displacement of the wire at position \(x\) and time \(t\) is given by the superposition of the two waves:

(69)\[\begin{split}\begin{align} \psi_{T}(x,t) &= \psi_{I}(x,t) + \psi_{R}(x,t)\nonumber \\ &= I\sin(kx - \omega t + \phi_I) + R\sin(-kx - \omega t + \phi_R), \label{start} \end{align}\end{split}\]

So, now we have one equation and five unknown parameters: \(I, k, \phi_I, R, \phi_R\). Note that I haven’t listed \(\omega\) as an unknown parameter since if we know \(v\) (which, again, is a property of the system) and \(k\), then we can calculate \(\omega\). Since I haven’t provided any numerical values, the problem boils down to being able to relate those different parameters to each other meaning that if we know \(I\), for example, we can calculate \(R\) etc. This is where the boundary constraints come in.

Since the wire is constrained to have zero amplitude at \(x=0\), then we can say that:

(70)\[\begin{equation} \psi_T(x=0, t) = I\sin(-\omega t + \phi_I) + R\sin(-\omega t + \phi_R) = 0 \end{equation}\]

Using the double angle formula \(\sin(a+b)=\sin(a)\cos(b)+\cos(a)\sin(b)\), and the fact that \(\sin(-x)=-\sin(x)\) and \(\cos(-x)=\cos(x)\), we can re-write the above as:

(71)\[\begin{split}\begin{align} 0 = &I(-\sin(\omega t)\cos(\phi_I) + \sin(\phi_I)\cos(\omega t) + \nonumber \\ &R(-\sin(\omega t)\cos(\phi_R) + \sin(\phi_R)\cos(\omega t)) \label{psitx0} \end{align}\end{split}\]

Now, I’m going to introduce another feature of \(\sin\) and \(\cos\) functions – that they are orthonormal. What this means is that there is no way that you can combine a bunch of \(\sin(x)\) functions to make \(\cos(x)\) (provided \(x\) is the same in each case). This means that if \(A\cos(x) + B\sin(x) = C\cos(x) + D\sin(x)\), then this implies \(A=C\) and \(B=D\). In general, if we have an equation all of whose terms are proportional to either \(\sin(\omega t)\) or \(\cos(\omega t)\), we may extract the terms that multiply each appearance of \(\sin(\omega t)\) and make an equation out of those terms. Similarly, we may do the same thing for the terms that multiply appearances of \(\cos(\omega t)\). This is called equating the coefficients of \(\sin(\omega t)\) and \(\cos(\omega t)\).

With orthonormality in mind, let’s group together the terms that multiply the \(\sin(\omega t)\)’s in Eqn. (71), then do the same for the terms that multiply the \(\cos(\omega t)\)’s. This gives:

(72)\[0 = -I\cos{\phi_I} - R\cos{\phi_R}\]

(73)\[0 = I\sin{\phi_I} + R\sin{\phi_R}.\]

Rearranging Eqns. (72) and (73) so that \(I/R\) is the subject in each case gives:

(74)\[\begin{align} \frac{I}{R} = -\frac{\cos(\phi_R)}{\cos(\phi_I)}=-\frac{\sin(\phi_R)}{\sin(\phi_I)}. \label{IR} \end{align}\]

Then using the fact that \(\tan{\theta}=\sin{\theta} / \cos{\theta}\), we get:

(75)\[\begin{align} \tan{\phi_I} = \tan{\phi_R} \end{align}\]

or, \(\phi_I = \phi_R + n\pi\), where \(n\) is any integer, including 0. Further, since we haven’t specified what the phase of the standing wave is at \(t=0\), we can take the easy approach and assume that \(\phi_I=0\). This therefore implies that \(\phi_R = n\pi\), including \(\phi_R =0\), too.

So, we’re getting there… we’ve now related the phase shifts of the incident and reflected waves meaning we have reduced the number of unknown parameters by one: if we know \(\phi_I\), then we also know \(\phi_R\). To relate the other parameters, we’ll now move on to consider the other boundary condition: that \(\psi_T(x=L, t) = 0\) (i.e., that the displacement is fixed at zero at \(x=L\)):

(76)\[\begin{equation} \psi_T(x=L, t) = I\sin(kL-\omega t) + R\sin(-kL -\omega t) = 0, \end{equation}\]

where I’ve used \(\phi_I = \phi_R = 0\). Again, using the double angle formulae, we can say:

(77)\[\begin{align} 0 = &I(\sin(kL)\cos(\omega t) - \sin(\omega t)\cos(kL)) + \nonumber\\ &R(-\sin(kL)\cos(\omega t) - \sin(\omega t)\cos(kL)) \end{align}\]

Again, using orthonormality to group the terms in \(\sin(\omega t)\) and \(\cos(\omega t)\), we get:

(78)\[0 = (I-R)\sin(kL)\]

(79)\[0 = (R+I)\cos(kL).\]

Equation (78) implies that \(I-R=0\) (i.e., \(I=R\)) and/or \(\sin(kL)=0\). If the former, then at \(x=0\), we have:

(80)\[\begin{equation} \psi_T(x=0, t) = I\sin(-\omega t) + I\sin(-\omega t) = 0, \end{equation}\]

which means that \(I=0\), and thus there is no displacement anywhere. This is known as the “trivial” solution, and is not considered of interest. In that case, the more interesting solution arises by considering that \(R+I=0\) in Eqn. (79). Since that implies that \(R=-I\), then \(I-R=2I\), meaning \(\sin(kL)\) in Eqn. (78) must be equal to zero. I’ll leave it as a fairly straightforward exercise for you to work out what that implies for the allowed values of \(k\).

So, now we have known values for the phase shift (i.e., \(\phi_I=0\), \(\phi_R=n\pi\)), the wavenumber, \(k\) (and, hence, \(\omega\) via \(\omega=vk\)), and how the incident and reflected amplitudes relate to one another (i.e., \(R=-I\)). Plugging all this into Eqn. (69) gives:

(81)\[\begin{align} \psi_T = & I\sin(kx-\omega t) - I\sin(-kx-\omega t) \nonumber\\ = &I(\sin(kx)\cos(\omega t) - \sin(\omega t)\cos(kx)) - \nonumber\\ &I(-\sin(kx)\cos(\omega t) - \sin(\omega t)\cos(kx) \nonumber\\ = &2I\sin(kx)\cos(\omega t)\nonumber\\ = &\psi_0\sin(kx)\cos(\omega t) \end{align}\]

where I’ve substituted \(\psi_0\) for \(2I\) in the last line. You’ll notice that the expression for \(\phi_T\) above is exactly the same (bar a phase shift) as that of the guitar string standing wave that we saw in Lecture 3. Unfortunately we haven’t got time in the lecture to go through the same derivation for the open-open and open-closed pipes we saw in Lecture 3. However, let me assure you that one can indeed obtain those same equations with the superposition of two waves travelling in opposite directions and applying the appropriate boundary conditions.