Transverse waves are waves where the displacement of the medium (e.g. the string) is perpendicular (or transverse) to the direction of wave propagation. So, for instance, the string is displaced in the y direction while the wave travels in the x direction.
Until now, we have not thought about how waves start and finish: we have assumed (implicitly) that they are infinite in time and space. However, this is not how real systems work: if a wave is to continue to propagate along a string (or down an optical fibre, say) then there must be some driving force. Let's consider a string stretched to a length $L$ with ends at $x=0$ and $x=L$. We'll place a driving mechanism of some kind at $x=0$.
What must the driving mechanism do ? It needs to produce a force which balances the transverse component of the tension in the string at $x=0$. This is a driving force:
\begin{equation} F_D = -T\left(\frac{\partial \psi}{\partial x}\right)_{x=0} = \frac{T}{c} \left(\frac{\partial \psi}{\partial t}\right)_{x=0} = Z_0 \left(\frac{\partial \psi}{\partial t}\right)_{x=0} \end{equation}We have used the travelling wave equation from Chapter 3 to simplify. The force at any instant (or instantaneous force) must be proportional to the transverse velocity of the string---which closely resembles a standard drag force in a damped harmonic oscillator. Why do we have this drag ? It arises from the energy being transported by the wave: if we want to keep a constant wave motion going, we must put energy into the system. Notice that the constant of proportionality is the characteristic impedance. It is a function of the system only (in this case the tension and mass density) and does not depend on the type of motion or its frequency.
We have just seen that creating or driving a wave along a stretched string (or, in fact, in any medium) requires energy to be put in to the system. But what about the other end of the string ? What happens there ? Let's think about a finite string, and how we can make it resemble an infinite string. We know what force we need at $x=0$ to create the wave. At the end of the string, $x=L$, we need a transverse force to balance tension in the string; if we match this tension then it will be as if the end was not there. This condition can be written:
\begin{equation} F_L = T\left(\frac{\partial \psi}{\partial x}\right)_{x=L} = -Z_0 \left(\frac{\partial \psi}{\partial t}\right)_{x=L} \end{equation}So if we have some form of drag to absorb the energy being transmitted along the string, then the wave will propagate along as if the string were infinite (assuming that we have set up the driver as described above). This idea is known as impedance matching and is important in many areas, particularly electromagnetism (where we must terminate, say, a power line or an aerial correctly). Remember that this all came about because we have a finite string (or medium in general) which we want to send a wave down, with the medium behaving as if it were infinite. This means that we must put energy in at one end, and take it out at the other.
If we do not provide the right force at the end, then something different will happen. In fact, these ideas can be extended to consider boundaries between different media (e.g. tying a light string to a rope, or light going from air into water or glass). The two media will have different impedances (in the case of the string and rope, probably different mass densities even if they're under the same tension) and something interesting will happen at the boundary.
Imagine a string under tension, with one end tied to a solid object (like a wall) and the other end free to be driven. If we send a pulse down the string (for example by moving the free end up and then down rapidly once), it will propagate along the string as a travelling wave. What will happen when it reaches the solid wall ? Intuition or experience tells us that when a wave reaches a solid object (i.e. an object with very large impedance) it tends to be reflected. A reflected wave in one dimension travels in the opposite direction to the incoming wave, so we will need both solutions for the wave equation (i.e. $\psi = \psi_i + \psi_r = f(x-ct) + g(x+ct)$, where $\psi_i$ is the incoming or incident wave and $\psi_r$ is the reflected wave). We can also deduce this need for both waves from a more general situation we'll see below.
Now, we need to work out the relationship between the two components of the wave, and we'll do this by thinking about the boundary conditions on the wave. Much of the work done in physics involves working out appropriate boundary conditions, and solutions of differential equations given appropriate boundary conditions. We will assume that the solid wall has an infinite impedance, so there can be no propagation of the wave (the velocity will be zero). The tension along the string is provided by the wall, but the tension transverse to the string at the wall must be zero (otherwise the wall would move up and down). We can therefore write that:
\begin{eqnarray} T\left( \frac{\partial \psi_i}{\partial x}\right)_{x=L} &+& T\left( \frac{\partial \psi_r}{\partial x}\right)_{x=L} = 0\\ \Rightarrow T\left( \frac{\partial \psi_i}{\partial x}\right)_{x=L} &=& - T\left( \frac{\partial \psi_r}{\partial x}\right)_{x=L}\\ \left( \frac{\partial \psi_i}{\partial x}\right)_{x=L} &=& -\left( \frac{\partial \psi_r}{\partial x}\right)_{x=L} \end{eqnarray}So there must be a 180$^\circ$ phase change on reflection; this is another way of saying that the reflected wave has the opposite sign to the incoming wave. But what happens now if, instead of a solid wall, we have another string with a different mass per unit length ? To be clear, we will assume that we have two strings of lengths $L_1$ and $L_2$ joined at $x=0$ held under tension T. The first string has mass density $\mu_1$, impedance $Z_1 = \sqrt{T\mu_1}$ and has a driver at its free end ($x=-L_1$). The second string has mass density $\mu_2$, impedance $Z_2 = \sqrt{T\mu_2}$ and has perfect termination (i.e. a drag term with impedance $Z_2$) at $x=L_2$.
Now consider an incident wave, starting at $x=-L_1$ and moving towards the joining point at $x=0$. As it propagates, there is a transverse force on the string given by:
\begin{equation} -T\left(\frac{\partial \psi}{\partial x} \right)= Z\left(\frac{\partial \psi}{\partial t} \right) \end{equation}We assume, generally, that $\psi = \psi_i + \psi_r, x\le 0$. Exactly at the join, the wavefunctions on the two sides must equal, so $\psi_i(0,t)
At the joining point, the drag exerted by the second string on the first string will be given by its impedance multiplied by the transverse velocity of the string:
\begin{equation} F_{\mathrm{drag}} = Z_2 \frac{\partial \psi_t}{\partial t} = Z_2 \frac{\partial }{\partial t}\left(\psi_i + \psi_r\right) = Z_2 \left(\frac{\partial \psi_i}{\partial t} + \frac{\partial \psi_r}{\partial t}\right) \end{equation}This force will be balanced by the transverse force from the string:
\begin{equation} -T\frac{\partial \psi}{\partial x} = -T\left(\frac{\partial \psi_i}{\partial x} + \frac{\partial \psi_r}{\partial x}\right) = Z_1 \left(\frac{\partial \psi_i}{\partial t} - \frac{\partial \psi_r}{\partial t}\right) \end{equation}Note that the change of sign between the partial derivatives comes when we use the travelling wave equation to relate the spatial and time derivatives: the incoming wave gives a minus sign (which cancels the minus sign outside the bracket) while the reflected wave gives a plus sign. We can now derive a condition relating the incident and reflected waves, by equating the two forces:
\begin{eqnarray} Z_1 \left(\frac{\partial \psi_i}{\partial t} - \frac{\partial \psi_r}{\partial t}\right) &=& Z_2 \left(\frac{\partial \psi_i}{\partial t} + \frac{\partial \psi_r}{\partial t}\right)\\ (Z_1 - Z_2)\left(\frac{\partial \psi_i}{\partial t}\right) &=& (Z_1 + Z_2) \left(\frac{\partial \psi_r}{\partial t}\right)\\ \left(\frac{\partial \psi_r}{\partial t}\right) &=& \frac{Z_1 - Z_2}{Z_1 + Z_2} \left(\frac{\partial \psi_i}{\partial t}\right) \end{eqnarray}If we integrate both sides with respect to time, we find that: \begin{equation} \psi_r(0,t) = \frac{Z_1 - Z_2}{Z_1 + Z_2} \psi_i(0,t) \end{equation} This gives the relation between $\psi_i$ and $\psi_r$ at the joining point of the two strings. As both waves are travelling waves, we can relate the value at one time and place to the value at another time and place: \begin{eqnarray} \psi_i(-l,t-l/c) &=& \psi_i(0,t)\\ \psi_r(-l,t+l/c) &=& \psi_r(0,t) = R\psi_i(0,t) = R \psi_i(-l,t-l/c)\\ R &=& \frac{Z_1 - Z_2}{Z_1 + Z_2} \end{eqnarray} In other words, the reflected wave at $x=-l$ and $t=t+l/c$ is the same as the incident wave at the same position but at a time $2l/c$ in the past, and scaled by $R$, which we call the reflection coefficient. Notice that the time delay is just the time it takes for the wave to travel along the string and back again.
We can also say something about the wave in the second string. We must have the following boundary condition:
\begin{equation} \psi_i(0,t) + \psi_r(0,t) = \psi_t(0,t), \end{equation}where $\psi_t$ is the transmitted wave. This must be so, otherwise there will be a discontinuity in the string. So the transmitted wave is related to the incident wave by the following formula:
\begin{eqnarray} \psi_t(0,t) &=& \psi_i(0,t) + R\psi_i(0,t)\\ &=& (1+R) \psi_i(0,t) = T \psi_i(0,t)\\ T &=& 1+R = \frac{2Z_1}{Z_1 + Z_2} \end{eqnarray}where $T$ is called the transmission coefficient. Notice that the reflection coefficient will take values between -1 and 1, while the transmission coefficient will take values between 0 and 2:
\begin{eqnarray} -1 &\le& R \le 1\\ 0 &\le& T \le 2 \end{eqnarray}If there is a single string terminated with a very large impedance (a solid wall, as before, so $Z_2 \gg Z_1$) then we will have $R = -Z_2/Z_2 = -1$ and we get the phase change we saw before (and no transmission). If the string is terminated with a very small impedance (a free end, for instance, so that $Z_1 \gg Z_2$) then we will have $R = Z_1/Z_1 = 1$ and no phase change in the reflected wave. When there are two strings, then in the limit of the second string having negligible impedance the transmitted wave will have twice the amplitude of the incident wave, and the reflected wave will have the same height as the incident wave. When the two impedances are matched, it will be as if there was only one string ($R=0, T=1$).
In the figure below, we plot a system with two impedances (it is assumed that these are two strings with different masses per unit length). You should try changing the impedances and rerunning the code to see the effect on the coefficients and behaviour of the system. When we move away from $t=0$ some interesting changes happen - but they are not wrong. Remember that a large amplitude in the transmitted wave may be because the impedance is small: it is easy to excite a wave on a string with small impedance.
fig = figure(figsize=[12,3])
# Add subplots: number in y, x, index number
ax = fig.add_subplot(121,autoscale_on=False,xlim=(-10,10),ylim=(-2,2))
ax2 = fig.add_subplot(122,autoscale_on=False,xlim=(-10,10),ylim=(-2,2))
# Set up limits on the graph: we will plot from -10 to 10, with the boundary at 0
xmin = -10
xmax = 10
# Create x arrays for x<0 and x>0
xL = linspace(xmin,0.0,1000)
xR = linspace(0.0,xmax,1000)
# System parameters: tension, masses per unit length, frequency, amplitude of incoming wave
T = 1.0
mu1 = 1.0
mu2 = 2.0
omega = 1.0
A = 1.0
# The time at which we visualise can have an important effect, and produce surprising results
time = 0.0
# Derive speeds, wavenumbers, impedances
c1 = sqrt(T/mu1)
c2 = sqrt(T/mu2)
k1 = omega/c1
k2 = omega/c2
Z1 = sqrt(T*mu1)
Z2 = sqrt(T*mu2)
# Ra and Ta are reflection and transmission coefficients (to avoid confusion with tension T above)
Ra = (Z1-Z2)/(Z1+Z2)
Ta = 1+Ra
print "Z1 and Z2 are: ",Z1,Z2
print "R and T are: ",Ra,Ta
# Create psi_i, psi_r and psi_t (include time - note that psi_r travels in negative x direction)
left_i = A * cos(k1*xL-omega*time)
left_r = A * Ra*cos(k1*xL+omega*time)
# left is the total wave for x<0
left = left_i + left_r
# right is the wave for x>0
right = A*Ta*cos(k2*xR-omega*time)
# Plot total waves in left plot
ax.plot(xL,left,label=r"$\psi_i+ \psi_r$")
ax.plot(xR,right,label=r"$\psi_t$")
ax.legend()
# Plot individual waves in right plot
ax2.plot(xL,left_i,label=r"$\psi_i$")
ax2.plot(xR,right,label=r"$\psi_t$")
ax2.plot(xL,left_r,label=r"$\psi_r$")
ax2.legend()
Z1 and Z2 are: 1.0 1.41421356237 R and T are: -0.171572875254 0.828427124746
<matplotlib.legend.Legend at 0x112dc0a10>
The phase velocities and wavelengths on either side of an interface between two media will be different, but the frequencies will be the same, provided that the interface has no mechanism for driving waves. We can understand by thinking about the effect of the interface on the second medium: it will act as a harmonic driving force at frequency $\omega$. Consider a join between two strings of different mass per unit length (but under the same tension) to be specific. Then if we send a wave with frequency $\omega$ down the first string, all points on the string will oscillate harmonically with frequency $\omega$. This must be true of the junction, which will then excite waves of the same frequency in the second string. So we have:
\begin{eqnarray} \omega_1 &=&\omega_2 = \omega\\ \nu_1 &=& \nu_2 = \frac{\omega}{2\pi}\\ c_1 &=& \sqrt{\frac{T}{\mu_1}}\\ k_1 &=& \frac{\omega_1}{c_1} = \frac{\omega}{c_1}\\ \lambda_1 &=& \frac{c_1}{\nu}\\ c_2 &=& \sqrt{\frac{T}{\mu_2}}\\ k_2 &=& \frac{\omega_2}{c_2} = \frac{\omega}{c_2}\\ \lambda_2 &=& \frac{c_2}{\nu}\\ \end{eqnarray}Later in the course we will encounter situations where the frequency and wavelength are not so simply related (called dispersive systems) but even for these materials frequencies are conserved across boundaries.
We can summarise the results on driving waves, terminating, and behaviour at interfaces as follows:
So far, we have considered a stretched string with a driver at one end and some form of impedance at the other end. Now we will consider a set-up where the string is fixed at both ends; we will assume that there is some means of driving a wave in the string (for instance a guitar pick or a violin bow). This is another example of boundary conditions which we can implement; again, we must have both forms of wave: $\psi = f(x-ct) + g(x+ct)$. If the string has length $L$ then we can write:
\begin{eqnarray} \psi(0,t) &=& \psi(L,t) = 0\\ \left(\frac{\partial \psi}{\partial t}\right)_{x=0} &=& \left(\frac{\partial \psi}{\partial t}\right)_{x=L} = 0 \end{eqnarray}Let us consider the general wave solution and assume that we will have a sinusoidal solution of some kind. Then we can write, quite generally,
\begin{equation} \psi(x,t) = Ae^{i(\omega t-kx)} + Be^{i(\omega t+kx)} \end{equation}Note that I've written the two so that the time variation shares the same sign and the x variation differs. This is just as general a solution of the wave equation, and makes the maths slightly easier. What can we learn from the boundary conditions ? Let's look at $x=0$ first.
\begin{eqnarray} \psi(0,t) &=& Ae^{i\omega t} + Be^{i\omega t} = 0\\ \Rightarrow A &=& -B\\ \psi(x,t) &=& Ae^{i\omega t}\left( e^{ikx} - e^{-ikx}\right) \\ &=& 2iAe^{i\omega t}\sin(kx) \end{eqnarray}So we've shown something quite important about the form of the wave just from the first condition; notice that the minus sign between the two components is exactly what we'd expect for a wave reflected from a solid wall. Now let's consider the boundary condition at $x=L$:
\begin{eqnarray} \psi(L,t) &=& 2iAe^{i\omega t}\sin(kL) = 0\\ \Rightarrow kL &=& n\pi \end{eqnarray}where $n$ is an integer, as we know that sin has zeroes for integer multiples of $\pi$. So we have a series of solutions for $k$ which will all fit the boundary conditions. We can write: \begin{eqnarray} k_n &=& \frac{n\pi}{L}\\ \lambda_n &=& \frac{2\pi}{k_n} = \frac{2L}{n}\\ n&=& 1, 2, 3, \dots \end{eqnarray} So there are only a certain set of wavelengths that will fit onto the string; this makes sense, as we have imposed a certain length scale on the system, and should expect the resulting solutions to fit into that length. In fact, if we take the idea of normal modes from Sec.~2.5 to its continuous limit, we find that the allowed solutions for the string are the normal modes of the system.
The first five standing waves for a string fixed at both ends are plotted below, though you should experiment with the parameters and see what happens.
fig = figure(figsize=[12,7])
# Set length scale
L = 5.0
# We will put the subplots into an array, ax
ax = []
# Define x coordinates and omega
x = linspace(0,L,1000)
omega = 1.0
# Loop over first five modes, n
for n in range(1,6):
# Create subplot; we want 5 in y and 1 in x, with incremented index - spn gives this
spn = 510+n
# Plot mode n
ax.append(fig.add_subplot(spn,autoscale_on=False,xlim=(0,L),ylim=(-1.1,1.1)))
# Plot at a number of times to show how the wave changes
for t in range(4):
ax[n-1].plot(x,sin(n*pi*x/L)*sin(omega*pi*t/2.0))
The patterns along the string are important, and we can define:
There are $n$ maxima or anti-nodes, and $n-1$ stationary points or nodes for mode $n$. Since the speed of the wave depends on the material properties of the system (the tension and mass density in a string), the frequencies of the waves must also be arranged in a series which depends on $n$: \begin{eqnarray} \omega_n &=& c k_n = \frac{n\pi c}{L}\\ \nu_n &=& \frac{\omega_n}{2\pi} = \frac{nc}{2L} \end{eqnarray} These are the normal frequencies which arise because of the boundary conditions imposed on the system. Notice that, for this system (which is rather special), the normal frequencies are integer multiples of the first frequency. In general, the integer multiples of a fundamental frequency are called the harmonics.
Notice that the standing waves arise because of interference between two waves moving in opposite directions. The formation of standing waves occurs because of constructive interference between waves moving in opposite directions.
A string fixed at both ends can support a series of different waves; the lowest frequency wave (with longest wavelength) is called the fundamental frequency. In the same way, most physical systems will have natural frequencies as we saw in the case of the harmonic oscillator. These frequencies are also known as resonant frequencies, as an excitation applied at this frequency will generate a resonance; as we mentioned briefly in Sec. 2.3, the amplitude of a damped, driven harmonic oscillator will be at a maximum when the driving frequency equals the natural or resonant frequency.
If a complex excitation is applied to an object, causing it to vibrate, it will tend to pick out its resonant frequencies, as these will respond with a large amplitude. A sharp impulse (like a kick or a whack from a stick) delivered to a simple oscillator will give a motion which is somewhat chaotic initially but will settle down to an oscillation at the natural frequency. The impulse consists of many frequencies (which can be investigated with Fourier analysis) but the oscillator will only respond strongly at its resonant frequency. Most objects which vibrate will have multiple resonant frequencies (depending on the boundary conditions which they impose).
The wave which is supported on a string fixed at both ends have a very different form to those we used above: the time variation has been separated from the spatial variation: \begin{equation} \psi(x,t) = \mathrm{Re}\left[2iAe^{i\omega t}\sin(kx)\right] = -2A \sin(\omega t) \sin(kx) \end{equation} So each point on the string is in phase with all other points on the string, and they undergo simple harmonic motion, with the amplitude of the motion depending on the position along the string. The points on the string where $kx = n\pi$ for $n$ an integer will have zero amplitude (as $\sin(n\pi) = 0$); these are the nodes. The points on the string where $kx = (n+\frac{1}{2})\pi$ for $n$ an integer will have maximum amplitude of $2A$.
The positions $x$ of the nodes and antinodes are found by substituting $k = 2\pi/\lambda$ into the expressions given above:
\begin{eqnarray} x_{\mathrm{node}} &=& \frac{n\pi}{2\pi/\lambda} = n\frac{\lambda}{2}\\ x_{\mathrm{antinode}} &=& \frac{\left(n + \frac{1}{2}\right)\pi}{2\pi/\lambda} = \left(n+\frac{1}{2}\right)\frac{\lambda}{2} \end{eqnarray}The nodes are separated from each other by half a wavelength (which is easy to see by calculating $x(n=1) - x(n=0)$), and the anti-nodes are also separated from each other by half a wavelength (this is by definition: these are the zeroes and extrema of a sinusoidal function and must be separated by half a wavelength). Each node is separated by $\lambda/4$ from the nearest anti-nodes.
As an example, consider a piano string (this is an oversimplification
length is typically around 0.01 kg/m and the tension is around 800N (equivalent to the gravitational force exerted by the mass of an average person). The velocity of waves on this string will be $\sqrt{800/0.01} = 283$m/s. If the wire is 0.6m long and the ends are fixed, then the lowest allowed wavelength will be 1.2m (with other waves with wavelengths 0.6m and 0.4m and so on also allowed). The frequencies of these waves will be:
\begin{eqnarray} f_1 &=& \frac{c}{\lambda_1} = 236 \mathrm{Hz}\\ f_2 &=& 472 \mathrm{Hz}\\ f_3 &=& 708 \mathrm{Hz} \end{eqnarray}The fundamental will be just below middle C (which is about 262 Hz).
The fundamental frequency for an ideal vibrating string will arise from the longest wavelength which can be supported. This is $\lambda_1 = 2L$ for a string of length $L$. We then write: \begin{eqnarray} f_1 &=& \frac{c}{\lambda_1} = \frac{c}{2L}\\ f_n &=& \frac{2}{\lambda_n} = \frac{nc}{2L} \end{eqnarray} for $n=1,2,3,\ldots$ where $c = \sqrt{T/\mu}$. A \emph{harmonic} is simply a vibration at an integer multiple of the fundamental frequency. Systems with the \emph{same} boundary conditions at both ends (e.g. fixed strings, open air columns etc.) will vibrate with all harmonics of the fundamental frequency, while systems with different boundary conditions (i.e. one end fixed and one free) will vibrate with only the odd harmonics. Most musical instruments will produce some harmonics when the fundamental frequency is excited (and it is also possible to excite the harmonics only, for instance by overblowing a wind instrument or inducing a node on a stringed instrument by touching the string lightly).
It is worth noting that there are other ways to set up standing waves. If a continuous wave (a sinusoidal wave, for example) is propagated along a semi-infinite string towards a fixed end, then it will reflect with a phase change of $\pi$ (or a change in sign). If we take the incoming wave to be $\psi_i = A \cos(kx - \omega t +\phi)$ the reflected wave will be $\psi_r = -A\cos(kx+\omega t +\phi)$. Then we will find:
\begin{eqnarray} \psi(x,t) &=& \psi_i(x,t) + \psi_r(x,t) \\ &=& A \cos(kx - \omega t) - A \cos(kx + \omega t)\\ &=& -2A\sin(kx)\sin(\omega t + \phi) \end{eqnarray}which is exactly the form we saw above.
This suggests that it is worth examining other boundary conditions on waves, to explore the full range of standing waves. Instead of a fixed end (or a node) we could allow the medium to have an antinode (or free end) at one or both ends. This is quite common when thinking about sound waves and we will consider pipes in detail in the section on standing waves in a fluid. For now, let us consider what an antinode at both ends would mean (think of holding a thin steel ruler lightly in the centre and wiggling it up and down: you impose a node at the centre but nothing at the ends). This is equivalent to setting $\partial \psi/\partial x = 0$ (think about the transverse component of the tension - we are setting it to zero as there is no constraint on the system at the ends).
We will take advantage of our knowledge of the system to suggest the form: \begin{eqnarray} \psi(x,t) = 2A\cos(kx)\sin(\omega t) \end{eqnarray} We will again impose the condition that $kx = n\pi$ at $x=0$ and $x=L$ for $n$ an integer, but this time we are forcing an antinode to be at the boundaries. The allowed series of wavelengths will then be given by: \begin{eqnarray} k_n &=& \frac{n\pi}{L}\\ \Rightarrow \lambda_n &=& \frac{2\pi}{k_n} = \frac{2L}{n} \end{eqnarray} which is the same set of wavelengths (really all we've done to the system is introduce a phase shift of $\pi/2$). We'll consider the effect of a free end and a fixed end when we look at sound waves in pipes.
We can plot these waves in exactly the same way as for the fixed boundary conditions, and this is done below.
fig = figure(figsize=[12,7])
# Set length scale
L = 5.0
# We will put the subplots into an array, ax
ax = []
# Define x coordinates and omega
x = linspace(0,L,1000)
omega = 1.0
# Loop over first five modes, n
for n in range(1,6):
# Create subplot; we want 5 in y and 1 in x, with incremented index - spn gives this
spn = 510+n
# Plot mode n
ax.append(fig.add_subplot(spn,autoscale_on=False,xlim=(0,L),ylim=(-1.1,1.1)))
# Plot at a number of times to show how the wave changes
for t in range(4):
ax[n-1].plot(x,cos(n*pi*x/L)*sin(omega*pi*t/2.0))