Waves!

Imagine a pulse propagating in a string. As the wave travels each segment of the string moves up and down perpendicular to the direction of propagation. At the macroscopic level, we observe a transverse wave that moves along the string, and the individual motion of the segments is not relevant. In contrast, at the microscopic level, we see discrete particles undergoing oscillatory motion perpendicular to the motion of the wave.

In a string, or a solid, or a fluid, the collective motion of the particles determine the velocity of sound in the medium, as well as the thermal transport properties.

Coupled oscillators

Coupled oscillators in one dimension.

Let us first consider a one-dimensional chain of $N$ particles of mass $m$ with equal equilibrium separation $a$. The particles are coupled to massless springs with force constant $k_{c}$, except for the first and last springs at the two ends of the chain which have spring constant $k$. the individual displacement of the particle $i$ from its equilibrium position along the $x$ axis is called $u_{i}$. The ends of the fist and last spring are assumed fixed: $$u_{0}=u_{N+1}=0.$$ Since the force of an individual mass is determined only by the compression or expansion of the adjacent springs, the equation of motion for particle $i$ is given by: $$\begin{eqnarray} m\frac{d^{2}u_{i}}{dt^{2}} &=&-k_{c}(u_{i}-u_{i+1})-k_{c}(u_{i}-u_{i-1}) \\ &=&-k_{c}(2u_{i}-u_{i+1}-u_{i-1}).\end{eqnarray}$$

The equations for particles $i=1$ and $i=N$ next to the walls are given by $$\begin{eqnarray} m\frac{d^2u_1}{dt^2}&=&-k_c(u_1-u_2)-ku_1, \\ m\frac{d^2u_N}{dt^2}&=&-k_c(u_N-u_{N-1})-ku_N.\end{eqnarray}$$

Note that for $k_c=0$ all the equations will decouple and the motion of the particles become independent of their neighbors. The above equations describe longitudinal oscillations, i. e. motion along the direction of the chain. The equations for transverse motion are equivalent.

Exercise 5.1

Write a generalization of the program for a mini solar system to describe the motion of $N$ coupled linear oscillators.

Exercise 5.2: two coupled oscillators

1. Use the previous program with $N=2$. Set the initial conditions $% u_{1}=0.5$ an $u_{2}=0$ and compute the time dependence of the displacements for $k=1,k_{c}=0.8$ and $k=k_{c}=1$. Remember that the mass of the particles has to be set to unity. Describe the qualitative behavior or the particles in each case. Is it possible to define a period of motion in the first case? What is the period in the second case?

2. Set $k=1,k_{c}=0.2$. Since $k_{c}<k$, we can describe the springs as being “weakly coupled”. Observe the time dependence of the displacement of particle one. Can you identify two kinds of oscillations superimposed on upon each other? What is the time between the oscillations of the amplitude? What is the time between the zeroes of the displacement? Compute the corresponding angular frequency of each oscillation. How does the displacement of particle 2 correspond to that of particle 1? Determine the qualitative changes in the frequencies of each oscillation for $k_{c}=1$.

3. Choose the initial conditions $u_{1}=u_{2}=0.5$ so that both particles have the same initial displacements, Set $k_{c}=0.1$ and $k=1$ and describe the motion. Compute the total energy of each particle as a function of time and describe its qualitative behavior. Does the period of motion depend on $k_{c}$? What if the dependence of the period on $k$?

4. Consider the initial conditions $u_{1}=-u_{2}=0.5$ so that both particles have equal but opposite initial displacements. Is there a simple sinusoidal oscillation in this case? Compute the period $T_{1}$ for ${% k=1,k_{c}=1}$, ${k=2,k_{c}=1}$, and ${k=1,k_{c}=2}$. Analyze your results for $\omega _{1}^{2}$ and determine its dependence on $k$ and $k_{c}$. What is the behavior of the energy of each particle as a function of time?

We will now calculate the power spectrum and determine the natural frequencies by using the numpy FFT module. The frequency interval is given by $$\Delta \omega = \frac{1}{N_{steps}\Delta t}$$

Exercise 5.3: response to an external force

Add an external driving force $F(t)=F_{0}\cos {\omega t}$ to particle 1. plot $u_{1}(t)$ and determine its maximum steady state amplitude $A(\omega )$ for each value of $\omega $. Confirm that near a resonance $A(\omega )$ exhibits a rapid increase with $\omega $, and $u_{1}(t)$ increases without bound at the resonant frequency. Determine the resonant frequencies for the pairs of $k$-values already considered in parts 2-4 in Exercise 5.2. How do these values of $\omega $ compare to those in the previous exercise?

Challenge 5.1:

• Complete exercise 5.3

Exercise 5.4: superposition of motion

The results of the previous exercises make plausible the assumption that an arbitrary motion of the system can be written as

$$\begin{eqnarray} u_1(t) &=& A_1\cos{(\omega _1t+\delta _1)} + A_2\cos{(\omega _2 t+\delta _2)} \\ u_2(t) &=& A_1\cos{(\omega _1t+\delta _1)} - A_2\cos{(\omega _2 t+\delta _2)}\end{eqnarray}$$

The values of these constants can be expressed in terms of the initial values of the displacement and velocities of each particle. determine these constants for $u_{1}=0.5,u_{2}=0,v_{1}=v_{2}=0$. Verify that the motion predicted by these equations is consistent with your measure values for $% u_{1}$ and $u_{2}$ for $k=1$ and $k_{c}=0.8$ in part 1 of Exercise 5.2. What is the periodicity of $u_{1}$ and $u_{2}$?

The effect of the spring $k_{c}$ is to couple the motions of two particles so that they no longer move independently. For special initial conditions, only one frequency of oscillation appears. The resulting motion is called a “normal mode” of the system, and the corresponding frequency a “normal mode frequency”. The higher frequency is given by $\omega _{1}^{2}=(k+2k_{c})/m$. In this mode, the two particles oscillate exactly out of phase with the displacements of opposite directions. The motion at the lowest frequency $\omega _{2}^{2}=k/m$ corresponds to the two particles oscillating exactly in phase.

The general motion is a superposition of the two normal modes. Unless there is a simple relation between the two frequencies, the general motion is a complicated function of time. However, if the coupling is small, $\omega _{1} $ and $\omega _{2}$ are nearly equal and $u_{1}$ and $u_{2}$ exhibit “beats”. In this case the displacements oscillate rapidly at the angular frequency $% 1/2(\omega _{1}+\omega _{2})$ with an amplitude that varies sinusoidally at the beat frequency $1/2(\omega _{1}-\omega _{2})$.

We also found that if we drive the system by an external force applied to either particle (or both), the system is in resonance if the frequency of the force corresponds to either of the normal modes. We use this method for determining the normal mode frequencies in the following exercises.

Exercise 5.5: three coupled oscillators

1. Run the program with $N=3$, $k_{c}=0.2$ and $k=1$ and arbitrary but nonzero initial displacements. Describe the time-dependence of the motion of the particles.

2. Consider the following initial conditions: ${u_{1}=u_{2}=u_{3}=0.5}$, ${u_{1}=0.5,u_{2}=-0.5,u_{3}=0.5}$, ${u_{1}=0.5,u_{2}=0,u_{3}=-0.5}$ (all the velocities are initially zero) If these initial conditions correspond to normal modes, determine the normal modes frequencies.

3. Add an external driving force to particle 1 and determine the normal mode frequencies. Compare your results with the frequencies you obtained in part 2. How many normal modes are there?

Exercise 5.6: $N$ coupled oscillators

1. Choose $k_{c}=k=1$ and $N=10$. Find the normal modes by applying an external force to one of the particles and determine the resonant frequencies. Drive the system for several periods of the external force and compute the steady state amplitude of the displacement of each particle for each value of $\omega $. Try values of $\omega $ in the range $0.2(k/m)^{1/2} $ to $3(k/m)^{1/2}$. If you think that you are close to a resonance, use several other values of $\omega $ to obtain a better estimate. How many normal modes are there?

2. Compare your results in part 1 with the analytic result $$\omega _{n}^{2}=\frac{4k}{m}\sin ^{2}\frac{n\pi }{2(N+1)},$$ where $N$ is the number of particles and the mode index is $n=1,2,...,N$.

Exercise 5.7: propagation speed in a linear chain

Another interesting property to analyze is the propagation of the energy. In this problem we’ll disturb the system determine the time that takes for the disturbance to travel a given distance.

1. Consider a linear chain of coupled oscillators at rest with $k=k_{c}=1 $. Create a disturbance giving particle 1 an initial displacement $u_{1}=1$. Determine the time it takes for particles $N/2$ and $N$ to satisfy the conditions $|u_{N/2}|\ge d$ and $u_{N}\ge d$. Choose $N=10$ and $d=0.3$ for your initial runs. Use your results to estimate $v$, the speed of propagation of the disturbance. Consider larger values of $N$ to ensure that your value is independent of $N$.

2. Do you expect the speed of propagation to be an increasing or decreasing function of the spring constant $k$? Do a simulation and estimate $v$ for different values of $k$.

3. Create a disturbance by applying an external force $F(t)=F_{0}\cos {% \omega t}$ to particle 1. Estimate the propagation speed of the disturbance as in part 1. Consider the value so $\omega =0.1$ and $\omega =1$. Explain why the propagation speed depends on $\omega $. Can a disturbance propagate for $\omega =4$? In what way does the system act as a mechanical filter? Explain the “filtering” property of the system in terms of the frequency of the normal modes.

Fourier analysis

In the previous section we found that the displacement of a particle an be described as a linear combination of normal modes, i.e. a superposition of sinusoidal terms. This decomposition of the motion into various frequencies is more general. It can be shown that any arbitrary periodic function $f(x)$ of period $T$ can be written as a Fourier series of sines and cosines: $$f(t)=\frac{1}{2}a_{0}+\sum_{n=1}^{\infty }(a_{n}\cos {n\omega t}+b_{n}\sin {% n\omega t}),$$ where $\omega _{0}$ is the fundamental angular frequency given by $$\omega _{0}=\frac{2\pi }{T}.$$ The sine and cosine terms represent the harmonics. The Fourier coefficients $% a_{n}$ and $b_{n}$ are given by $$\begin{eqnarray} a_{n} &=& \frac{2}{T}\int_{-T/2}^{T/2}{f(t)\cos{n\omega _0 t}dt}\\ b_{n} &=& \frac{2}{T}\int_{-T/2}^{T/2}{f(t)\sin{n\omega _0 t}dt}\\\end{eqnarray}$$ The constant term $1/2a_{0}$ is the average value of $f(t)$. (all the oscillating contributions vanish in average)

In general, an infinite number of terms is needed to represent an arbitrary function $f(t)$. In practice, a good approximation can usually be obtained by including a relatively small number of terms.

The sawtooth function studied in Exercise 5.8

Exercise 5.8: Fourier analysis

1. Consider the following series $f(t)=\frac{2}{\pi }\sum_{n=1}^{\infty }\sin {nt}.$Plot $f(t)$ retaining only the first 3 terms. Increase the number of terms until you are satisfied that you are converging to $f(t)$ with some arbitrary but sufficient accuracy. What is the function represented by the sum?

2. Use the analytical expression for the Fourier coefficients and calculate the integrals using the “sawtooth” function depicted in Fig 5.2 to show that they are effectively given by $a_{n}=0$ and $b_{n}=(1/n\pi )(-1)^{n-1}$

3. What function os represented by the Fourier series with coefficients $% a_{0}=0$ and $a_{n}=b_{n}=1/n^{2}$ for $n\ne 0$?

Waves on a string

A stretched string of length $l$ with the ends fixed

The difference between waves and oscillatory motion is the scale. Waves are the “continuum” limit of the problem, or in the jargon “the long wavelength” limit. This is because in this limit, all the microscopic details are “washed out” and only the long distance behavior survives. In order to understand the transition between these two limits we have to perform a change of scale. The discrete equations of motion, as shown previously, can be written as: $$\frac{d^{2}u}{dt^{2}}=-\frac{k}{m}(2u_{i}-u_{i+1}-u_{i-1}).$$ We consider the limits $$N\rightarrow \infty ,a\rightarrow 0.$$ with the length of the chain kept constant. The main result is that in this limit, the discrete equations of motion can me replaced by the continuous wave equation: $$\frac{\partial ^{2}u(x,t)}{\partial t^{2}}=\frac{1}{v^{2}}\frac{\partial ^{2}u(x,t)}{\partial x^{2}}, $$ where $v$ has the dimension of velocity and it is given by $$v=\sqrt{k/\rho },$$ where $k$ is the string tension and $\rho $ is the linear density. Observe that the displacement $u$ of the string is the dependent variable, and that the position along the string $x$ and the time $t$ are the independent variables. The existence of two independent variables makes this a Partial Differential Equation (PDE).

There are many solutions to this equation. Examples are: $$\begin{eqnarray} u(x,t) &=&A\cos \frac{2\pi }{\lambda }(x\pm vt), \\ u(x,t) &=&A\sin \frac{2\pi }{\lambda }(x\pm vt).\end{eqnarray}$$ In fact, it is easy to show that any function of the form $f(x\pm vt)$ is a solution. Since the differential equation is a linear equation and hence satisfies the superposition principle, we can understand the behavior of a wave of arbitrary shape using the Fourier’s theorem to represent its shape as a sum of sinusoidal waves.

Because both ends of the string are tied down, the boundary conditions are: $$u(0,t)=u(l,t)=0.\,\,\,\mathrm{(boundary\,\,condition)}$$

Since this is a second order PDE, we still need to determine the initial distortion $u(x,t=0)$ and velocity $\partial u/\partial t(x,t=0)$. If the string is released from rest, this reduces to $$\begin{eqnarray} u(x,t) &=&f(x),\,\,\,\,\mathrm{(initial\,\,condition\,\,1)} \\ \frac{\partial u}{\partial t}(x,t &=&0)=0.\,\,\,\mathrm{(initial\,% \,condition\,\,2)} \end{eqnarray}$$

Numerical solution: finite differences

Finite differences grid for the vibrating string

To solve the equation ([wave]) as a function of position and time we need to discretize the $(x,t)$ space in a rectangular grid (see Fig [grid]). In the present case, the horizontal axis represents the position $x$ along the string, and the vertical axis represent time. We convert the equation to a finite difference equation expressing the second derivatives in terms of differences

$$\begin{eqnarray} \frac{\partial ^{2}u(x,t)}{\partial t^{2}} &\simeq &\frac{u(x,t+\Delta t)+u(x,t-\Delta t)-2u(x,t)}{(\Delta t)^{2}}, \\ \frac{\partial ^{2}u(x,t)}{\partial x^{2}} &\simeq &\frac{u(x+\Delta x,t)+u(x-\Delta x,t)-2u(x,t)}{(\Delta x)^{2}}.\end{eqnarray}$$

After substituting into ([wave]) we obtain the discrete equations: $$u(x,t+\Delta t)=2u(x,t)-u(x,t-\Delta t)+\frac{v^{2}}{C^{2}}\left[ u(x+\Delta x,t)+u(x-\Delta x,t)-2u(x,t)\right] ,$$ with $C=\Delta t/\Delta x$ is a constant with the dimension of velocity.

As shown in Fig. [grid], this is a recurrence relation that propagates the wave from the two earlier times $t-\Delta t$ and $t$, and the three nearby positions $x-\Delta x$, $x$, and $x+\Delta x$, to a later time $% t+\Delta t$, and a single position $x$. We can see right a way that this is not a self starting algorithm, in the sense that we need to know the position for two earlier times to start the iteration. However, we can use a simple trick to overcome this difficulty. Rewriting the initial conditions ([initial1]) and ([initial2]) in the finite differences form, we obtain: $$\begin{eqnarray} \frac{\partial u(x,t=0)}{\partial t} &=&0\Rightarrow \frac{u(x,\Delta t)-u(x,-\Delta t)}{2\Delta t}=0, \\ &\Rightarrow &u(x,-\Delta t)=u(x,\Delta t).\end{eqnarray}$$ Using this condition for the first iteration we obtain

$$u(x,\Delta t)=u(x,0)+\frac{1}{2}\frac{v^2}{C^2} \left[ u(x+\Delta x,0)+u(x-\Delta x,0)-2u(x,0)\right] .$$

The success of this method depends on the relative sizes of the time and space steps. This stability criterion says that the finite difference algorithm is stable if $$v\leq \frac{\Delta x}{\Delta t}=C. $$ This means that the solution gets better with smaller time steps, but worse with smaller space steps!

Exercise 5.9: Finite differences for the wave equation

Write a program to solve the wave equation using finite differences. Assume that the string has a length $l=1$m, a linear density $\rho =0.01$kg/m, and tension $k=40$N.

1. Assume as initial conditions that the string is “plucked”: $$u(x,t=0)=\left\{ \begin{array}{ll} 1.25x/l, & \mathrm{for\,\,}x\leq 0.8l, \\ 5.0(1-x/l), & \mathrm{for\,\,}x>0.8l, \end{array} \right.$$ Plot the displacement $u$ as a function of $(x,t)$ using a space step $% \Delta x=0.01$m, and choosing the time step such that the solution is stable.

2. Explore the use of different steps $\Delta x$ and $\Delta t$ and determine at which values the solution becomes unstable. Does your assessment agree with the condition ([stability])?

3. Change the initial conditions to $$u(x,t=0)=\left\{ \begin{array}{ll} x/l, & \mathrm{for\,\,}0\leq x\leq 0.5, \\ -x/l, & \mathrm{for\,\,}0.5\leq x\leq 0.1, \end{array} \right.$$ and compare results with the previous simulation.

4. Consider a string plucked at two points: $$u(x,t=0)/0.005=\left\{ \begin{array}{ll} 0, & 0.0\leq x\leq 0.1, \\ 10x-1, & 0.1\leq x\leq 0.2, \\ -10x+3, & 0.2\leq x\leq 0.3, \\ 0, & 0.3\leq x\leq 0.7, \\ 10x-7, & 0.7\leq x\leq 0.8, \\ -10x+9, & 0.8\leq x\leq 0.9, \\ 0 & 0.9\leq x\leq 1.0. \end{array} \right.$$ Solve and observe wether the pulses move or oscillate up and down.

5. Explore what happens when the string is places in a “normal mode”, for example:

   $$u(x,t=0)=0.001\sin (2\pi x).$$

   Try other modes. See if the sum of two modes gives “beating”.

Exercise 5.10: String with friction

The effect of the friction on an element of string $(x,x+\Delta x)$ is to oppose its motion. Assume that the force of friction is proportional to the vertical velocity $\partial u/\partial t$ of the string element. This changes the wave equation to $$\frac{\partial ^{2}u}{\partial t^{2}}+2\kappa \frac{\partial u}{\partial t}=% \frac{\partial ^{2}u}{\partial x^{2}},$$ where the constant $\kappa $ is proportional to the viscosity of the medium in which the string is vibrating, and is inversely proportional to the density of the string. Generalize the algorithm for the wave equation to include friction and observe the change in the wave behavior. Use the same parameters of the pervious example, with $\kappa =5$.

Challenge 5.2:

Exercise 5.10