\documentclass[12pt]{article} \usepackage{graphicx} %\usepackage{epsfig} %\topmargin=+0.0in %\oddsidemargin=0.25in \textwidth=6.0in %\textheight=8in %\footskip=6ex %\footheight=2ex \begin{document} \begin{center} {\bf Notes on Oscillations and Mechanical Waves} \end{center} \bigskip The topics for the second part of our physics class this quarter will be oscillations and waves. We will start with periodic motion for the first two lectures, with our specific examples being the motion of a mass attached to the end of a spring, and the pendulum. The last six lectures will be devoted to mechanical waves and their properties.\\ \centerline{\bf Periodic Motion} \bigskip Periodic motion is motion that repeats itself. For example, a small object oscillating at the end of a spring, a swinging pendulum, the earth orbiting the sun, etc. are examples where the objects motion "approximately" keeps repeating itself. If an objects motion is periodic, then there is a characteristic time: the time it takes for the motion to repeat itself. This is called the period (of the periodic motion) and is usually given the symbol $T$:\\ \noindent {\bf Period} ($T$): The time for one complete cycle of the periodic motion.\\ \noindent For example, the period of the rotation of the earth about its axis is one day. During the quarter, when classes are in session, the period of our activities is one week. We can also speak of the number of cycles repeated per unit time. This is called the frequency of the periodic motion:\\ \noindent {\bf frequency} ($f$): The number of cycles per unit time.\\ \noindent A common unit for frequency is one cycle per second. This is defined as one Hertz (Hz). 1 Hz $\equiv$ 1 cycle/sec. If the periodic motion occurs $f$ times per second, then the time for one cycle is $1/f$, so \begin{equation} T = {1 \over f} \end{equation} \noindent or equivalently \begin{equation} f = {1 \over T} \end{equation} In this class we will mainly be describing periodic motion that occurs in one dimension. For example an object will execute periodic motion along the x- or y-axis. Suppose that the motion of an object occurs along the x-axis. Then the position of the object is given by its position function $x(t)$. If the motion is periodic, it means that: \begin{equation} x(t+T) = x(t) \end{equation} \noindent where $T$ is the period of the motion. The function $x(t)$ can in general have any shape as long as $x(t) = x(t+T)$. In some cases, $x(t)$ has a simple oscillating shape, that of a pure sinusoidal form. If the motion is perfectly sinusoidal, i.e. a sin or cos function, then motion is called "simple harmonic motion".\\ \bigskip \centerline{\bf Simple Harmonic Motion} \bigskip If the motion of an object is a perfect single sinusiodal function, we term the motion "simple harmonic". Let's review the mathematics of simple trig functions. We need just three parameters to describe any single sinusoidal function: the period ($T$), the amplitude ($A$), and a parameter related to the position at time $t=0$. The amplitude of the motion is the maximum value from equilibrium that the object travels. Suppose the object starts off initially at $x=0$, has some initial speed in the +x direction and an amplitude $A$. In this case, the motion of the object is described by: \begin{equation} x(t) = A sin({{2 \pi t} \over T}) \end{equation} \noindent where $T$ is the period of the motion. One often defines $2 \pi / T$ to be the variable $\omega$: \begin{equation} \omega \equiv {{2 \pi} \over T} \end{equation} \noindent So one often sees the equation for this case of simple harmonic motion written as $x(t) = A sin(\omega t)$. At time $t=0$, the position is at $x=0$ since $sin(0) = 0$. The velocity of the particle at any time $t$ is found by differentiating $x(t)$: \begin{equation} v(t) = {{dx} \over {dt}} = A \omega cos(\omega t) \end{equation} \noindent The initial velocity is just $v(0) = A \omega$, and is in the + direction. Actually $A \omega$ is the largest speed that the object ever acquires. The above equation only applies to simple harmonic motion for which the object starts at $x=0$. How do we generalize this to the case where the object starts off at $x=x_0$, where $x_0 \neq 0$? This can be done by adding a phase shift, $\phi$ measured in radians, to the argument of the sin function: \begin{equation} x(t) = A sin(\omega t + \phi) \end{equation} \noindent If $\phi <0$ then the sin function is shifted to the right. If $\phi > 0$ the sin function is shifted to the left. By adjusting the phase angle one can shift the sin function to match the initial condition. If the initial position is defined as $x(0) \equiv x_0$, then \begin{equation} x_0 = A sin(\phi) \end{equation} \noindent or \begin{equation} \phi = sin^{-1}({x_0 \over A}) \end{equation} The function $x(t) = A sin(\omega t + \phi)$ can be expressed in an alternative form by expanding the sin function. Using the angle addition formula for sin: \begin{equation} sin(\omega t + \phi) = sin (\omega t) cos \phi + cos(\omega t) sin \phi \end{equation} \noindent Substituting into the equation for $x(t)$ gives: \begin{equation} x(t) = (A cos \phi) sin (\omega t) + (A sin \phi) cos(\omega t) \end{equation} \noindent This form for $x(t)$ can be written in a more convenient way. At $t=0$, $x_0 \equiv x(0) = A sin \phi$: \begin{equation} x(t) = (A cos \phi) sin (\omega t) + x_0 cos(\omega t) \end{equation} \noindent Differentiating the above equation gives: \begin{equation} v(t) = A \omega cos\phi cos(\omega t) - x_0 \omega sin(\omega t) \end{equation} \noindent The initial velocity $v_0 \equiv v(0) = A \omega cos\phi$. This gives $A \cos\phi = v_0/\omega$. Substituting into the equation for $x(t)$ and rearrainging the terms gives a nice form for $x(t)$ in terms of the initial position and velocity: \begin{equation} x(t) = x_0 cos(\omega t) + {v_0 \over \omega} sin(\omega t) \end{equation} This concludes our review of the two different forms of a sinusoidal function. Now for the Physics! \bigskip \noindent {\bf What kind of Force Produces Simple Harmonic Motion?} \bigskip Suppose an object, of mass $m$, is undergoing simple harmonic motion along the x-axis. This means that the object's position follows a single sinusiodal function, whose general form is $x(t) = A sin(\omega t + \phi)$. What kind of force produces this kind of motion? Since we know $x(t)$, we can find $a(t)$. The force is just $F(t) = m a(t)$. The velocity is found by differentiating $x(t)$: \begin{equation} v(t) = A \omega cos(\omega t + \phi) \end{equation} \noindent Differentiating again for $a(t)$: \begin{equation} a(t) = - A \omega^2 sin(\omega t + \phi) \end{equation} \noindent However, notice that $x(t) = A sin(\omega t)$, which simplifies to: \begin{equation} {{d^2 x} \over {dt^2}} = - \omega^2 x(t) \end{equation} \noindent or \begin{equation} a(t) = - \omega^2 x(t) \end{equation} \noindent or simply $a = - \omega^2 x$ at all times and all positions $x$. This is a simple result: the acceleration is proportional to $x$, the objects displacement from the equilibrium position. The negative sign means that if $x>0$ then $a<0$, and if $x<0$ then $a>0$. This means that the acceleration is always towards the origin. Since $F = ma$, we have \begin{equation} F = - m \omega^2 x \end{equation} \noindent This result is also quite simple. The force is always towards the origin, and is proportional to $x$ the distance from the origin. If the object is moved twice as far away from the equilibrium position, it feels twice the force pushing it back. This type of force is called a "linear restoring force". It is a restoring force because of the minus sign. For positive $x$ the force is in the negative direction, and for negative $x$ the force is in the positive direction. It tries to "restore" the object back to $x=0$ the equilibrium position. It is a linear force because its magnitude is proportional to $x$ the displacement from equilibrium. Thus we see that {\bf a linear restoring force produces motion that is a single sinusoidal function (simple harmonic motion)}. Let's consider some forces to see if they produce simple harmonic motion, and if so, how the period depends on the parameter of the force.\\ \noindent {\bf The simple spring} \bigskip We are all familiar with a spring. Suppose an object of mass $m$ is attached to one end, while the other end of the spring is held fixed. The object will have an equilibrium position, call it $x=0$. If the object is displaced away from $x=0$ the spring pulls (or pushes) it back to the equilibrium position ($x=0$). Thus, the spring produces a restoring force. But is the restoring force linear? That is, if we pull the object twice as far away from $x=0$ is the force twice as much? This question is answered by doing the experiment. You will do measurements in lab that show that {\bf the restoring force of a common spring is approximately linear}. That is: \begin{equation} F_{spring} \approx - k x \end{equation} \noindent The proportionality constant $k$ is called the spring constant for the particular spring. If $k$ is large, then the spring is stiff and produces a lot of force for a small displacement. If $k$ is small, then the spring is "loose" and doesn't pull back with much force. The spring constant $k$ can be measured by applying a force ($F_0$) and measuring how much the spring stretches ($d$). Then $k = F_0/d$. $k$ has units of force/distance (N/m). If time permits, we will discuss different spring examples in class. Having established that the simple spring produces a linear restoring force (approximately), we can determine period of the resulting periodic (simple harmonic) motion. From the equation above, $F = - m \omega^2 x$. Since $F_{spring} = - k x$ we have: \begin{equation} k = m \omega^2 \end{equation} \noindent or \begin{equation} \omega = \sqrt{{k \over m}} \end{equation} \noindent Since the period $T = 2 \pi / \omega$, the period for the objects motion for the spring force is: \begin{equation} T = 2 \pi \sqrt{{m \over k}} \end{equation} Although this formula was derived for the simple spring, a similar formula will result in general for a linear restoring force. If the force on an object (moving in one dimension) is of the form $F_x = -C x$, then the motion will be simple harmonic, $x(t) = A sin(\omega t + \phi)$ where $\omega = \sqrt{C/m}$. Notice that in the equation for the period $T$ for the linear restoring force that $T$ only depends on the mass and the restoring force constant. It does not depend on the amplitude $A$ of the motion. Thus, oscillations with large amplitudes will have the same period as oscillations with small amplitudes. This phenomena is a nice characteristic of the linear restoring force. The final equation of motion for the object at the end of the spring is: \begin{equation} x(t) = A sin(\sqrt{{k \over m}} t + \phi) \end{equation} \bigskip \noindent {\bf The Simple Pendulum} \bigskip A simple pendulum is just a pendulum made out of a small object attached to a light string. Ideally, it is a "point" particle attached to a massless string which is fixed to a pivot point. If the pendulum is displaced from equilibrium, it swings back and forth, and its motion is periodic. The questions we want to consider are: is the motion simple harmonic, and what is the equation for the period $T$ of the motion? To answer these questions, one starts with the equation relating forces and motion. I am going to use a different variable than the textbook. I will specify the position of the particle by the distance, along an arc, that the particle is from the equilibrium position. See the figure at the end of this section. If the length of the pendulum is $l$, the ratio $x/l$ is the angle $\theta$ (in radians) that the string makes with the vertical. Most textbooks use the angle $\theta$ to specify the location of the particle, so $x = l \theta$ is the connection between the text's $\theta$ and our $x$. I think this will be easier for us to make reference to the spring equation and avoid using torque. When the pendulum is hanging vertical, $x=0$, right displacement is positive $x$ and left displacement is negative $x$. If an object is displaced a distance $x$ along the arc, the component of gravity in the direction of the arc is $-mg sin(\theta)$ or $-mg sin(x/l)$. The negative sign means that the force of gravity is a restoring force. If $x>0$, $sin(x/l)>0$ and the force is in the negative direction. If $x<0$, $sin(x/l)<0$ and the force is in the positive direction. Thus, \begin{equation} F_x = - mg sin({x \over l}) \end{equation} \noindent For a force to produce simple harmonic motion, the force must be proportional to $-x$. The equation for the simple pendulum is NOT of this form. Here the force is proportional to $-sin(x/l)$. Thus, {\bf the motion of a simple pendulum is NOT not simple harmonic}. Don't get confused with the "sin" function. {\bf A sinusoidal restoring force does not produce perfect sinusoidal motion}. Only a linear restoring force gives perfect sinusoidal motion. This is easy to demonstrate. For simple harmonic motion, the period does not depend on the amplitude. Pendula with a larger amplitude (larger $x_{max}$) have a longer period than ones with a smaller amplitude (smaller $x_{max}$). \bigskip \noindent {\it Approximate Sinusoidal Motion} \bigskip If $x/l$ is small, less than $0.1$ , then $sin(x/l) \approx x/l$. This can be seen from the series expansion of $sin(x/l)$. The expansion for $sin(\theta)$ is $sin(\theta) = \theta - \theta^3/3! + \theta^5 /5! - \dots$ (for $\theta$ in radians). Replacing $\theta = x/l$ gives: $sin(x/l) = x/l - (x/l)^3/3! + (x/l)^5 /5! - \dots$ If $(x/l) < 0.1$, then the second term is $0.0016$ times the first, so to a pretty good approximation $sin(x/l) \approx x/l$. In this case, we have \begin{equation} F_x \approx - {{mgx} \over l} \end{equation} \noindent Thus, for small angles the restoring force is proportional to displacement, and the resulting motion is simple harmonic. The proportionality constant $C$ is $mg/l$. Since the proportionality constant equals $m \omega^2$, we have $m \omega^2 = mg/l$, or \begin{equation} \omega = \sqrt{g \over l} \end{equation} \noindent For small oscillations, the position $x$ along the arc of the path is just $x(t) = A sin(\sqrt{{g \over l}} t + \phi)$. \noindent Since $T = 2 \pi/\omega$ we have \begin{equation} T \approx 2 \pi \sqrt{l \over g} \end{equation} \noindent for a simple pendulum swinging at small displacements $x$ such that $x_{max}/l$. In terms of the angle that the string makes with the vertical the maximum angle $\theta_{max}$ should be less than $0.1$ radians. The differential equation for $x(t)$ can be obtained by using $F_x = m d^2x/dt^2$: \begin{equation} m {{d^2 x} \over {dt^2}} = -mg sin({x \over l}) \end{equation} \noindent or \begin{equation} {{d^2 x} \over {dt^2}} = -g sin({x \over l}) \end{equation} \noindent Using the small angle approximation, we have \begin{equation} {{d^2 x} \over {dt^2}} \approx - {g \over l} x \end{equation} \noindent To compare with the equations in the book, just substitute $x = l \theta$. \bigskip \centerline{\bf Energy considerations for Simple Harmonic Motion} \bigskip To obtain an expression for the total mechanical energy that a simple harmonic oscillator has, we need an expression for the potential energy for the force acting. Let's consider the case of the simple spring. The work done by the spring force when an object moves from the position $x_1$ to the position $x_2$ is: \begin{equation} W_{1 \rightarrow 2} = \int_{x_1}^{x_2} (-kx) dx \end{equation} \noindent One needs to integrate since the force is not constant, but rather changes with position. The integral is: \begin{equation} W_{1 \rightarrow 2} = {k \over 2} x_1^2 - {k \over 2} x_2^2 \end{equation} \noindent From the work-energy theorum: $W_{net} = \Delta(K.E.)$. If the only force acting on the object is the force of the spring, then the net work is the work done by the spring. In this case we have: \begin{equation} {k \over 2} x_1^2 - {k \over 2} x_2^2 = {m \over 2} v_2^2 - {m \over 2} v_1^2 \end{equation} \noindent Rearrainging the terms gives: \begin{equation} {k \over 2} x_1^2 + {m \over 2} v_1^2 = {k \over 2} x_2^2 + {m \over 2} v_2^2 \end{equation} \noindent All the terms on the left are from position $1$, and all the terms on the right are from position $2$. Since these positions are arbitrary, we have that the expression $ {k \over 2} x^2 + {m \over 2} v^2 $ is a constant of the motion. In this expression, $x$ is the position of the object and $v$ is the speed of the object (at the position $x$). As the object moves back and forth, $x$ and $v$ both change but the special combination $ {k \over 2} x^2 + {m \over 2} v^2 $ remains constant. We recognize the first term to be the potential energy and the second term to be the kinetic energy of the object. The sum, which is the total mechanical energy, remains constant. \begin{equation} E_{tot} = {k \over 2} x^2 + {m \over 2} v^2 \end{equation} \noindent To determine the total mechanical energy for simple harmonic motion in general, we need to substitute for $x(t)$ and $v(t)$. In general $x(t) = A sin(\omega t + \phi)$. The velocity is given by $v(t) = dx/dt = \omega A cos(\omega t + \phi)$. Substituting into the above equation, we have: \begin{equation} E_{tot} = {k \over 2} A^2 sin^2(\omega t + \phi) + {m \over 2} \omega^2 A^2 cos^2 (\omega t + \phi) \end{equation} \noindent However, since $m \omega^2 = k$, we have \begin{equation} E_{tot} = {k \over 2} A^2 (sin^2(\omega t + \phi) + cos^2 (\omega t + \phi)) \end{equation} \noindent Since $sin^2 + cos^2 = 1$, the equation simplifies to \begin{equation} E_{tot} = {k \over 2} A^2 \end{equation} \noindent The total mechanical energy of a mass-spring system is $k/2$ times the square of the amplitude of the motion. In general, if the proportionality constant is $C$ ($F_x = - C x$), the total mechanical energy is ${C \over 2} A^2$. \bigskip \centerline{\bf "Mechanical Waves": Waves in a Medium} \bigskip In our previous discussion, we considered oscillations, or periodic motion, in one dimension. That is, an object that moves back and forth in one dimension (e.g. a mass on a spring). Now we want to consider oscillations that take place in a medium. By a {\bf medium} we mean a material like a gas, liquid or solid. {\bf Oscillations that take place in a medium are called mechanical waves}. The word mechanical just means that there is a medium that is oscillating. What is meant by the expression that a medium is oscillating? It means that the constituents of the medium (the atoms and molecules) are moving back and forth about an average equilibrium position. For the single particle undergoing periodic motion, we described its motion by its position $x(t)$ on the x-axis. Now we have a much different situation to describe. Each point in the medium will have its own displacement as a function of time. I am going to use the symbol $d$ to stand for the displacement of the medium. (The text uses the symbol $y$ which I find confusing.) Each point of the medium will have a displacement that usually varies with time:\\ \noindent d($\vec{r},t)$ is defined as the displacement of the medium at the position $\vec{r}$ and at the time $t$.\\ \noindent There are two "distances" involved here. One is the position $\vec{r}$ in the medium. The other is $d$, the displacement of the medium at the position $\vec{r}$. $d$ depends on position and time. It is actually a "field" quantity, a quantity that depends on position and time. To make the analysis simple, here we will only consider changes in the position $\vec{r}$ in one dimension. For example, suppose the medium is a taut string, like a guitar string. Let the x-axis lie along the string. $d(x,t)$ refers to how much the guitar string is displaced above its equilibrium position at the position $x$ at time $t$. If the string is not vibrating, then $d(x,t) =0$ for all $x$ and $t$. Suppose the string is vibrating. If $x(5 \; cm, 2 \; sec) = 0.5 \; cm$, it means that at the position $5$ cm down the string, the string is displaced $0.5$ cm at the time $2$ seconds. $d$ is different at other positions on the string and as the time changes so will $d$. It is important to understand what $d(x,t)$ means. We will use the function $d(x,t)$ to describe waves, and will derive an equation that $d(x,t)$ satisfies. One cannot make a "simple" graph of $d(x,t$ in two dimensions, since $d$ itself depends on $x$ and $t$. However, two graphs can be used to describe the displacement. One would be a "snapshot" of the wave at a certain time $t$, say $t=0$. In this graph, one would plot $d(x,0)$ versus $x$. It would show a picture of {\it the disturbance at a moment in time}. The other graph would be a graph of the displacement as a function of time at at certain location, say $x=0$. Here, one would plot $d(0,t)$ as a function of time, and the graph would display the displacement of the medium as a function of time at the position $x=0$. \bigskip \noindent {\bf Traveling Waves} \bigskip Under certain conditions, the disturbance (or wave) can move with a definate velocity $v$. One part of the medium "pushes" on the part next to it, and the disturbance propagates down the medium. We will discuss later why this can occur, but lets first consider how such a traveling wave can be described mathematically. Suppose the wave has the functional form $d = f(x)$ at time $t=0$ If the wave {\bf travels to the right} with speed $v$, what is $d$ for times $t>0$? One just replaces $x$ with $x-vt$: \begin{equation} d(x,t) = f(x-vt) \end{equation} \noindent for all $x$ and $t$. If the disturbance (or wave) is {\bf traveling to the left}, then $x$ is replaced with $x+vt$: \begin{equation} d(x,t) = f(x+vt) \end{equation} \noindent Let's consider an example to better understand this concept. Suppose at $t=0$ the displacement is \begin{equation} d(x,0) = {1 \over {x^2 + 1}} \end{equation} \noindent A graph of this function is given on the figures page. It looks like a mountain that has its largest value of $1$ at $x=0$. To make this shape move to the right with speed $2$, we just replace $x$ with $(x-2t)$: \begin{equation} d(x,t) = {1 \over {(x-2t)^2 + 1}} \end{equation} \noindent To see that this shape moves to the right, we can graph $d(x,1)$ the displacement at time $t=1$: \begin{equation} d(x,1) = {1 \over {(x-2)^2 + 1}} \end{equation} \noindent This graph has the same "mountain" shape as the previous one, but is shifted two units to the right. It has a maximum of $1$ at $x=2$, since at $x=2$ the term $(x-2)$ is zero.\\ \noindent Note that the function $f(x)$ can (in principle) be anything. $f(x)$ does not need to be periodic. Periodic traveling waves are a special case, which we consider next.\\ \noindent {\bf Periodic Traveling Waves} \bigskip If the function $f(x)$ is periodic, then the traveling wave is periodic in position $x$. If one takes a "snapshot" of the wave at any time $t$, the disturbance repeats as $x$ is increased (or decreased). The {\bf shortest distance that repeats is called the wavelength} and is usually given the symbol $\lambda$. An interesting property of traveling waves that are periodic in $x$ (at a fixed time $t$), is that they are also periodic in time for a fixed position $x$. $d(x,t)$ for $x$ fixed oscillates periodically in time. As with the spring and pendulum, we define the number of oscillations per unit time as the frequency $f$. Thus, a periodic traveling wave has a definate wavelength $\lambda$ and a definate frequency $f$. These two quantities are related to the wave's velocity. If $f$ cycles occur every second, and for each cycle the wave travels a distance $\lambda$, then the velocity of the wave is equal to $f \lambda$: \begin{equation} v = f \lambda \end{equation} \noindent This equation is universal for periodic traveling waves. It holds for sound waves, water waves, waves on a string, and waves on a spring to name a few. It even is true for non-mechanical waves like electro-magnetic radiation (light). Periodic traveling waves have three quantities associated with them: a velocity, wavelength and frequency. General traveling waves only have a velocity associated with them, since the shape need not be periodic. A periodic traveling wave has some interesting mathematical properties: $d(x \pm \lambda,t) = d(x,t)$ for all times $t$, and $d(x,t \pm T) = d(x,t)$ for all positions $x$ along the wave. These properties define the periodic traveling wave, the wavelength $\lambda$, and the period $T$ of the wave. Combining the two equations gives: \begin{equation} d(x \pm \lambda, t) = d(x, t \pm T) \end{equation} \noindent Since a traveling wave can be written in the form $d(x \pm vt)$, the above equation yields: \begin{equation} (x \pm \lambda) - vt = x - v(t \pm T) \end{equation} \noindent for a wave traveling to the right. The $x$ and $vt$ cancel on both sides, and we are left with \begin{equation} \lambda = vT \end{equation} \noindent using the $+$ sign for $\lambda$ and the $-$ sign for $T$. Since $f=1/T$ we have \begin{equation} v = f \lambda \end{equation} \bigskip Before we talk about a specific kind of periodic traveling wave, the sinusoidal traveling wave, let's discuss some general properties of traveling waves. \bigskip \noindent {\bf Longitudinal versus Transverse Disturbances} \bigskip The velocity of a traveling wave will have a certain direction. Let's suppose it is traveling in the +x direction. The displacement of the medium can be in any direction. It can be in the x-, the y-, or the z-direction, or a combination of the three directions. We have special names if the displacement is in the direction of the velocity (x-direction) or perpendicular to the direction of the velocity (y- or z-direction).\\ \noindent {\bf Longitudinal Waves} are traveling waves whose displacements are in the direction of the velocity.\\ \noindent {\bf Transverse Waves} are traveling waves whose displacements are perpendicular to the direction of the velocity.\\ \noindent Some waves are a combination of both longitudinal and transverse displacements. Let's list some of these properties for some common waves: \begin{center} \begin{tabular}{c||c|c|c|c} Property & string & sound & surface water & E-M (light)\\ \hline medium: & string & gas, liquid, solid & water & NONE (space and time)\\ \hline type: & transverse & longitudinal & combination of both & transverse in free space\\ \hline speed & $\sqrt{F/ \mu}$ & $\sqrt{B/\rho}$ & depth of water & $3 \times 10^8$m/s for all observers\\ \hline frequency & & pitch & & color for visible light \\ \hline \end{tabular} \end{center} \bigskip \noindent {\bf Developing a Wave Equation} \bigskip We would like to understand the propagation of a traveling wave in terms of some level of fundamental physics. Some questions we would like to answer are: a) how can the disturbance of a medium travel in space? b) what does the speed of the traveling wave depend upon? We will answer these questions for a simple case: transverse displacements on a taut string. There are basically only three parameters that describe the string: its mass, its length, and the tension in the string. One could use unit analysis to get an idea how the mass $m$, length $l$, and tension affect the speed of the wave. We'll use the symbol $F$ for the tension in the string ($T$ is used for period). $F$ has units of Newtons = mass(length)/(time)$^2$. Since we can only use mass and length, we must divide by mass, multiply by length, and take the square root to obtain a velocity: $\sqrt{Fl/m}$. Thus, the speed can only depend on the tension $F$ and the ratio of mass to length $m/l$. The mass per length is a {\bf linear mass density} and is given the symbol $\mu$. So $v \sim \sqrt{F/\mu}$. Unit analysis can only take us this far. We do not know if there is a unitless factor in front of $\sqrt{F/ \mu}$. We need to use more physics. Consider a small segment of the string, which starts at $x \equiv x_1$ and has a length of $\Delta x$ and a mass $\Delta m$. See the figure on the last page. The right side of the segment is at $x + \Delta x \equiv x_2$. From the figure, the force on the segment at $x_2$ equals $F sin \theta_2$ and is away from equilibrium. The force on the segment at $x_1$ equals $F sin \theta_1$ and is towards equilibrium. The net force on the segment is therefore $F_{net} = F sin \theta_1 - F sin \theta_2$. If the angle is small, $sin \theta \approx tan \theta$. But, $tan \theta$ is equal to the derivative of $d$ with respect to $x$. So the net force on the segment is \begin{equation} Net \; Force \approx F {{d"d"} \over {dx}}|_{x+\Delta x} - F {{d"d"} \over {dx}}|_x \end{equation} \noindent Equating Net Force to mass times acceleration gives: \begin{equation} \Delta m {{d^2"d"} \over {dt^2}} = F {{d"d"} \over {dx}}|_{x+\Delta x} - F {{d"d"} \over {dx}}|_x \end{equation} \noindent Dividing by $\Delta x$ on both sides gives \begin{equation} {{\Delta m} \over {\Delta x}} {{d^2"d"} \over {dt^2}} = F {{{{d"d"} \over {dx}}|_{x+\Delta x} - {{d"d"} \over {dx}}|_x} \over {\Delta x}} \end{equation} \noindent Taking the limit as $\Delta x \rightarrow 0$, the equation becomes \begin{equation} {{d^2"d"} \over {dt^2}} = {F \over \mu} {{d^2 "d"} \over {dx^2}} \end{equation} This equation is called a "wave equation". It contains derivatives in space and time for the function $d(x,t)$ This looks like a difficult differential equation to solve, but surprizingly, the general solution is fairly simple. The solution can be written in the form: $d(x,t) = g(u)$ where $u=x \pm vt$ and $g(u)$ is any function of $u$ that has a non-zero second derivative with respect to $u$. The "-" produces a disturbance (or wave) traveling to the right, and a "+" is a wave traveling to the left. Let's show why $g(u)$ is a solution. I will let $g' \equiv dg/du$ and $g'' \equiv d^2g/du^2$. Consider the case for which $u=x-vt$. $d"d"/dx$ is given by \begin{equation} {{d"d"} \over {dx}} = {{dg} \over {du}} {{du} \over {dx}}= g' \end{equation} \noindent using the chain rule. The second derivative is determined in the same way: \begin{equation} {{d^2"d"} \over {dx^2}} = {{dg'} \over {du}} {{du} \over {dx}}= g'' \end{equation} \noindent We can use the chain rule in a similar way with the time derivatives: \begin{equation} {{d"d"} \over {dt}} = {{dg} \over {du}} {{du} \over {dt}}= (-v) g' \end{equation} \noindent The $-v$ comes from differentiating $u=x-vt$ with respect to $t$. Taking the second derivative with respect to time gives: \begin{equation} {{d^2"d"} \over {dt^2}} = {{dg'} \over {du}} {{du} \over {dt}}= v^2 g'' \end{equation} \noindent Comparing the second derivative equations for $x$ and $t$, we see that \begin{equation} {{d^2 "d"} \over {dt^2}} = v^2 {{d^2 "d"} \over {dx^2}} \end{equation} \noindent Thus, any function of the form $d=g(x-vt)$ which has non-zero second derivatives will satisfy the wave equation, with the velocity of the wave equal to $v = \sqrt{F/ \mu}$.\\ Let's consider a specific example to better understand this concept. Since $g(u)$ can be anything, let's let $g(u)=u^3$. This means that $d(x,t)$ is $(x-vt)^3$: \begin{equation} d(x,t) = (x-vt)^3 \end{equation} \noindent If we take the second derivative of $d$ with respect to x we get $6(x-vt)$ using the chain rule. The second derivative of $d$ with respect to $t$ is $6v^2(x-vt)$. Thus, \begin{equation} {{d^2 "d"} \over {dt^2}} = 6 v^2 (x-vt) \end{equation} \noindent However, $6 (x-vt) = d^"d"/dx^2$. Substituting this into the above equation, we have \begin{equation} {{d^2 "d"} \over {dt^2}} = v^2 {{d^2 "d"} \over {dx^2}} \end{equation} \noindent Thus, $d(x,t) = (x-vt)^3$ is a solution to the wave equation with $v=\sqrt{F/ \mu}$. We would have had the same result if we had used any function $g(u)$ with $u=x-vt$.\\ This is a nice result and deserves some comments:\\ \noindent 1. We applied Newton's law of motion to the segment of string, with the resulting motion that of a traveling wave: if a pulse is formed on the string, it keeps its shape and travels down the string. It is not obvious that this should occur. What was the key property of the medium that lead to a "fixed shape" traveling down the string? From the wave equation, we see that it was important that the second derivative of $d$ in space is proportional to the second derivative in time. The second derivative in time is the acceleration of the segment which is proportional to the force. So the key property is that the force on a segment of the medium is proportional to the second derivative of $d$. We shall see this same property when we analyze a sound wave.\\ \noindent 2. The velocity only depends on the tension $F$ and the linear mass density $\mu$ of the string. {\bf The velocity does not depend on the frequency of the disturbance}. If a medium has this property it is called a {\bf non-dispersive medium}. In a non-dispersive medium a pulse can travel down the medium without its shape changing. $f$ times $\lambda$ equals $v$, which is the same for all frequencies, as long as the tension and density do not change. Thus if the frequency increases, the wavelength decreases with the product being constant. Air is non-dispersive for sound waves. This is fortunate, because if the speed of sound depended on the pitch of the note, you wouldn't want to listen to music.\\ \bigskip \noindent {\bf Sinusoidal Traveling Waves} \bigskip Sinusoidal traveling waves are traveling waves that have a perfect sinusoidal function. Their shape is given by $d=A sin(2 \pi u/\lambda)$ where $\lambda$ is the wavelength and $A$ is the amplitude of the wave. If they are traveling to the right, then $u=x-vt$. If they are traveling to the left, then $u=x+vt$. Substituing these expressions for $u$ into the sinusoidal function we have \begin{equation} d(x,t) = A sin({{2 \pi} \over \lambda}(x \pm vt)) \end{equation} \noindent where the + (-) refers to waves traveling to the left (right). Since $v/\lambda=f$, this can be written as: \begin{equation} d(x,t) = A sin({{2 \pi} \over \lambda}x \pm 2 \pi f t)) \end{equation} \noindent The expression can be made simplier if we define $\omega \equiv 2 \pi f$ and $k \equiv 2 \pi / \lambda$. k is called the {\bf wave number}. \begin{equation} d(x,t) = A sin(kx \pm \omega t) \end{equation} \noindent. If $d(0,0)$ is not zero, then we need an additional constant to match the initial conditions. As in the case of simple harmonic motion in one dimension, we use a phase $\phi$ to "put" the sine function in the correct position at $t=0$: \begin{equation} d(x,t) = A sin(kx \pm \omega t + \phi) \end{equation} \noindent The "-" corresponds to waves traveling to the right (+x direction) and the "+" to waves traveling to the left (-x direction). \bigskip \noindent {\bf Sound Waves} \bigskip The derivation of the wave equation for sound waves is similiar to the string. In the case of sound waves the speed is equal to $v_{sound} = \sqrt{B/\rho}$, where $B$ is the bulk modulus and $\rho$ is the density of the medium. The speed of sound is quite different in gases, liquids, and solids. For example, in air $v_{sound} \approx 340$ m/s, in water $v_{sound} \approx 1500$ m/s, and in solids it can be as large as $5000$ m/s. Of the many topics we could cover on sound, we will discuss two: the Doppler effect and Sound Level Intensity.\\ \noindent {\bf The Doppler Effect for Sound Waves}\\ Suppose there is a source of sound which emits waves of a definate frequency $f_s$, with the $s$ standing for source. If the source is moving in the medium, the crests of the sound waves will be closer together in front of the source and further apart behind the source than if the source were at rest in the medium. An observer stationary in the medium will hear a frequency different than $f_s$. This same effect will occur if the source is at rest in the medium and the observer is moving. A change in the frequency of the observed sound from $f_s$ is called the Doppler effect, named after Doppler who first observed it.\\ \noindent {\bf Moving Source}\\ Let $f_s$ be the frequency of the source of the sound and $f_o$ be the frequency observed by an observer at rest in the medium. Let $v_s$ be the velocity of the source with respect to the medium, and $v$ the velocity of sound in the medium. Suppose the source is moving towards an observer who is stationary in the medium. The frequency that the observer will hear will be higher than $f_s$. We will derive in class that: \begin{equation} f_o = {f_s \over {1 - v_s/v}} \end{equation} \noindent The minus sign in the denominator makes $f_o > f_s$. If the source is moving away from the observer the result is \begin{equation} f_o = {f_s \over {1 + v_s/v}} \end{equation} \noindent and $f_o < f_s$. Both of these equations can be combined into one: \begin{equation} f_o = {f_s \over {1 \pm v_s/v}} \end{equation} \noindent The key idea for deriving this equation is to note that the wavelength of the sound in the medium equals the "normal" wavelength minus $v_s T$ where $T$ is the period of the wave.\\ \noindent {\bf Moving Observer}\\ Suppose the source is at rest in the medium and an observer is moving towards the source. The observer will observe a higher frequency than $f_s$. Let $v_o$ be the velocity of the observer. It will be shown in lecture that \begin{equation} f_o = f_s (1 + v_o/v) \end{equation} \noindent for an observer moving towards a source at rest in the medium. If the observer is moving away from the source \begin{equation} f_o = f_s (1 - v_o/v) \end{equation} \noindent In this case $f_o < f_s$. These two equations can be combined into one: \begin{equation} f_o = f_s (1 \pm v_o/v) \end{equation} \bigskip \noindent {\bf General Doppler Shift Equation} \bigskip Both situations of moving source and observe can be expressed in one equation: \begin{equation} f_o = f_s {{1 \pm v_o/v} \over {1 \pm v_s/v}} \end{equation} \noindent The key to using this equation properly is to use all variables {\bf in the reference frame of the medium}. That is, in the frame with the medium at rest. When determining $f_o$, the frequency heard by the observer, I substitute into the equation for $f_s$, the frequency of the source, $v_o/v$, and $v_s/v$ without assigning a "+" or "-" in the numerator or demonimator. Then I think:\\ \noindent a) If the {\it observer is moving} towards the source, $f_o$ should increase $\rightarrow$ a "+" in the numerator. If the {\it observer is moving} away from the source, $f_o$ should decrease $\rightarrow$ a "-" in the numerator.\\ \noindent b) If the {\it source is moving} towards the observer, $f_o$ should increase $\rightarrow$ a "-" in the denominator. If the {\it source is moving} away from the observer, $f_o$ should decrease $\rightarrow$ a "+" in the denominator.\\ \noindent If both are moving with respect to the medium, then apply both a) and b). above. \bigskip \noindent {\bf Sound Intensity} \bigskip Sound intensity, $I$, is a measure of the Energy per Area per sec of the sound wave. The metric units of sound intensity is Joules/m$^2$/sec. The human ear is rather amazing. It can detect sound intensity as small as $10^{-12}$ W/m$^2$ and as large as $1$ W/m$^2$. That is $12$ orders of magnitude!! For this reason it is useful to have a different way to assign a number to sound intensity. It is called the sound level and the unit given is the decibel. The sound level in dB is defined in terms of the sound intensity $I$ as: \begin{equation} Sound \; Level \;(dB) \; \equiv 10 log({I \over {10^{-12}}}) \end{equation} \noindent where the sound intensity $I$ is in units of W/m$^2$, and the log is in base $10$. The table below gives some examples of sound levels for various everyday activities: \begin{center} \begin{tabular}{c|c|c|c|c} Activity & $I$ W/m$^2$ & $I/10^{-12}$ & $log(I/10^{-12}) $ & dB \\ \hline Threshold & $10^{-12}$ & $1$ & 0 & 0\\ of hearing & & & & \\ \hline Whisper at 1 Meter & $10^{-10}$ & $10^2$ & 2 & 20 \\ \hline Classroom & $10^{-7}$ & $10^5$ & 5 & 50 \\ \hline Jackhammer & $10^{-3}$ & $10^9$ & 9 & 90 \\ \hline Rock Band & $10^{-1}$ & $10^{11}$ & 11 & 110 \\ \hline Pain & $1$ & $10^{12}$ & 12 & 120 \\ \hline \end{tabular} \end{center} \bigskip \noindent {\bf Inverse Square Law} \bigskip Usually sound sources produce a certain amount of sound energy per unit time. That is their output is usually measured in terms of their power. Power is energy per unit time, with the metric unit being Watts. One Watt equals one Joule/sec. In general, the further away one is from a source of sound, the smaller is the sound intensity. If the source of sound is a "small" source and the waves travel away equally in three dimensions the sound intensity is easy to determine. If the observer is a distance $r$ away from the source, the energy distributes itself equally on a spherical surface of radius $r$. Thus, the sound intensity from a source of power $P$ is just: \begin{equation} I = {P \over {4 \pi r^2}} \end{equation} \noindent If one moves twice as far away from a sound source, the intensity decreases by a factor of $2^2$ or $4$. \bigskip \noindent {\bf Superposition Principle} \bigskip Suppose there are two or more sources of mechanical waves. What is the net displacement of the medium at a particular position? As verified by experiment, the net displacement is the sum of the displacements from the individual sources. This property of nature is called the superposition principle. Simply stated it is:\\ \noindent {\bf When two or more waves are present simultaneously at the same plece, the resulting displacement is the sum of the displacements from the individual sources.}\\ In this course, we will consider the case when there are only two sources, and each source emits pure sinusoidal waves of equal amplitude. \bigskip \noindent {\bf Two Sources of Sinusoidal Waves} \bigskip For two sources of waves, the net displacement will be the sum of the displacements from the two sources: $d_{net} = d_1 + d_2$. If the sources emit sinusoidal waves, $d_{net}$ will be the sum of two sinusoidal waves. We will need to use the formula for the sum of two sin functions with equal amplitude: \begin{equation} sin(\alpha) + sin(\beta) = 2 sin({{\alpha + \beta} \over 2}) cos({{\alpha - \beta} \over 2}) \end{equation} \noindent In this course we will consider three different cases with two sources: a) two sources separated in space, b) two sources of slightly different frequency, and c) standing waves. \bigskip \noindent {\bf Two Sources separated in Space with the Same Frequency} \bigskip Suppose there are two sources of sinusoidal waves, each having the {\bf same frequency} and in phase with each other. Let $D$ be the separation of the two sources. The displacement at time $t$, and at a position located a distance $r_1$ away from the first source is: \begin{equation} d_1(r_1,t) = A(r_1) sin(kr_1 - \omega t) \end{equation} \noindent where the amplitude $A(r)$ will decrease with distance from the source. Similarly, the displacement due to the second source is \begin{equation} d_2(r_2,t) = A(r_2) sin(kr_2 - \omega t) \end{equation} \noindent where $r_2$ is the distance the observer is away from source 2. If an observer receives both sources, the net displacement at the position of the observer is the sum of the displacements due to each source separately (Superposition Principle): \begin{equation} d_{net} = d_1 + d_2 \end{equation} \noindent In general, this sum can be complicated. However, lets assume that $A(r_1)$ and $A(r_2)$ are approximately equal: $A(r_1) \approx A(r_2) \equiv A$. This is a good approximation if $D<