\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<