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Analyzing Waves on a String
Michael Fowler 3/26/07
From Newton
Everything there is to know about waves on a uniform string
can be found by applying 
, to one tiny bit of the string. Well, at least this is true of the small
amplitude waves we shall be studying—we’ll be assuming the deviation of the
string from its rest position is small compared with the wavelength of the
waves being studied. This makes the math
simpler, and is an excellent approximation for musical instruments, etc. Having said that, we’ll draw diagrams, like
the one below, with rather large amplitude waves, to show more clearly what’s
going on.
An animated version of this diagram is available here!
Let’s write down for the small length
of string between x and x + Dx in the diagram above.
Taking the string to have mass density 
kg/m, we have 
The forces on the bit of string (neglecting the tiny force
of gravity, air resistance, etc.) are the tensions T at the two ends. The
tension will be uniform in magnitude along the string, but the string curves if
it’s waving, so the two 
vectors at opposite
ends of the bit of string do not quite cancel, this is the net force 
we’re looking for.
Bearing in mind that we’re only interested here in small amplitude waves, we can see from
the diagram (squashing it mentally in the y-direction)
that both vectors will be close
to horizontal, and, since they’re pointing in opposite directions, their
sum—the net force —will be very close to vertical:
The vertical component of the tension at the x + Dx end of the bit of string is 
, where 
is the angle of slope
of the string at that end. This slope is of course just 
, or, more precisely, 
.
However, if the wave amplitude is small, as we’re assuming,
then is small, and we can
take 
, and therefore take the vertical component of the tension
force on the string to be 
. So the total vertical
force from the tensions at the two ends becomes
the equality becoming exact in the limit 
.
At this point, it is necessary to make clear that y is a function of t as well as of x: 
. In this case, the
standard convention for denoting differentiation with respect to one variable
while the other is held constant (which is the case here—we’re looking at the
sum of forces at one instant of time) is to replace 
.
So we should write:
.
The final piece of the puzzle is the acceleration of the bit
of string: in our small amplitude approximation, it’s only moving up and down,
that is, in the y-direction—so the acceleration
is just 
, and canceling Dx between the
mass 
and , gives:
.
This is called the wave equation.
It’s worth looking at this equation to see why it is
equivalent to . Picture the graph , showing the position of the string at the instant t.
At the point x, the
differential 
is the slope of the
string. The second differential, 
, is the rate of change of the slope—in other words, how much
the string is curved at x.
And, it’s this curvature that ensures the ’s at the two ends of a bit of string are pointing along
slightly different directions, and therefore don’t cancel. This force, then,
gives the mass´acceleration
on the right.
Solving the Wave Equation
Now that we’ve derived a wave equation from analyzing the
motion of a tiny piece of string, we must check to see that it is consistent
with our previous assertions about waves, which were based on experiment and
observation. For example, we stated that
a wave traveling down a rope kept its shape, so we could write 
. Does a general
function 
necessarily satisfy
the wave equation? This f is a function of a single variable,
let’s call it 
. On putting it into
the wave equation, we must use the chain rule for differentiation:
and the equation becomes
so the function will always satisfy the wave equation provided
.
All traveling waves move at the same speed—and the speed is determined by the tension and the mass per unit length. We could have figured out the equation for v2 dimensionally, but there would have been an overall arbitrary constant. We need the wave equation to prove that constant is 1.
Incorporating the above result, the equation is often written:
Of course, waves can travel both ways on a string: an
arbitrary function 
is an equally good solution.
The Principle of Superposition
The wave equation has a very important property: if we have two solutions to the equation, then the sum of the two is also a solution to the equation. It’s easy to check this:
.
Any differential equation for which this property holds is
called a linear differential equation: note that 
is also a solution to
the equation if a, b are constants. So you can add
together—superpose—multiples of any two solutions of the wave equation to find
a new function satisfying the equation.
Harmonic Traveling Waves
Imagine that one end of a long taut string is attached to a simple harmonic oscillator, such as a tuning fork—this will send a harmonic wave down the string,
.
The standard notation is
where of course
.
More notation: the wavelength of this traveling wave is 
, and from the form 
, at say 
,
.
At a fixed x, the
string goes up and down with frequency given by 
, so the frequency f
in cycles per second (Hz) is
Now imagine you’re standing at the origin watching the wave go by. You see the string at the origin do a complete up-and-down cycle f times per second. Each time it does this, a whole wavelength of the wave travels by. Suppose that at t = 0 the wave, coming in from the left, has just reached you. Then at t = 1 second, the front of the wave will have traveled f wavelengths past you—so the speed at which the wave is traveling
Energy and Power in a Traveling Harmonic Wave
If we jiggle one end of a string and send a wave down its length, we are obviously supplying energy to the string—for one thing, as the wave moves down, bits of the string begin moving, so there is kinetic energy. And, there’s also potential energy—remember the wave won’t go down at all unless there is tension in the string, and when the string is waving it’s obviously longer than when it’s motionless along the x-axis. This stretching of the string takes work against the tension T equal to force times distance, in this case equal to the force T multiplied by the distance the string has been stretched. (We assume that this increase in length is not sufficient to cause significant increase in T. This is usually ok.)
For the important case of a harmonic wave traveling along a string, we can work out the energy per unit length exactly. We take
If the string has mass per unit length, a
small piece of string of length 
will have mass 
, and moves (vertically) at speed 
, so has kinetic energy 
, from which the kinetic energy of a length of string is
For the harmonic wave 
and since the average value 
, for a continuous harmonic wave the average K.E.
per unit length
To find the average potential energy in a meter of string as the wave moves through, we need to know how much the string is stretched by the wave, and multiply that length increase by the tension T.
Let’s start with a small length of string, and suppose
that the change in y from one end to
the other is 
:
The string (red) is the hypotenuse of this right-angled
triangle, so the amount of stretching 
of this length is how much longer the hypotenuse is than the base .
So
Remembering that we’re only considering small amplitude
waves, 
is going to be small,
so we can expand the square root using the result
to find
To find the total stretching of a unit length of string, we
add all these small stretches, taking the limit of small 
to find
Now, just as for the kinetic energy discussed above, since , the average potential energy per meter of string is
That is to say, the average potential energy is the same as the average kinetic energy. This is a very general result: it is true for all harmonic oscillators (excepting the case of heavy damping).
Finally, the power in a wave traveling down a string is the rate at which it delivers energy at its destination. Adding together the kinetic and potential energy contributions above,
Now, if the wave is traveling at v meters per second, and being totally absorbed at its destination (the end of the string) the energy delivered to that end in one second is all the energy in the last v meters of the string. By definition, this is the power: the energy delivered in joules per second, That is,
Standing Waves from Traveling Waves
An amusing application of the principle of superposition is adding together harmonic traveling waves moving in opposite directions to get a standing wave:
.
You can easily check that 
is a solution to the wave equation (provided , of course) and it is always zero at points x satisfying 
, so for a string of length L, fixed at the two ends, the appropriate k are given by 
.
The longest wavelength standing wave for a string of length L fixed at both ends has wavelength 
, and is termed the fundamental.
The x-dependence
of this wave, sin kx, is clearly 
, so 
The radial frequency of the wave is given by , so 
and the frequency in
cycles per second, or Hz, is
(This is the same as the frequency 
of a traveling wave
having the same wavelength.)
Here’s a realization of the superposition of two traveling waves to form a standing wave using a spreadsheet:
Here the red wave is and moves to the
right, the green 
moves to the left, the black is the sum of the two and its
oscillations stay in place.
But this represents just one instant! To see the full development in time—which you need to do to get real insight into what’s going on—download the spreadsheet from http://galileo.phys.virginia.edu/classes/152.mf1i.spring02/WaveSum.xls ,then click and hold at the end of the slider bar to animate.
Exercise: what do you think the black wave will look like if the red and green have different amplitudes? Try it on the spreadsheet.