If the string is vibrating in its fundamental mode, give the equation of motion for a particle at a point .36 meter from one of the ends, at a point 1/3 of the way from an end to the point of maximum amplitude, and at the point of maximum amplitude, assuming that
At different times the string will have different shapes. Tell what function you would graph to show a 'snapshot' of the string at each of the following times, and describe each graph:
Repeat the above for the first overtone (i.e., the second harmonic) of the wave, given that it has amplitude .06885 m.
Describe how you would use the principle of superposition to construct the graph of the wave at some instant, if you had the graphs of its fundamental and its first overtone.
Give the function which describes the wave form at time t, assuming a fundamental harmonic with amplitude .1 m and a first overtone with amplitude .09 m, again assuming the two oscillations to be in phase and the string to be in its equilibrium position at t=0.
Repeat for the wave whose fundamental harmonic has amplitude .42 and whose first three overtones have amplitudes .4, .22 and .16 m, with assumptions analogous to those made in the previous question.
The shape of the string at its maximum positive displacement is a portion of the 'sine wave' function for displacement y vs. particle position x:
where k is adjusted to the distance between adjacent nodes and A is the maximum displacement of an antinode.
Since the adjacent nodes are at either end of the string, k must be adjusted so that nodes appear 6 meters apart; since the wavelength is double the distance between nodes, k must therefore give us a wavelength of
It follows that k ( 12 m) must correspond to a complete cycle of the reference circle, so that
with the result that
Since the amplitude is .17 meters, at maximum amplitude a position x is given by the function
which describes wave positions when the wave is at its maximum extension.
A point .36 meter from one of the ends will therefore have
Its motion will thus be SHM with amplitude .0318 m and frequency 160 Hz; since when t=0 the point is at equilibrium moving in the positive direction, there will be no phase shift of the sine function. Denoting the angular frequency 160 Hz (2 `pi rad) = 320 `pi rad/s by `omega, the displacement y of the point will be given by the equation of motion
For the point 1/3 of the way to the maximum-amplitude point the phase of the yMax = A sin(kx) function will be 1/3 of the way to `pi /2, `pi/2 being the phase corresponding to maximum amplitude. Thus for the point 1/3 of the way to the maximum-amplitude point, yMax = A sin(1/3 * `pi /2) = .5 A = .5( .17 meters) = .085 m. The SHM of this point will therefore be
Note that this same result could have been obtained by observing that the maximum-amplitude point is at the x = 3 meter position, so that 1/3 of this distance is at x = 1 meters. Substituting x = 1 meters into yMax= amplitude m * sin( .5233 m (x)) we obtain the same value, .085 m, for yMax.
At the maximum-amplitude point, the amplitude of motion is of course .17 m, so yMax = .17 m and
The maximum-amplitude point (or antinode point), as noted above, thus has time dependence that can be expressed in function notation as
This function determines how the wave changes in time:
The string shape is governed throughout by the function sin(kx) = sin( .5233/m * x); when the amplitude at the antinode is yAntinode(t), the function for displacement y at position x is
At t = .0005208, yAntinode(t) = .17 m * sin( 320 `pi * .0005208) = .17 m sin( .5235 `pi ) = .17 m sin( `pi /6) = .17 m ( .4999) = .085 m. Thus we have y vs. x function
This function and its graph describes a sine wave with amplitude .08498 meters and nodes at x = 0, 6, 12, `thirdNode, ... meters. Since the actual wave is confined to the interval from x = 0 to x = 12m, only the first half-cycle (corresponding to the length of the string) is relevant here. This half-cycle begins with displacement y = 0 when x = 0, proceeds through a maximum displacement of .08498 m when x = 3 meters, and returns to 0 when x = 6 meters.
When the maximum-amplitude point is halfway to its maximum positive displacement of .17 m, it is at .08498 m, so yAntinode(t) = .08498 m. It is not necessary to know t in order to determine this fact, though t could be found easily enough. The y vs. x function will therefore be
This function describes a sine wave with displacement 0 at each node and displacement .08498 m at the antinode.
At maximum negative displacement, yAntinode(t) = -.1701 m, so the desired function is
This function describes a sine wave with displacement 0 at each node and displacement -.1701 m at the antinode; note that this wave form is on the opposite side of the equilibrium line from the previous functions.
At time t, the function is
Since yAntinode(t) = .17 m sin( 320 `pi *t), we have
The bracketed factor indicates the changing displacement of the antinode point as it undergoes SHM of amplitude .17 m and frequency 160 Hz. The factor sin( .5233/m * x) ensures that the wave form has nodes at x = 0 and x = 6 m, and an antinode halfway between.
`This function can be thought of as having two variables, position x and time t. This function is the product of a sine function of t, which as we have seen gives the maximum amplitude at a given instant, and a function of x, which through the choice of k controls the spacing of nodes and antinodes.
If the maximum displacement of a point on the string is .2 A, then the displacement vs. position function is
This function indicates a y vs. x sine wave with nodes at 0 and 6 m and an antinode point with displacement .2 A, which can be regarded as .2 of the maximum antinode amplitude A. That is, the string has bowed out .2 of the way to its maximum position.
If the maximum displacement is r * A, indicating a proportion r of the maximum displacement A of the maximum-amplitude point (that is, the maximum displacement observed anywhere and anytime on the string), we have
This function indicates a y vs. x sine wave with nodes at 0 and 6 m and an antinode point with displacement r A, which can be regarded as proportion r of the maximum antinode amplitude A. That is, the string has bowed out a proportion r of the way to its maximum position.
The first overtone differs from the fundamental harmonic in that its wavelength is half as great and its frequency twice as great. That is, the wavelength will be 6 m and the frequency 320 Hz.
It follows that y = A sin(kx) with
while
The maximum amplitude at position x is therefore given by
The wave form will now have two antinode points, with equal displacements but on opposite sides of the equilibrium postion.
At x = .36 m, we have
Note that the amplitude at the x = .36 m point is greater in proportion the the maximum amplitude than for the fundamental; this happens because the antinode occurs closer to the x=0 end in this case, so as x increases the maximum amplitudes must increase more quickly than before.
The resulting SHM will be described by the equation
The argument made previously for the point 1/3 of the way to the maximum amplitude can be repeated almost verbatim, with identical results, except for omega = angular frequency = 640 `pi rad/s instead of 320 `pi rad/s. An analogous statement is true for the maximum-amplitude point.
The first antinode point for the first overtone will have time dependence
Thus for t = .0005208 s,
The wave form will then be given by
Halfway to the maximum displacement the first antinode point will again be at y = .03442 meters; the second antinode point will at that time be at displacement -.03443 meters. The corresponding wave form is described by
When the first antinode point is at maximum negative displacement, the waveform is
The second antinode point will be on the opposite side of the equilibrium line, at
At time t the displacement at position x is
When the amplitude is .2 A, and when it is r A, the displacements will be
Given the graphs of the fundamental and first overtone, the resulting graph is obtained by adding the displacements indicated by these graphs. This process is called `superposition.
In the present example we have the fundamental waveform
and the first overtone
These waveforms give the displacements from equilibrium at position x and time t corresponding to the fundamental and first overtone. The sum of these displacements will be the displacement of the wave defined by these waveforms.
The resulting displacement at time t and position x will therefore be
= .17 sin( 320 `pi t) sin( .5233 /m * x) + .06885 sin( 640 `pi t) sin( 1.046 /m * x).
The motion corresponding to this function will consist of a fundamental half-wavelength with amplitude .17 m and frequency 160 Hz, on which is superimposed a full wavelength with amplitude .06885 meters and twice the frequency of the fundamental. The shorter wave can be seen as oscillating about the longer fundamental with double the frequency of the fundamental.
If the amplitudes of the fundamental and first harmonic are respectively .1 m and .09 m, then the combined wave has the form and time dependence defined by
If the amplitudes of the fundamental and the first three harmonics are .48, .34, .25 and .17 m, then the wave is described in position and time by
This combination of these amplitudes will produce a recognizably different type of sound than any other combination. It is the combination of the amplitudes of the various harmonics that gives different instruments their distinctive sounds. It is the relative amplitudes of the harmonics that make a piano sound different than a flute or a trumpet, and which make different voices recognizable.
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