#$&*
course phy122
2/2411
013. `Query 11
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Question: `query introductory set six, problems 15-18
How do we determine the energy of a traveling wave given the amplitude and frequency of the wave, and the mass of the segment of string in question?
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Your solution:
#15
A standing wave in a string 8 meters long has a fundamental harmonic with frequency 100 Hz. and amplitude .19 meters. At t = 0 the string is at its equilibrium position and moving in the positive direction. If the string is vibrating in its fundamental mode, give the equation of motion for a particle at a point .64 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.
As the string ocsillates it will have different shapes at different times.
Give the function which describes the shape at t = .0008333 seconds, at the instant the maximum-amplitude point is halfway to its maximum positive displacement from equilibrium, at the instant each point reaches its maximum negative displacement, at the instant when the maximum displacement of any point on the string is .25 * A, where A is the amplitude of the wave, at the instant when the maximum displacement of any point on the string is r * A and at the instant when the time is t. Describe the graph of each function.
Repeat the above for the first overtone (i.e., the second harmonic) of the wave, given that it has amplitude .0627 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.
What would be the resulting displacement at position x at time t, for the fundamental and first overtone specified above, with respective amplitudes .19 m and .0627 m, if we assume that the two oscillations are in phase and the string is in its equilibrium position at t=0?
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 .14 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 .55 and whose first three overtones have amplitudes .34, .23 and .17 m, with assumptions analogous to those made in the previous question.
Solution (copied from introset page)
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:
yMax = A sin( kx ),
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 8 meters apart; since the wavelength is double the distance between nodes, k must therefore give us a wavelength of
wavelength = 2* 8 m = 16 m.
It follows that k ( 16 m) must correspond to a complete cycle of the reference circle, so that
k ( 16 m) = 2 `pi ,
with the result that
k = 2 `pi /( 16 m) = .3925 /m.
Since the amplitude is .19 meters, at maximum amplitude a position x is given by the function
yMax = .19 m * sin( .3925/m (x) )
which describes wave positions when the wave is at its maximum extension.
A point .64 meter from one of the ends will therefore have
yMax = .19 m * sin( .3925/m ( .64 m) ) = .04721 m.
Its motion will thus be SHM with amplitude .04721 m and frequency 100 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 100 Hz (2 `pi rad) = 200 `pi rad/s by `omega, the displacement y of the point will be given by the equation of motion
y = yMax sin( `omega*t ) = .04721 m sin( 200 `pi rad/s * t).
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( .19 meters) = .095 m. The SHM of this point will therefore be
y = .095 m sin( 200 `pi rad/s * t).
Note that this same result could have been obtained by observing that the maximum-amplitude point is at the x = 4 meter position, so that 1/3 of this distance is at x = 1.333 meters. Substituting x = 1.333 meters into yMax= amplitude m * sin( .3925 m (x)) we obtain the same value, .095 m, for yMax.
At the maximum-amplitude point, the amplitude of motion is of course .19 m, so yMax = .19 m and
y = .19 m sin( 200 `pi rad/s * t).
The maximum-amplitude point (or antinode point), as noted above, thus has time dependence that can be expressed in function notation as
yAntinode(t) = .19 m sin( 200 `pi *t).
This function determines how the wave changes in time:
At t = 0, when yAntinode(t) = 0, the string is at equilibrium at its maximum-amplitude point, or antinode, and is therefore straight.
At a later time (specifically, when 200 `pi *t = `pi /2; you can easily solve for t if you wish to determine the exact time) the displacement of the antinode point becomes .19 meters, and the string forms a sine wave between 0 displacement at its ends and the .19 meter displacement of the antinode point. Between t = 0 and this time the string's antinode point went through 1/4 cycle of SHM, with the string shape following sine waves of corresponding amplitudes.
The antinode point continues following the SHM defined by y = .19 m sin( 200 `pi rad/s * t), with the rest of the wave following along.
The string shape is governed throughout by the function sin(kx) = sin( .3925/m * x); when the amplitude at the antinode is yAntinode(t), the function for displacement y at position x is
y = yAntinode(t) sin( .3925/m * x).
At t = .0008333, yAntinode(t) = .19 m * sin( 200 `pi * .0008333) = .19 m sin( .5235 `pi ) = .19 m sin( `pi /6) = .19 m ( .4999) = .095 m. Thus we have y vs. x function
y = .09498 m sin( .3925/m * x).
This function and its graph describes a sine wave with amplitude .09498 meters and nodes at x = 0, 8, 16, `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 .09498 m when x = 4 meters, and returns to 0 when x = 8 meters.
When the maximum-amplitude point is halfway to its maximum positive displacement of .19 m, it is at .09498 m, so yAntinode(t) = .09498 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
y = .09498 m sin( .3925/m * x).
This function describes a sine wave with displacement 0 at each node and displacement .09498 m at the antinode.
At maximum negative displacement, yAntinode(t) = -.1901 m, so the desired function is
y = -.1901 m sin( .3925/m * x).
This function describes a sine wave with displacement 0 at each node and displacement -.1901 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
y = yAntinode(t) sin( .3925/m * x).
Since yAntinode(t) = .19 m sin( 200 `pi *t), we have
y(t,x) = [ .19 m sin( 200 `pi *t)] sin( .3925/m * x).
The bracketed factor indicates the changing displacement of the antinode point as it undergoes SHM of amplitude .19 m and frequency 100 Hz. The factor sin( .3925/m * x) ensures that the wave form has nodes at x = 0 and x = 8 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 .25 A, then the displacement vs. position function is
y = .25 A sin( .3925/m * x).
This function indicates a y vs. x sine wave with nodes at 0 and 8 m and an antinode point with displacement .25 A, which can be regarded as .25 of the maximum antinode amplitude A. That is, the string has bowed out .25 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
y = rA sin( 200 `pi *t).
This function indicates a y vs. x sine wave with nodes at 0 and 8 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 8 m and the frequency 200 Hz.
It follows that y = A sin(kx) with
k = 2 `pi / ( 8 m) = .785 / m
while
`omega = 2 `pi rad ( 200 Hz) = 400 `pi rad/sec.
The maximum amplitude at position x is therefore given by
yMax = .0627 m sin( .785 / m * x).
The wave form will now have two antinode points, with equal displacements but on opposite sides of the equilibrium postion.
At x = .64 m, we have
yMax = .0627 m sin( .785/m * .64 m) = .01558 m
Note that the amplitude at the x = .64 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
y = .01558 m sin( 400 `pi * t).
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 = 400 `pi rad/s instead of 200 `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
yAntinode(t) = .0627 m sin( 400 `pi rad/s * t).
Thus for t = .0008333 s,
yAntinode(t) = .0627 m sin( 400 `pi rad/s * .0008333 s) = .0205 m.
The wave form will then be given by
y = .0205 m sin( .785/m * x).
Halfway to the maximum displacement the first antinode point will again be at y = .03135 meters; the second antinode point will at that time be at displacement -.03136 meters. The corresponding wave form is described by
y = .03135 m sin( .785/m * x).
y = - .03135 m sin( .785/m * x) at the second.
When the first antinode point is at maximum negative displacement, the waveform is
y = -.1901 m sin( .785/m * x).
The second antinode point will be on the opposite side of the equilibrium line, at
y = .19 m sin( .785/m * x).
At time t the displacement at position x is
y = [ .19 m sin( 400 `pi t) ] (sin( .785/m * x)).
When the amplitude is .25 A, and when it is r A, the displacements will be
y = .25 A sin( .785/m * x)) and
y = r A sin( .785/m * x)).
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
yFundamental = .19 sin( 200 `pi t) sin( .3925 /m * x)
and the first overtone
yFirstOvertone = .0627 sin( 400 `pi t) sin( .785/m * x).
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
y = yFundamental + yFirstOvertone
= .19 sin( 200 `pi t) sin( .3925 /m * x) + .0627 sin( 400 `pi t) sin( .785 /m * x).
The motion corresponding to this function will consist of a fundamental half-wavelength with amplitude .19 m and frequency 100 Hz, on which is superimposed a full wavelength with amplitude .0627 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.
Note that the time dependence in each case is of the form A sin(`omega * t), because of the assumptions made regarding the initial phases of the oscillation (i.e., that they are identical). This is not necessarily the case--in fact with most oscillating objects is not the case, since each of the two harmonics may be started at any desired instant. A more general oscillation is therefore possible with equation y = .19 sin( 200 `pi t + `phi1) sin( .3925 /m * x) + .0627 sin( 400 `pi t + `phi2) sin( .785 /m * x), where `phi1 and `phi2 can be any initial angular position on a reference circle.
If the amplitudes of the fundamental and first harmonic are respectively .1 m and .14 m, then the combined wave has the form and time dependence defined by
y = .1 sin( 200 `pi t) sin( .3925 /m * x) + .14 sin( 400 `pi t) sin( .785 /m * x).
If the amplitudes of the fundamental and the first three harmonics are .42, .33, .28 and .15 m, then the wave is described in position and time by
y = .42 sin( 200 `pi t) sin( .3925/ * x) + .33 sin( 400 `pi t) sin( .785 /m * x) + .28 sin( 600 `pi t) sin( 1.177 /m * x) + .15 sin( 800 `pi t) sin( 1.57 /m * x).
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.
#16
A standing wave in a string 6 meters long has a fundamental harmonic with frequency 70 Hz. and amplitude .28 meters.
If the string is vibrating in its fundamental mode, give the equation of motion for a particle at each of the following positions, assuming that t = 0 when the string is at its equilibrium position and moving in the positive direction:
a point .36 meter from one of the ends
a point 1/3 of the way from an end to the point of maximum amplitude
the point of maximum amplitude.
What function would you graph to show a 'snapshot' of the string at each of the following times, and describe each graph:
t = .00119 seconds
the instant the maximum-amplitude point is halfway to its maximum positive displacement from equilibrium
the instant each point reaches its maximum negative displacement
the instant when the maximum displacement of any point on the string is .26 * A, where A is the amplitude of the wave.
the instant when the maximum displacement of any point on the string is r * A
the instant when the time is t.
Repeat the above for the first overtone (i.e., the second harmonic) of the wave, assuming that its amplitude is .1344 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.
What would be the resulting displacement at position x at time t, for the fundamental and first overtone specified above, with respective amplitudes .28 m and .1344 m, and assuming the two oscillations to be in phase, and the string to be in its equilibrium position at t=0?
Give the function which describes the wave form at time t, assuming a fundamental harmonic with amplitude .11 m and a first overtone with amplitude .06 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 .37 and whose first three overtones have amplitudes .33, .29 and .12 m, with assumptions analogous to those made in the previous question.
Solution (copied from introset page)
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:
yMax = A sin( kx ),
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
wavelength = 2* 6 m = 12 m.
It follows that k ( 12 m) must correspond to a complete cycle of the reference circle, so that
k ( 12 m) = 2 `pi ,
with the result that
k = 2 `pi /( 12 m) = .5233 /m.
Since the amplitude is .28 meters, at maximum amplitude a position x is given by the function
yMax = .28 m * sin( .5233/m (x) )
which describes wave positions when the wave is at its maximum extension.
A point .36 meter from one of the ends will therefore have
yMax = .28 m * sin( .5233/m ( .36 m) ) = .05238 m.
Its motion will thus be SHM with amplitude .05238 m and frequency 70 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 70 Hz (2 `pi rad) = 140 `pi rad/s by `omega, the displacement y of the point will be given by the equation of motion
y = yMax sin( `omega*t ) = .05238 m sin( 140 `pi rad/s * t).
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( .28 meters) = .14 m. The SHM of this point will therefore be
y = .14 m sin( 140 `pi rad/s * t).
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, .14 m, for yMax.
At the maximum-amplitude point, the amplitude of motion is of course .28 m, so yMax = .28 m and
y = .28 m sin( 140 `pi rad/s * t).
The maximum-amplitude point (or antinode point), as noted above, thus has time dependence that can be expressed in function notation as
yAntinode(t) = .28 m sin( 140 `pi *t).
This function determines how the wave changes in time:
At t = 0, when yAntinode(t) = 0, the string is at equilibrium at its maximum-amplitude point, or antinode, and is therefore straight.
At a later time (specifically, when 140 `pi *t = `pi /2; you can easily solve for t if you wish to determine the exact time) the displacement of the antinode point becomes .28 meters, and the string forms a sine wave between 0 displacement at its ends and the .28 meter displacement of the antinode point. Between t = 0 and this time the string's antinode point went through 1/4 cycle of SHM, with the string shape following sine waves of corresponding amplitudes.
The antinode point continues following the SHM defined by y = .28 m sin( 140 `pi rad/s * t), with the rest of the wave following along.
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
y = yAntinode(t) sin( .5233/m * x).
At t = .00119, yAntinode(t) = .28 m * sin( 140 `pi * .00119) = .28 m sin( .5233 `pi ) = .28 m sin( `pi /6) = .28 m ( .4997) = .14 m. Thus we have y vs. x function
y = .1399 m sin( .5233/m * x).
This function and its graph describes a sine wave with amplitude .1399 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 .1399 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 .28 m, it is at .1399 m, so yAntinode(t) = .1399 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
y = .1399 m sin( .5233/m * x).
This function describes a sine wave with displacement 0 at each node and displacement .1399 m at the antinode.
At maximum negative displacement, yAntinode(t) = -.2801 m, so the desired function is
y = -.2801 m sin( .5233/m * x).
This function describes a sine wave with displacement 0 at each node and displacement -.2801 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
y = yAntinode(t) sin( .5233/m * x).
Since yAntinode(t) = .28 m sin( 140 `pi *t), we have
y(t,x) = [ .28 m sin( 140 `pi *t)] sin( .5233/m * x).
The bracketed factor indicates the changing displacement of the antinode point as it undergoes SHM of amplitude .28 m and frequency 70 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 .26 A, then the displacement vs. position function is
y = .26 A sin( .5233/m * x).
This function indicates a y vs. x sine wave with nodes at 0 and 6 m and an antinode point with displacement .26 A, which can be regarded as .26 of the maximum antinode amplitude A. That is, the string has bowed out .26 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
y = rA sin( 140 `pi *t).
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 140 Hz.
It follows that y = A sin(kx) with
k = 2 `pi / ( 6 m) = 1.046 / m
while
`omega = 2 `pi rad ( 140 Hz) = 280 `pi rad/sec.
The maximum amplitude at position x is therefore given by
yMax = .1344 m sin( 1.046 / m * x).
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
yMax = .1344 m sin( 1.046/m * .36 m) = .02514 m
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
y = .02514 m sin( 280 `pi * t).
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 = 280 `pi rad/s instead of 140 `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
yAntinode(t) = .1344 m sin( 280 `pi rad/s * t).
Thus for t = .00119 s,
yAntinode(t) = .1344 m sin( 280 `pi rad/s * .00119 s) = .04394 m.
The wave form will then be given by
y = .04394 m sin( 1.046/m * x).
Halfway to the maximum displacement the first antinode point will again be at y = .0672 meters; the second antinode point will at that time be at displacement -.06721 meters. The corresponding wave form is described by
y = .0672 m sin( 1.046/m * x).
y = - .0672 m sin( 1.046/m * x) at the second.
When the first antinode point is at maximum negative displacement, the waveform is
y = -.2801 m sin( 1.046/m * x).
The second antinode point will be on the opposite side of the equilibrium line, at
y = .28 m sin( 1.046/m * x).
At time t the displacement at position x is
y = [ .28 m sin( 280 `pi t) ] (sin( 1.046/m * x)).
When the amplitude is .26 A, and when it is r A, the displacements will be
y = .26 A sin( 1.046/m * x)) and
y = r A sin( 1.046/m * x)).
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
yFundamental = .28 sin( 140 `pi t) sin( .5233 /m * x)
and the first overtone
yFirstOvertone = .1344 sin( 280 `pi t) sin( 1.046/m * x).
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
y = yFundamental + yFirstOvertone
= .28 sin( 140 `pi t) sin( .5233 /m * x) + .1344 sin( 280 `pi t) sin( 1.046 /m * x).
The motion corresponding to this function will consist of a fundamental half-wavelength with amplitude .28 m and frequency 70 Hz, on which is superimposed a full wavelength with amplitude .1344 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.
Note that the time dependence in each case is of the form A sin(`omega * t), because of the assumptions made regarding the initial phases of the oscillation (i.e., that they are identical). This is not necessarily the case--in fact with most oscillating objects is not the case, since each of the two harmonics may be started at any desired instant. A more general oscillation is therefore possible with equation y = .28 sin( 140 `pi t + `phi1) sin( .5233 /m * x) + .1344 sin( 280 `pi t + `phi2) sin( 1.046 /m * x), where `phi1 and `phi2 can be any initial angular position on a reference circle.
If the amplitudes of the fundamental and first harmonic are respectively .11 m and .06 m, then the combined wave has the form and time dependence defined by
y = .11 sin( 140 `pi t) sin( .5233 /m * x) + .06 sin( 280 `pi t) sin( 1.046 /m * x).
If the amplitudes of the fundamental and the first three harmonics are .45, .33, .25 and .12 m, then the wave is described in position and time by
y = .45 sin( 140 `pi t) sin( .5233/ * x) + .33 sin( 280 `pi t) sin( 1.046 /m * x) + .25 sin( 420 `pi t) sin( 1.569 /m * x) + .12 sin( 560 `pi t) sin( 2.093 /m * x).
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.
#17
If a string of mass density 3.9 grams / meter carries a traveling wave with frequency 400 Hz, wave velocity 67 m/s and amplitude .45 meters, how much energy is there in 60 meters of the string? How much energy is there in the string?
College and University Physics: How much power does it take to maintain this wave?
Solution (copied from introset page)
Each particle in the string undergoes SHM with amplitude .45 meters and frequency 400 Hz. From the mathematic of SHM, or simply from the circular model, the maximum velocity of each particle is therefore
vMax = `omega * A = ( 400 cycles/sec * 2 `pi rad / cycle) * .45 meters = 1130 m/s.
You will recall that the total energy of any mass in SHM is constant throughout the motion, with all the energy potential at the extreme points of the motion and all the energy kinetic at the equilibrium point, where the velocity is maximum. Since every particle is in SHM with the same frequency and amplitude, though different particles are at different points and have differing mixes of KE and PE, the total energy of every particle is equal to the KE it has at its maximum velocity. We can therefore state that the total energy of the wave motion in the string is the energy its total mass would have at the maximum velocity:
total energy in wave motion of string = .5 * mass of string * vMax^2,
where vMax is the maximum SHM velocity of a particle.
The total mass of the string is simply 3.9 grams/meter * 60 meters * (1 kg / (1000 grams) ) = .234 kg. We have already found vMax. Thus the total energy of the string is
total energy in wave motion of string = .5 * .234 kg * ( 1130 m/s) ^ 2 = 149397.3 Joules.
The power required to maintain the wave is the energy supplied per unit of time. Each time the wave travels the length of the string, the 149397.3 Joules in the string will pass on beyond the end of the string (perhaps to another string, perhaps into some mechanism that makes use of the energy, perhaps dissipated into some medium--it doesn't matter where it goes, it just goes).
It takes 60 meters / ( 67 meters / sec) = .8955224 sec for the wave to travel the length of the string.
We must therefore supply 149397.3 Joules every .8955224 seconds, so the power required is
power = 149397.3 Joules / ( .8955224 sec) = 166827 Joules / sec = 166827 watts.
University Physics students note: To be completely rigorous we should consider a Riemann sum of the energies over small increments. A typical length increment `dx of the string will have its SHM defined by a sample point xi in the ith interval. Its total energy will be equal to that of the mass 3.9 g/m * `dx in the interval, and will hence be .5( 3.9 g/m * `dx ) * ( 1130 m/s)^2 = 4979910 kg / s^2 * `dx. When this quantity is summed over all intervals the 4979910 kg/s^2 factors out, the remaining increments `dx add up to the length of the string and we obtain 4979910 kg/s^2 * 60 meters = 149397.3 Joules. This result does not depend on the increments `dx, but the approximation that all the mass in `dx moves with the same SHM is not completely valid. However, as the interval length `dx approaches zero, the approximation becomes more and more valid, and becomes exact in the limit
#18
Two strings are attached to the navel ring of a volunteer. The strings are held under identical 5.6 Newtons tensions and have mass density 13.2 grams/meter.
The far ends of the strings are attached to the same simple harmonic oscillator. The oscillator can be driven at 11.78 Hz, 94.29 Hz, 70.72 Hz or 35.36 Hz. If one string is 9 cm longer than the other, then at a given amplitude of oscillation, which frequency would be most likely to cause the volunteer discomfort, and which the least?
Solution
If the peaks of the two waves arrive at the ear of the volunteer simultaneously, whatever discomfort they cause will be maximized, since they will reinforce one another. If a 'peaks' of one wave arrive simultaneously with the 'valleys' of the other, the two will 'cancel out' and cause minimum discomfort.
If the difference in the distances traveled by the two waves is 1, 2, 3, ... complete wavelengths then, since the two start in phase, they will arrive in phase. If the difference in distances is 1/2, 3/2, 5/2, ... complete wavelengths then the peaks of one will arrive along with the valleys of the other.
We calculate the wavelengths corresponding to the four given frequencies:
We first calculate the wave velocity, which is v = `sqrt( T / `mu ), where T is the tension 5.6 N and `mu the mass density 13.2 grams/ meter * (1 kg / 1000 grams). A simple calculation shows that v = 424.2 m/s.
For each frequency f the wavelength is `lambda = v / f, so we obtain wavelengths
`lambda1 = 424.2 m/s / ( 11.78 cycles/sec) = 36.01 meters,
`lambda2 = 424.2 m/s / ( 94.29 cycles/sec) = 4.498 meters,
`lambda3 = 424.2 m/s / ( 70.72 cycles/sec) = 5.998 meters,
`lambda4 = 424.2 m/s / ( 35.36 cycles/sec) = 11.99 meters.
We next calculate the number of wavelengths in the distance 9 meters:
For 11.78 Hz we see that the wave on the longer string lags the other by 9 meters / ( 36.01 meters/cycle) = .2499 cycle(s).
For 94.29 Hz we see that the wave on the longer string lags the other by 9 meters / ( 4.498 meters/cycle) = 2 cycle(s).
For 70.72 Hz we see that the wave on the longer string lags the other by 9 meters / ( 5.998 meters/cycle) = 1.5 cycle(s).
For 35.36 Hz we see that the wave on the longer string lags the other by 9 meters / ( 11.99 meters/cycle) = .7506 cycle(s).
From these results it is clear which frequency gives us a whole number (e.g., 1, 2, 3, ... ) of wavelengths of path difference, which gives an integer plus a half-integer number (e.g, .5, 1.5, 2.5, ...), and which give integer plus or minute quarter-integer numbers (e.g., .25, .75, 1.25, 1.75, ...). The first will reinforce and cause maximum effect on the ear; the second will cancel and cause minimum effect; and the last will cause an intermediate effect.
Generalized Solution (copied from introset page)
If a string of mass density `mu is held at tension T, transverse waves will propagate in the string at velocity v = `sqrt( T / `mu ).
If harmonic traveling waves start out in phase along two such strings of differing length and terminate at the same point, then the wave traveling in the longer string will lag the wave traveling in the shorter string by the length difference `dL, which will also be referred to as the 'path difference'.
If the path difference is a whole number of wavelengths the waves will reinforce, and we will say that they undergo 'constructive interference'.
If the path difference is a whole number of wavelengths plus half a wavelength the waves will cancel one another, and we will say that they under 'destructive interference'.
If the wave is driven by a simple harmonic oscillator at frequency f, then the waves will have wavelength
`lambda = v / f = 1/f * `sqrt(T / `mu).
The number of wavelengths in the path difference is `dL / `lambda. We test to see if this number is a whole number or a whole number plus half, which would correspond to constructive or destructive interference respectively. The closer to these values the more nearly the behavior of the joining waves is, respectively, constructive or destructive.
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Given Solution:
`aSTUDENT ANSWER AND INSTRUCTOR RESPONSE: Energy = 2*pi^2*m*f^2*A^2
INSTRUCTOR RESPONSE:
** You should understand the way we obtain this formula.
We assume that every point of the string in in SHM with amplitude A and frequency f. Since the total energy in SHM is the same as the maximum potential or the max kinetic energy, all we need to do is calculate the max potential energy or kinetic energy of each point on the string and add up the results.
Since we know mass, frequency and amplitude, we see that we can calulate the max kinetic energy we can get the result we desire. Going back to the circular model, we see that frequency f and amplitude A imply reference point speed = circumference / period = circumference * frequency = 2 `pi A f. The oscillator at its maximum speed will match the speed of the reference point, so the maximum KE is .5 m v^2 = .5 m (2 `pi A f)^2 = 2 `pi^2 m f^2 A^2. **
STUDENT QUESTION
I found the equation above in the book and understand, but my answer was based on problem 17.
I dont understand why the solution is different?
INSTRUCTOR RESPONSE
Your solution was in terms of omega; the equation you quote was an intermediate step in the solution process. That equation is accurate, but it doesn't express the result in terms of the quantities given here. The given information didn't include omega, but rather gave the frequency of the oscillation.
So putting everything in terms of f, A and mass m:
omega = 2 pi f, so vMax = omega * A = 2 pi f * A and
total energy = .5 m vMax^2 = .5 m * ( 2 pi f * A)^2,
which when expanded is equal to the expression in the given solution.
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Question: `qIf the ends of two strings are driven in phase by a single simple harmonic oscillator, and if the wave velocities in the strings are identical, but the length of one string exceeds that of the other by a known amount, then how do we determine whether a given frequency will cause the 'far ends' of the strings to oscillate in phase?
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Your solution:
We need to find the wavelength corresponding to the given frequency, which need not be a harmonic frequency. Any frequency will give us a wavelength; any wavelength can be divided into the difference in string lengths to determine whether the extra length is an integer number of wavelengths.
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Given Solution:
`a** the question here is whether the far ends of the strings are at the same phase of motion, which occurs only if their lengths differ by exactly one, two, three, ... wavelengths. So we need to find the wavelength corresponding to the given frequency, which need not be a harmonic frequency. Any frequency will give us a wavelength; any wavelength can be divided into the difference in string lengths to determine whether the extra length is an integer number of wavelengths.
Alternatively, the pulse in the longer string will be 'behind' the pulse in the shorter by the time required to travel the extra length. If we know the frequency we can determine whether this 'time difference' corresponds to a whole number of periods; if so the ends will oscillate in phase **
STUDENT QUESTION:
The pulse of the longer string will take obviously longer than the shorter string but if the
frequency is the same they will be oscillating at the same rate. Im not sure if I truly understand.
INSTRUCTOR RESPONSE:
If the strings are of the same length then, given the specified conditions, their ends will oscillate in phase. When a peak arrives at the end of one string, a peak will arrive simultaneously at the end of the other.
If you trim a little bit off the end of one of the strings, this won't be the case. When a peak arrives at the end of the untrimmed string, the peak will have already passed the end of the trimmed string, which is therefore oscillating ahead of the phase of the end of the untrimmed string.
The ends of both strings will therefore be oscillating with the same frequency, but out of phase.
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Question: Openstax: How many times a minute does a boat bob up and down on ocean waves that have a wavelength of 40.0 m and a propagation speed of 5.00 m/s?
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Your Solution:
40 m/s /(5 s)=8 sec
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Given Solution: Crests of the wave travel at 5 m/s and are separated by one wavelength, or 40 m. Thus a crest will pass the boat every 40 m / s (5 s) = 8 seconds.
Your Self-Critique:
Your Self-Critique Rating:
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Question: Openstax: A car has two horns, one emitting a frequency of 199 Hz and the other emitting a frequency of 203 Hz. What beat frequency do they produce?
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Your Solution:
203-199=4Hz
The beat frequency will therefore be 4 Hz
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Given Solution: A wave with frequency 199 Hz will come in and out of phase with a wave having frequency 203 Hz four times every second. So the beat frequency is 4 Hz.
Your Self-Critique:
Your Self-Critique Rating:
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Question: Openstax: (a) Seismographs measure the arrival times of earthquakes with a precision of 0.100 s. To get the distance to the epicenter of the quake, they compare the arrival times of S- and P-waves, which travel at different speeds. Figure 16.48) If S- and P-waves travel at 4.00 and 7.20 km/s, respectively, in the region considered, how precisely can the distance to the source of the earthquake be determined? (b) Seismic waves from underground detonations of nuclear bombs can be used to locate the test site and detect violations of test bans. Discuss whether your answer to (a) implies a serious limit to such detection. (Note also that the uncertainty is greater if there is an uncertainty in the propagation speeds of the S- and P-waves.)
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Your Solution:
The P wave travels 3.20 km/s faster than the S wave. The time between the arrival of the two waves will be dt=ds
Dt=ds/(3.20km/s)
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Given Solution:
One wave travels 3.20 km/s faster than the other. If the distance to the epicenter is `ds, then the time between the arrival of the two waves will be `dt = `ds / (3.20 km/s).
If `dt is uncertain by 0.10 second, then `ds will be uncertain by 3.20 km/s * 0.10 s = .32 km, or 320 meters.
Your Self-Critique:
I do not understand the uncertainty factor. Can you explain in a little more detail professor? If dt is uncertain by say .20 second, then ds will be uncertain by 3.20km/s*0.20s=0.64 km. Am I understanding this correctly?
Your Self-Critique Rating:
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Yes. At that speed, in .20 second, the wave travels 320 meters. Since the time interval is uncertain by .20 second, the distance traveled during that interval is uncertain by 320 meters.
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Question: `qGeneral College Physics and Principles of Physics 11.38: AM 550-1600 kHz, FM 88-108 mHz. What are the wavelength ranges?
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Your solution:
I am not sure what this problem is asking for. I will look below for further guidance.
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Given Solution:
`a At 3 * 10^8 m/s:
a frequency of 550 kHz = 550 * 10^3 Hz = 5.5 * 10^5 Hz will correspond to a wavelength of 3 * 10^8 m/s / (5.5 * 10^5 cycles / sec) = 545 meters.
a frequency of 1600 kHz = 1.6* 10^6 Hz will correspond to a wavelength of 3 * 10^8 m/s / (1.6 * 10^6 cycles / sec) =187 meters.
The wavelengths for the FM range are calculated similarly.
a frequency of 88.0 mHz= 88.0 * 10^6 Hz = 8.80 * 10^7 Hz will correspond to a wavelength of 3 * 10^8 m/s / (8.80 * 10^7 cycles / sec) = 3.41 meters.
The 108 mHz frequency is calculated similarly and corresponds to a wavelength of 2.78 meters.
STUDENT QUESTION
I dont understand where the v came from? How did you get the velocity to work the problem?
INSTRUCTOR RESPONSE
This is electromagnetic radiation. Its propagation velocity is the speed of light.
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Self-critique (if necessary):
I think that I may understand this but its still a little foggy in my mind. If I am looking at AM frequency the 550 Hz will indicate 550 * 10 ^3 Hz ( I understand that part) I dont however understand where the 3 8 10^8 m/s came from. Can you please explain this in a little more detail?
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This should be problem 11.38 in your text.
In any case the AM wave has frequency between 550 * 10^3 cycles / second and 1600 cycles / second.
The wave is an electromagnetic wave, and electromagnetic waves travel at 3 * 10^8 m/s.
What therefore is the wavelength of the 550 * 10^3 cycle/second wave?
Similar questions would be answers in a similar manner for the other waves.
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Question: `qGeneral College Physics and Principles of Physics 11.52: What are the possible frequencies of a violin string whose fundamental mode vibrates at 440 Hz?
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Your solution:
After looking below for guidance I can see that the second, third, and fourth frequencies can be found by simply multiplying 2, 3, and 4 by the number of Hz:
Second frequency is 2 * 440 Hz= 880 Hz
Third frequency is 3 * 440 Hz= 1320 Hz
Fourth frequency is 4 * 440 Hz= 1760 Hz
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Given Solution:
`aThe fundamental mode for a string fixed at both ends fits half a wavelength onto the string and therefore has a wavelength equal to double its length. The next three harmonics fit 2, 3 and 4 half-wavelengths into the length of the string and so have respectively 2, 3 and 4 times the frequency of the fundamental. So the first 4 harmonics are
fundamental frequency = 440 Hz
First overtone or second harmonic frequency = 2 * 440 Hz = 880 Hz
Second overtone or third harmonic frequency = 3 * 440 Hz = 1320 Hz
Third overtone or fourth harmonic frequency = 4 * 440 Hz = 1760 Hz
STUDENT QUESTION
Where does the 4th come from? Didnt see that in the notes?
INSTRUCTOR RESPONSE
The pattern of the frequencies should be clear from the way the wavelengths are determined. The pattern follows the obvious pattern of the first three harmonics.
The nth harmonic will contain n half-wavelengths in the distance covered by a single half-wavelength of the fundamental. So its frequency will be n times higher than the fundamental. The frequency of the nth harmonic is therefore n times that of the fundamental.
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Question: `qGeneral College Physics Problem: Earthquake intensity is 2.0 * 10^6 J / (m^2 s) at 48 km from the source. What is the intensity at 1 km from the source?
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Your solution:
Intensity= power/ surface area
Intensity at 1 km is (48km/1km)^2=2300
Intensity at 1 km will be 2300*2.0*10^6J *( m^2 s)= 4.6 x 10 ^9 J m^2 s
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Given Solution:
The wave is assumed spherical so its surface area increases as the square of its distance.
Its intensity, which is power / surface area, therefore decreases as the square of the distance.
So the intensity at 1 km will be (48 km / 1 km)^2 = 2300 times as great.
2300 times the original intensity is
intensity at 1 km = 2300 * 2.0 * 10^6 J / (m^2 s) = 4.6 * 10^9 J/(m^2 s).
In symbols, we express the fact that the intensity is inversely proportional to the area by indicating that the ratio of intensities is the reciprocal of the ratio of areas:
I2 / I1 = 1 / (A2 / A1)
Since 1 / (A2 / A1) = A1 / A2 we can express the same thing as follows:
I2 / I1 = A1 / A2
Now the ratio of the areas of two sphere is equal to the square of the ratio of their radii: A1 / A2 = (r1 / r2)^2. It follows that
I2 / I1 = (r1 / r2)^2.
In the present case, then, we easily obtain
I2 = I1 * (r1 / r2)^2
Since r1 = 48 km and r2 = 1 km we get
I2 = I1 * (48 km / (1 km))^2 = 2300 * I1, approx..
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Question: `qAt what rate did energy pass through a 5.0 m^2 area at the 1 km distance?
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Your solution:
4.6 x 10^9 J/(m^2 s) x 5.0 m^2= 2.3 x 10^10 J/s
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Given Solution:
`aThrough a 5 m^2 area the rate of energy passage is therefore 4.6 * 10^9 J / (m^2 s) * 5.0 m^2 = 2.3 * 10^10 J / s, or 23 billion watts.
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&#Good responses. See my notes and let me know if you have questions. &#