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course phy 122
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012. `Query 10
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Question: `q**** Query introductory set six, problems 11-13 **** given the length of a string how do we determine the wavelengths of the first few harmonics?
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Your Solution:
# 11
Suppose that you are holding one end of a heavy flexible cable and the other end is attached to a wall. The cable is held at some constant tension. A transverse pulse in the cable requires .5 seconds to travel its length.
With what frequency must you drive the end of the cable to create a wave with nodes at its two ends and only one antinode? Assume that you are driving the end in low-amplitude SHM, which effectively makes your end a node.
What would be the frequency of your SHM if the wave, with nodes at both ends, contained exactly two antinodes?
Answer the same question for a wave consisting of two antinodes.
Answer again for three antinodes, and for four antinodes.
What is the wavelength of each mode of vibration, in terms of the length of the cable (e.g., 1/3 of the length, 5 times the length, etc.)?
What is the ratio sequence corresponding to the sequence of frequencies?
Solution: (copied lengthy answer from introset solution)
To obtain the 2-node standing wave, you must be moving so as to reinforce each pulse when it returns to you. Then you will be in resonance with the standing wave: each of your movements will contribute to the energy of the wave, and the wave's amplitude will increase until it is dissipating energy (through internal friction, work against air resistance, etc.) as fast as you supply it.
When you send a 'positive' pulse down the cable (if you wish, you may regard 'positive' as meaning 'to the right'; your positive pulse would then be created by moving your end to the right; a negative pulse would of course be to the left), it remains positive until it reaches the wall, then it returns as a negative pulse. It is as if there is a 'negative' phantom pulse coming at you from the wall, with its phase precisely matched to your positive pulse so that the two pulses cancel at the wall, with the result that the wall end of the cable remains fixed (as it must).
When the negative pulse again reaches your end, it will tend to reflect in the analogous manner and travel back down the cable as a positive pulse. If you are moving your end in the positive direction at the instant the negative pulse arrives, you will reinforce its natural tendency to reflect as a positive pulse. You will be in resonance, as described above. You need not move the cable much; whatever energy you add pulse by pulse will accumulate in the wave. Your end can therefore be virtually a node.
The time required for the pulse to travel down the cable and back is 2( .5 sec) = 1 sec. If you are originally at equilibrium and you begin moving your end in SHM, initially in the positive direction, then 1 seconds later the pulse will return to you. The pulse will be reinforced if you are at this instant again at the equilibrium position and moving in the positive direction. This pulse will again be reinforced on its next return if at that instant you are once more passing through equilibrium in the positive direction. If your timing is such that this precise reinforcement occurs with every return, you will remain in phase with the wave's natural behavior, and you will achieve the maximum amplitude for the SHM you are producing.
Your SHM must therefore have a period of 1 sec. The corresponding frequency is 1/( 1 sec) = 1 Hz.
We need to understand the relationship between the spacing of nodes and antinodes and wavelength.
The wavelength of a pulse is the distance between its peaks.
Imagine that a very great length was added to your cable, while maintaining the same tension with the same driving SHM. Your disturbance would travel down the cable as a sine wave, with a series of peaks and valleys, and with 'antinodes' in between ('antinodes' is in quotes because antinodes are really defined for standing waves and not for traveling waves; however the meaning of the term is clear and it will be employed here). The wavelength is the distance between peaks. If you sketch a sine wave you will see that there are two 'antinodes' between consecutive peaks.
The wavelength is thus twice the distance between 'antinodes'.
Since the disturbance you send out into your original cable has exactly two antinodes in the length of the cable, the length of the cable must correspond to half of a peak-to-peak distance, or to half of a wavelength. The wavelength must therefore be twice the length of the cable.
The standing wave so produced is called the 'fundamental harmonic' of the cable. The fundamental harmonic of a standing wave can be thought of as the naturally reinforcing mode of oscillation in which the number of nodes is a minimum.
Digression (note for reference in a later problem): Note that to maintain the wave work must be done.
You can feel that when the cable returns to your end, it tends to pull you in the direction of its pulse, pulling in the negative direction when you encounter a negative pulse and in the positive direction when the pulse is positive. Since the pulse tends to reverse itself at your end, the force you exert must be in a direction opposite to that of the pulse, and you must therefore perform work as you apply this force through the SHM cycle.
If rather than using a cable you use a long heavy cable, and use a relatively large amplitude for the SHM, you will come to appreciate the fact that you are indeed exerting a force (you will feel that in your joints and muscles) and that you are doing work (which will manifest itself as the fatigue in your joints and muscles). The cable itself will have considerable kinetic and potential energy due to its motion and its displacement from equilibrium against the effects of the cable's tension.
As with SHM, the kinetic energy is at its maximum when the cable moves through its equilibrium position, and zero when the cable stops for an instant at its extreme position; the potential energy is zero at the equilibrium position and maximized at the extreme position (it requires work to move the cable to this position against the effects of tension); and in the steady state where energy is dissipated at a rate equal to that at which you are doing work, the sum of the kinetic and potential energies is the same in one position as in another.
To create a 3-node wave (which will have 2 antinodes) you must send two positive pulses down the cable during the time required for the initial pulse to return.
In this manner your positive pulse will be followed by a negative pulse in precisely the time required for the pulse to travel the length of the cable. At the instant you begin a new cycle, the reflected cycle of your preceding pulse will be returning to you. These pulses will meet and cancel one another out when they are both halfway to their respective ends of the cable. The timing is just right so that, at the midpoint of the cable, the reflected pulse will at every instant cancel your pulse. The result is a node at the middle point; the standing wave therefore has three nodes, one at each end and one in the middle.
When you create such a wave you will be timing your movements in such a way that the ends of the cable are moving in opposite directions, so you will probably focus on the motion of the half of the cable on the other end and simply move in the opposite direction.
It is therefore clear that you must complete each cycle in precisely the time required for the wave to travel the distance of the cable. The period of your SHM is therefore .5 seconds and its frequency is 1/( .5 sec) = 2 Hz.
This 2-antinode wave has nodes in the middle as well as at the ends. The nodes are separated by half the length of the cable, and this node-to-node distance is half the wavelength. The wavelength is therefore equal to the length of the cable.
Any motion created by timing your SHM in such a way that your cycle is opposite in phase to that of the return cycle will result in the same sort of resonance, with one antinode corresponding to each half wavelength of your cyclic pulse. Thus:
To create a 4-node wave (with 3 antinodes) you will send three positive pulses in the time required for the initial pulse to return.
Thus three cycles must be completed in the 1 seconds required for the round trip, and the period of motion is 1 sec/ 3 = .3333 sec; this corresponds to 1 cycle / ( .3333 sec) = 3 cycles/second; that is, the frequency is 3 Hz.
The cable will have 4 nodes so will have 3 node-to-node distances along its length. This corresponds to 3 half-wavelengths, so 3/2 (wavelength) = cable length and the wavelength is therefore 2/3 the length of the cable.
To create a 5-node wave (with 4 antinodes) you will send four positive pulses in the time required for the initial pulse to return.
Thus four cycles must be completed in the 1 seconds required for the round trip, and the period of motion is 1 sec/ 4 = .25 sec; this correspondence to 1 cycle / ( .25 sec) = 4 cycles/second; that is, the frequency is 4 Hz.
The cable will have 5 nodes so will have 4 node-to-node distances along its length. This corresponds to 4 half-wavelengths, so 4/2 (wavelength) = cable length and the wavelength is therefore 2/4 or the length of the cable.
The frequencies are 1, 2, 3, and 4 Hz. These frequencies form the ratio sequence
2/ 1 = 2
3/ 2 = 3/2
4/ 3 = 4/3.
It is easy to see that the frequency ratios will continue as 5/4, 6/5, ..., n/(n-1), ....
These frequencies and their corresponding wavelengths will arise naturally if the cable is plucked. A random pluck has the potential to give rise to a wide range of frequencies and wavelengths, but only those that 'fit' in the cable will reinforce themselves and after a few trips back and forth along the cable the only wavelengths and frequencies that are still observable will be those of the natural modes of vibration.
`The naturally arising frequencies of a Slinky, a chain, a string, a resonant column of air, or any other vibrating object are called 'harmonics'.
The harmonic with the fewest nodes is the 'fundamental harmonic', and the others are called 'overtones'. The frequency of vibration of the fundamental harmonic is called the 'pitch' of the resonator, and is perceived as a high or a low sound, depending on how high or low the frequency is.
Musical instruments typically reflect certain vibrations of these strings or air columns back to the strings or air columns themselves and hence tend to reinforce those vibrations they reflect. This results in different combinations of the harmonics, and gives each instrument its characteristic sound. The shapes of the resonant cavities in your head, and the structure of your vocal cords and the tension you create in them, give your voice its distinctive character.
#12
A 13-meter length of a Slinky has a mass of .1 kg and is stretched out between two points under a constant tension of 49 Newtons. If both ends are fixed to rigid supports and the Slinky is simultaneously plucked at several points, what are the first three frequencies that you would expect to measure if you performed a Fourier analysis on a set of force vs. time data at one end of the object?
Solution
A Fourier analysis is used to determine the frequencies present in a set of data. For example if a certain harmonic 'jerks' the end of the object 30 times per second, a Fourier analysis will clearly reveal this frequency, even if many other frequencies and a lot of random 'noise' are present.
Any disturbance introduced by the pluck which does not reinforce itself will end up interfering with its own reflections, and will therefore disappear relatively quickly. Only those disturbances which are very close to the natural harmonics of the object will maintain any significant presence much beyond a couple of reflection cycles.
The harmonics for this object are determined by the fact that there must be nodes at the ends, separated by the 13- meter length of the object. The first three harmonics therefore have 2, 3, and 4 nodes in the 13 meters, so they have 1, 2, and 3 half wavelengths in the same distance. So we have:
harmonic 1: 1/2 wavelength = 13 meters, so wavelength = 26 meters
harmonic 2: 2/2 wavelength = 13 meters, so wavelength = 13 meters
harmonic 3: 3/2 wavelength = 13 meters, so wavelength = 8.666 m.
Knowing the wavelengths, we can obtain the frequencies if we know the velocity of wave propagation. This velocity is given by
v = `sqrt( tension / mass per unit length) = `sqrt( 49 N / ( .1 kg / 13 m) ) = `sqrt( 49 N / .007692 kg/m) = 79.81 m/s.
The first wave has wavelength 26 m.
Therefore if the wave moves at 79.81 m/s, a 1-second segment 79.81 meters long would contain 79.81/ 26 = 3.069 wavelengths each 26 meters long, so the frequency would be 3.069 Hz.
Reasoning similarly, we find that the second and third harmonics (these are the first and second overtones) will have frequencies of 6.139 Hz and 9.209 Hz.
#13
What are the frequencies of the first four harmonics of a hole bored into the side of a hill, 9 meters long with its opening clear? Assume that sound travels in the air inside at 347 m/s.
Solution
The longitudinal pressure wave created in the hole is analogous to, but different from the transverse wave created when a string, free at one end, is randomly plucked.
The main difference is that the pressure wave is longitudinal, making the process of wave formation similar to that of the hanging Slinky.
The similarity is that the pressure wave in the hole has an antinode at one end and a node at the other. Since there is a quarter-wavelength between a node and its adjacent antinode, there can be 1, 3, 5, 7, ... quarter wavelengths between the ends of the hole. From this we can obtain the wavelengths of the first four harmonics. Then from the wavelengths and speed of wave propagation we can calculate the frequencies of these harmonics.
Since a full cycle of a wave goes from node to (positive) antinode to node to (negative) antinode to node, the full cycle consists of four node-antinode distances. (Note that we are using the terms 'node' and 'antinode' a bit loosely here, as if they could apply to a traveling wave; these terms really only apply to a standing wave, but the meaning here should be clear).
We reason out the wavelengths of the first four harmonics::
We see that the first harmonic has just a single node-antinode distance in the 9 meters between its ends. This corresponds to 1/4 wavelength, so the wavelength of this harmonic must be 4 * 9 meters = 36 meters.
The second harmonic goes from node to antinode to node to antinode, a distance of 3/4 wavelength. Thus 3/4 wavelength = 9 meters and the wavelength is 4/3( 9 m ) = 12 meters.
Similarly the third and forth harmonics span 5/4 and 7/4 of a wavelength, respectively, and there respective wavelengths are thus 4/5( 9 m) = 7.2 m and (4/7)( 9 m) = 5.142 m.
Since sound moves 347 m in each second, there will be 347/ 36 = 9.638 peaks in each second of the fundamental harmonic. The frequency is therefore 9.638 Hz.
Similarly the three overtones will have frequencies of
( 347 m/s) / ( 12 m) = 28.91 Hz,
( 347 m/s) / ( 7.2 m) = 48.19 Hz and
( 347 m/s) / ( 5.142 m) = 67.48 Hz.
These frequencies could have been found from the fundamental frequency 9.638 Hz and the harmonic ratios 3/1, 5/3 and 7/5 for a standing wave with a node at one end and an antinode at the other.
In this manner we would obtain the corresponding frequencies (3/1)( 9.638 Hz) = 28.91 Hz; (5/3)( 28.91 Hz) = 48.19 Hz and (7/5)( 48.19 Hz) = 67.48 Hz.
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Given Solution: `q** As wavelength decreases you can fit more half-waves onto the string. You can fit one half-wave, or 2 half-waves, or 3, etc..
So you get
1 half-wavelength = string length, or wavelength = 2 * string length; using `lambda to stand for wavelength and L for string length this would be
1 * 1/2 `lambda = L so `lambda = 2 L.
For 2 wavelengths fit into the string you get
2 * 1/2 `lambda = L so `lambda = L.
For 3 wavelengths you get
3 * 1/2 `lambda = L so `lambda = 2/3 L; etc.
Your wavelengths are therefore 2L, L, 2/3 L, 1/2 L, etc..
FOR A STRING FREE AT ONE END:
The wavelengths of the first few harmonics are found by the node - antinode distance between the ends. The first node corresponds to 1/4 wavelength. The second harmonic is from node to antinode to node to antinode, or 4/3. the third and fourth harmonics would therefore be 5/4 and 7/4 respectively. **
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Question: `q**** Given the wavelengths of the first few harmonics and the velocity of a wave disturbance in the string, how do we determine the frequencies of the first few harmonics?
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Your solution:
Frequency is equal to the number of crests passing per unit of time. If we imagine a one second chunk of the wave divided into segments each equal to the wavelength. The number of peaks is equal to the length of the entire chunk divided by the length of a 1 wavelength segment. This is the number of peaks passing per second.
Frequency=wave velocity/wavelength.
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Given Solution:
`a** The frequency is the number of crests passing per unit of time.
We can imagine a 1-second chunk of the wave divided into segments each equal to the wavelength. The number of peaks is equal to the length of the entire chunk divided by the length of a 1-wavelength segment. This is the number of peaks passing per second.
So frequency is equal to the wave velocity divided by the wavelength. **
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Question: `q **** Given the tension and mass density of a string how do we determine the velocity of the wave in the string?
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Your solution:
We divide the tension/mass unit length. V=sqrt (tension/mass length)
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Given Solution:
`a** We divide tension by mass per unit length:
v = sqrt ( tension / (mass/length) ). **
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Question: `q**** gen phy explain in your own words the meaning of the principal of superposition
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Your solution:
When 2 different waveforms meet or are present in a medium, the displacements of the two waveforms are added at each point to create the waveform that will be seen. The text reads When two or more waves are present simultaneously at the same place, the resultant disturbance is the sum of the disturbances from the individual waves.
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Given Solution:
`a** the principle of superposition tells us that when two different waveforms meet, or are present in a medium, the displacements of the two waveforms are added at each point to create the waveform that will be seen. **
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Question: `q **** gen phy what does it mean to say that the angle of reflection is equal to the angle of incidence?
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Your solution:
The angle of incidence with a surface is the angle with the perpendicular to that surface. When a ray comes in at a given angle of incidence, it reflects at an equal angle on the other side of that perpendicular.
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Given Solution:
`a** angle of incidence with a surface is the angle with the perpendicular to that surface; when a ray comes in at a given angle of incidence it reflects at an equal angle on the other side of that perpendicular **
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