Query 11

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course phy122

Your solution, attempt at solution:If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.

011. `Query 10

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Question: **** Query introductory set six, problems 11-14 **** 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 1.3 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 (long lengthy answer copied from introset and pasted to this page)

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( 1.3 sec) = 2.6 sec. If you are originally at equilibrium and you begin moving your end in SHM, initially in the positive direction, then 2.6 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 2.6 sec. The corresponding frequency is 1/( 2.6 sec) = .3846 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 1.3 seconds and its frequency is 1/( 1.3 sec) = .7692 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 2.6 seconds required for the round trip, and the period of motion is 2.6 sec/ 3 = .8666 sec; this corresponds to 1 cycle / ( .8666 sec) = 1.153 cycles/second; that is, the frequency is 1.153 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 2.6 seconds required for the round trip, and the period of motion is 2.6 sec/ 4 = .65 sec; this correspondence to 1 cycle / ( .65 sec) = 1.538 cycles/second; that is, the frequency is 1.538 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 .3846, .7692, 1.153, and 1.538 Hz. These frequencies form the ratio sequence

• .7692/ .3846 = 2

• 1.153/ .7692 = 3/2

• 1.538/ 1.153 = 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 11 meter length of uniform rope is stretched out between two points under constant tension of 27 newtons. If both ends are fixed to rigid supports and the rope is simultaneously plucked at several points, what are the first 3 frequencies that you would expect the system to exhibit? The rope has a mass of 1.6 kg

Solution: a fourier analysis is used to determine the frequencies present in a set of data. For example if certain harmonic jerks at 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 11 meter length of the object. The first three harmonics therefore have 2, 3, and 4 nodes in the 11 meters, so they have 1, 2, and 3 half wavelengths in the same distance.

So we have:

Harmonic one: ½ wavelength=11 meters, so wavelength= 22 meters

Harmonic 2: 2/2/wavelength = 11 meters, so wavelength is 11 meters

Harmonic 3: 3/2 wave length= 11 meters/1.5=7.333

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( 27N/1.6 kg/11m)=sqrt (27N/.1454kg/m)=13.62 m/s

The first wave has a wavelength of 22 m:

Therefore if the wave moves at 13.62 m/s a 1 sec segment 13.62 meters long would contain 13.62/11=.619 Hz

The second wave has a wavelength of 11 meters, therefore if the wave moves 13.62 m/s a 1 sec segment 13.62 meters long would contain 13.62/11= 1.24 wavelengths each 11 m long, so the frequency would be 1.24 hz.

The third wave has a wavelength of 7.333 meters. Therefore if the wave moves at 13.62 m/s at one second segment 13.62 m long would contain 13.62/7.333=1.86 wavelengths each 7.333 long, so the frequency will be 1.86 Hz.

#13: A hollow organ pipe, open at one end, is set into resonant vibrations, by means of a mechanical oscillator at the open end. The hollow cavity of the pipe is 6 m long, sound travels in the air inside at 328 m/s. What are the frequencies of the first 4 harmonics?

Solution:

The longitudinal pressure wave created in the pipe by the blow from the stick is analogous too, 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 pipe 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 pipe. From this we can obtain the wavelengths of the first 4 harmonics. Then from the wavelengths and speed of the wave propagation we can calculate the frequencies of these harmonics.

Since a full cycle of wave goes from a node to an antinode to node, the full cycle consists of 4 node to antinode distances. We reason out the wavelengths of the first four harmonics.

The first harmonic has just a single node to antinode distance in the 6 meters between its ends. This corresponds to ¼ wavelength, so the wavelength of this harmonic must be 4 x 6m= 24 meters

The second harmonic goes from node to antinode to node to antinode, a distance of ¾ a wavlenth. Thus ¾ wavelength= 6 m and the wavelength is 4/3(6m)=8m

Similarly he 3rd and 4th harmonics span 5/4 and 7/4 of a wavelength, respectively and their respective wavelengths are thus 4/5(6m)=4.8 m and 4/7(6m)=3.428m

Since sound moves 328 m in each second, there will be 328/24=13.66 peaks in each second of the fundamental harmonic. The frequency is therefore 13.66 Hz

The 3 overtones will have frequencies of:

328/8m=41 hz

328/4.8=68.3 hz

328/3.428=95.68 hz

In this matter we would obtain the corresponding frequencies:

(3/1)(13.66 hz)=41 hz

(5/3)(41 hz)= 68.33 hz

(7/5)(68.33 hz)=95.68 hz

#14 A piano is tuned so that each key gives a frequency 2^(1/12)=1.0594 times that of the previous. Show that an interval of 7 keys will achieve a ratio of 2/1 (a doubling of frequency, the ratio of an octave)

Determine how many keys we must count up from a given key to come as close as possible to the ratios 4/3 (called a fourth) 5/4 (called a third) and 6/5 (called a minor third)

How close do the ratios based on the ratio 2^ (1/12) come to mimicking the natural ratios of the harmonics of a string?

Use your calculator to find the ratio that would correspond to 38 subdivisions of the 2/1 ratio and determine how nearly the natural ratios could be mimicked using multiples of this ratio.

It turns out that the next number of divisions that actually improves on the 12 is 43. This ratio is used in some eastern music. How much better can we do with a division of the ration 2 into 43 equal subdivisions, as compared to 12 subdivisions?

Solution:

Counting up to 7 keys:

The first will have a ratio 2^(1/12)

The second a ratio of 2^(1/12)= 2^(2/12)= 2^(1/6)

The third ratio of 2^(1/12)= 2^(3/12)= 2^(1/4)

The fourth ratio of 2^(1/12)= 2^(4/12)= 2^(1/3)

The fifth ratio of 2^(1/12)= 2^(5/12)= 2^(5/12)

The sixth ratio of 2^(1/12)= 2^(6/12)= 2^(1/2)

The seventh ratio of 2^(1/12)= 2^(7/12)= 2^(7/12)

Other ratios are similarly calculated:

Counting up 12 keys we could obtain a ratio of 2^(12/12)= 2(1)=2

I am not going to continue going on with the process, I have a good general idea of how this is calculated.

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Given Solution:

** 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 node-antinode distance corresponds to 1/4 wavelength, so the wavelength is 4 times the length of the string.

The second harmonic is from node to antinode to node to antinode, or 3/4 of a wavelength. So 3/4 of this wavelength is equal to the length of the string, and the wavelength is therefore 4/3 the length of the string.

The third and fourth harmonics would therefore be 5/4 and 7/4 the length of the string, respectively. **

STUDENT QUESTION (instructor comments in bold)

In the explanation, I don’t understand why the wavelengths were halved [L = 1 * 1/2(‘lambda)].

As indicated in the given solution, you can fit an even number of half-wavelengths onto a string fixed at both ends.

• If you have a single half-wavelength, then the length of the string is 1/2 wavelength; hence L = 1 * (1/2 lambda).

• If you have two half-wavelengths, then the length of the string is 2 * 1/2 wavelength; hence L = 2 * (1/2 lambda).

• etc.

I get the explanation at the bottom were the 1st harmonic is 1/4 the wavelength and the 2nd is 3/4 the wavelength, etc….. but where does that come into play when determining the actual wavelength. I can’t tell if both of the explanations say the same things, or if it’s a 2-part explanation.

I believe you are referring to the solution for a string which is free at one end. For the string free at one end, the first harmonic isn't 1/4 of the wavelength. The first harmonic has a wavelength, which is related to the length of the string.

• For the first harmonic there is a single node, at one end, and a single antinode, at the other. The length of the string is therefore a single node-antinode distance. Since the node-antinode distance is 1/4 of the wavelength, the length of the string is 1/4 wavelength. (It would follow that the wavelength is 4 times the length of the string).

• For the second harmonic three node-antinode distances are spread along the wave, so the wavelength is 4/3 the length of the string, as indicated in the given solution.

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Question: **** 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:

The frequency is the number of crests passing per unit of time.

We can imagine a 1 sec 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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Given Solution:

** 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: **** 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 tension by mass per unit length. V=sqrt(tension/ (mass/length)

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Given Solution:

** We divide tension by mass per unit length:

v = sqrt ( tension / (mass/length) ). **

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Question: **** gen phy explain in your own words the meaning of the principal of superposition

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Your Solution:

This principle tells us that 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.

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Given Solution:

** 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: **** 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:

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:

** 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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Question: query univ phy problem 15.52 / 15.50 11th edition 15.48 (19.32 10th edition) y(x,t) = .75 cm sin[ `pi ( 250 s^-1 t + .4 cm^-1 x) ] What are the amplitude, period, frequency, wavelength and speed of propagation?

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Given Solution:

** y(x, t) = A sin( omega * t + k * x), where amplitude is A, frequency is omega / (2 pi), wavelength is 2 pi / k and velocity of propagation is frequency * wavelength. Period is the reciprocal of frequency.

For A = .75 cm, omega = 250 pi s^-1, k = .4 pi cm^-1 we have

A=.750 cm

frequency is f = 250 pi s^-1 / (2 pi) = 125 s^-1 = 125 Hz.

period is T = 1/f = 1 / (125 s^-1) = .008 s

wavelength is lambda = (2 pi / (.4 * pi cm^-1)) = 5 cm

speed of propagation is v = frequency * wavelength = 125 Hz * 5 cm = 625 cm/s.

Note that v = freq * wavelength = omega / (2 pi) * ( 2 pi ) / k = omega / k. **

STUDENT COMMENT:

2*pi is one full cycle, but since the function is cos, everything is multiplied by pi. So does this mean that the cos function only represents a half a cycle?

It's late and I might well be missing something, but I don't think the cosine is mentioned in this problem. Let me know if I'm wrong.
I believe that only the sine function is mentioned. However there isn't much difference between the two (only a phase difference of pi/2) and everything below would apply to either:
pi is there because of the periodicity of the sine and cosine functions, as described below.
A sine or cosine function completes a full cycle as its argument changes by 2 pi.
The argument might be a function of clock time, or of position.
If the argument is of the form omega * t, then a period is completed every time omega * t changes by 2 pi. This occurs when t changes by 2 pi / omega. So the period of the function is 2 pi / omega.
If the argument is of the form k x, then it changes by 2 pi every time x changes by 2 pi / k, so the wavelength of the function is 2 pi / k.

STUDENT COMMENT

the book has this as position = Acos(k*x+ -omega*t)

and velocity omega*A*sin(k*x-omega*t)

this is no big deal, they mean the same as the student sort of mentions, that the sine is 2pi shifted which is in the wave number.

INSTRUCTOR RESPONSE

It doesn't make a lot of difference. The sine or cosine is a valid function to use, and whether you use k x - omega t, kx + omega t, -kx + omega t or -kx - omega t you get the equation of a traveling harmonic wave that satisfied the wave equation y_tt = 1/c * y_xx, where for example _tt represents second derivative with respect to t, and c represents the propagation velocity of the wave.

The general form of a traveling harmonic wave can be taken as

y(x, t) = A cos(k x - omega t + phi)

where phi is the initial phase. Any initial phase is possible. If phi = -pi/2 then since sin(theta) = cos(theta - pi/2) the equation could be

y(x, t) = A sin(kx - omega t).

If phi = 3 pi / 2 then since sin(theta) = - sin(theta + pi) = sin(-theta) we have

y(x, t) = A sin(omega t - k x).

Recall that the graph of y = f( x - h) is identical in shape to the graph of y = f(x), but translated h units in the horizontal direction.

Now kx - omega t can be written as k ( x - omega / k * t), and f(k ( x - omega / k * t) ) can therefore be seen as a horizontal translation of the graph of y = f(kx), the amount of the translation being omega / k * t . This translation is positive, so as t increases the horizontal translation increases, and the graph progresses to the right. The rate of progression with respect to t is omega / k, which is therefore the speed of propagation.

Thus the argument k x - omega t guarantees that the function y = f( k x - omega t) represents an unchanging shape that moves to the right with speed omega / k.

If f is a sine or cosine function, or any superposition of sine and cosine functions, we have a harmonic wave.

Another interpretation is that k x - omega t = -omega ( t - k / omega * x). In this case k / omega * x is regarded as the time required for the wave to propagate from position 0 to position x. The function f(-omega * t) that describes the motion of the point x = 0 in time is translated in space, so that the same motion occurs at position x after a 'time lag' of k / omega * x.

Similar analysis shows that a function of the form y = f( omega t + k x) represents a wave traveling in the negative x direction.

You'll probably have more questions, which I'll welcome.

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Question: **** Describe your sketch for t = 0 and state how the shapes differ at t = .0005 and t = .0010.

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Your Solution:

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Given Solution:

** Basic precalculus: For any function f(x) the graph of f(x-h) is translated `dx = h units in the x direction from the graph of y = f(x).

The graph of y = sin(k * x - omega * t) = sin(k * ( x - omega / k * t) ) is translated thru displacement `dx = omega / k * t relative to the graph of sin(k x).

At t=0, omega * t is zero and we have the original graph of y = .75 cm * sin( k x). The graph of y vs. x forms a sine curve with period 2 pi / k, in this case 2 pi / (pi * .4 cm^-1) = 5 cm which is the wavelength. A complete cycle occurs between x = 0 and x = 5 cm, with zeros at x = 0 cm, 2.5 cm and 5 cm, peak at x = 1.25 cm and 'valley' at x = 3.75 cm.

At t=.0005, we are graphing y = .75 cm * sin( k x + .0005 omega), shifted -.0005 * omega / k = -.313 cm in the x direction. The sine wave of the t=0 function y = .75 cm * sin(kx) is shifted -.313 cm, or .313 cm left so now the zeros are at -.313 cm and every 2.5 cm to the right of that, with the peak shifted by -.313 cm to x = .937 cm.

At t=.0010, we are graphing y = .75 cm * sin( k x + .0010 omega), shifted -.0010 * omega / k = -.625 cm in the x direction. The sine wave of the t = 0 function y = .75 cm * sin(kx) is shifted -.625 cm, or .625 cm left so now the zeros are at -.625 cm and every 2.5 cm to the right of that, with the peak shifted by -.625 cm to x = +.625 cm.

The sequence of graphs clearly shows the motion of the wave to the left at 625 cm / s. **

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Your Solution:

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Given Solution:

**** If mass / unit length is .500 kg / m what is the tension?

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Question: ** Velocity of propagation is

v = sqrt(T/ (m/L) ). Solving for T:

v^2 = T/ (m/L)

v^2*m/L = T

T = (6.25 m/s)^2 * 0.5 kg/m so

T = 19.5 kg m/s^2 = 19.5 N approx. **

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Question: **** What is the average power?

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Your Solution:

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Given Solution:

** The text gives the equation Pav = 1/2 sqrt( m / L * F) * omega^2 * A^2 for the average power transferred by a traveling wave.

Substituting m/L, tension F, angular frequency omeage and amplitude A into this equation we obtain

Pav = 1/2 sqrt ( .500 kg/m * 19.5 N) * (250 pi s^-1)^2 * (.0075 m)^2 =

.5 sqrt(9.8 kg^2 m / (s^2 m) ) * 62500 pi^2 s^-2 * .000054 m^2 =

.5 * 3.2 kg/s * 6.25 * 10^4 pi^2 s^-2 * 5.4 * 10^-5 m^2 =

54 kg m^2 s^-3 = 54 watts, approx..

The arithmetic here was done mentally so double-check it. The procedure itself is correct. **

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QUESTION FROM STUDENT

My question is on the following question I was trying during a practice test:

Analyze the pressure vs. volume of a 'bottle engine' consisting of 8 liters of an ideal gas as it operates between minimum

temperature 200 Celsius and maximum temperature 360 Celsius, pumping water to half the maximum possible height.

Sketch a pressure vs. volume graph from the original state to the maximum-temperature state and use the graph to

determine the useful work done by the expansion. Then, assuming a diatomic gas, determine the thermal energy required

to perform the work and the resulting practical efficiency of the process.

I understand the pressure vs volume graph. Does this question basically mean that I should find d'Q(v) and d'Q(p) and if

so, what temperatures do I use for the temperature change. Then once i find d'Q, how do i find work. It was stated that

it is the area under the curve, but is this the same as the equation d'Q(p)-d'Q(v). Also efficiency is found by taking Max

temp-min temp/max temp, so I know how to do that, but why would this change with the amount of energy needed to

perform work. I am very confused on this problem.

INSTRUCTOR RESPONSE

Your questions are well posed and very relevant.

Note also that the Bottle Engine is addressed fairly extensively in Class Notes between #08 and #12, and is the subject of the two video experiments to be viewed as part of Assignment 11.

For the situation in question the maximum pressure possible, operating the system between 200 C and 360 C, is about

T_max / T_min * P_min = 623 K / (473 K) * 100 kPa = 132 kPa.

This would allow us to support a column of water which exerts a pressure of 32 kPa. This column would be about 3.2 meters high (easily found using Bernoulli's equation).

To raise water to half this height would require a temperature of about 280 C, after which the gas would expand by factor

623 K / (553 K) = 1.14.

The volume would change by this factor by displacing an equal volume of water, which would be raised to the 1.6 meter height.

The temperature of the gas would be raised from 200 C to 280 C at constant volume, then from 280 C to 360 C at constant pressure.

The efficiency we calculate here is the 'practical efficiency', which is the ratio of the mechanical work done to the thermal energy added to the system. The mechanical work is the raising of the water, so is equal to the PE change of the water which is raised to the 1.6 m height. The thermal energy added is the energy required to heat the gas, first at constant volume then at constant pressure.

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