Query 11

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course Phy 122

3/17 1

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:

Assuming the string is fixed at both ends:

Each harmonic corresponds to a half wave. For example, the first harmonic is ½ wave, the second is 2, ½ waves, the third is 3, ½ waves, etc. For the first harmonic, the length of the string (L) is equal to ½ the wavelength. For the second harmonic, the length of the string is equal to the wavelength. For the third harmonic, the length of the string is equal to 3/2 the wavelength. The fourth harmonic, the length of the string is equal to twice the wavelength, etc.

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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 equal to the 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:

The velocity of the wave in the string is the square root of the tension of a string divided by its mass and length. So:

Velocity = sqrt (tension/mL)

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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:

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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:

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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 15.50 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. **

If your solution to stated problem does not match the given solution, you should self-critique per instructions at http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm.

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.

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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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