Query 11

course Phy 232

Something was messed up in the query form or mayber i just copy and pasted it wrong.

011. `Query 10*********************************************

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: a wavelength = 2 * string length so we would have 2L = lambda for 2 wavelengths L=lambda for 3 wavelengths lambda = 2/3 L and so on. 2L,L,2/3L,1/2L,..

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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 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: **** 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. Which would mean frequency is the wave velocity/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 take tension/mass per unit length. To find velocity we take sqrt(tension/mass per unit 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: query univ phy problem 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?

confidence rating:A we get .750cm. For frequency we get 125 Hz by taking 250 pi/2pi =125 s^-1 = 125 Hz. For the period we take 1/125 which would be 0.008 seconds. Wavelength would be 2 pi/.4*pi = 5cm. Speed of propagation would be 125 * 5 = 625 cm/sec.

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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 / x 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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Your Solution:to what question?

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Given Solution:

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

Your Self-Critique: The answer is a question something is messed up

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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:Pav=1/2 sqrt(m/L*F)*omega^2*A^2 but then we substitute getting Pav=1/2 sqrt(.500 *19.5)*(250)^2*(0.0075)^3 = approx. 5.5 watts

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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 =

5.5 kg m^2 s^-3 = 5.5 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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&#Good responses. Let me know if you have questions. &#