Asst 10

course Phy 202

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I have inserted a message within the query, but I had been doing an un-updated version of the queries and had already answered these questions in asst 9 query. I downloaded the new version today and had already completed asst 10 query on the old query for asst 9. I apologize and I am back on track with the appropriate program now. Thank you. Is it okay if I go ahead and perfore the Bottle Experiment before I take test 1?

It's a good idea to perform the bottle experiment before taking the test.

I just within the past half hour posted a new update to the Query. No difference in content, but the formatting is much better, so download it when you get a chance.

12:31:23 **** 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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RESPONSE --> I just checked blackboard today and downloaded the new query link. I did asst 9 and it had the same questions. I will proceed if there are any new questions and answer those. For asst 11 I will use the newly downloaded query generator. I am sorry. I had to go through problems 11-14 for asst 9 because I had not done them during the asst and I see now what happened. I apologize, I am back on track now.

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12:31:26 ** 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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12:31:57 **** 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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12:32:14 ** We divide tension by mass per unit length:

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

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RESPONSE --> Those were both in asst 9.

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12:32:23 **** gen phy explain in your own words the meaning of the principal of superposition

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RESPONSE --> asst 9

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12:32:27 ** 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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12:32:27 ** 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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12:32:32 **** gen phy what does it mean to say that the angle of reflection is equal to the angle of incidence?

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RESPONSE --> asst 9

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12:32:36 ** 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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12:32:41 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?

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12:32:43 ** 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. **

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12:32:45 **** Describe your sketch for t = 0 and state how the shapes differ at t = .0005 and t = .0010.

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12:32:48 ** 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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12:32:50 **** If mass / unit length is .500 kg / m what is the tension?

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12:32:52 ** 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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12:32:54 **** What is the average power?

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12:32:57 ** 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 * 195 N) * (250 pi s^-1)^2 * (.0075 m)^2 = .5 sqrt(98 kg^2 m / (s^2 m) ) * 62500 pi^2 s^-2 * .000054 m^2 = .5 * 9.9 kg/s * 6.25 * 10^4 pi^2 s^-2 * 5.4 * 10^-5 m^2 = 17 kg m^2 s^-3 = 17 watts, approx..

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

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