course PHY 122 June 23 @ 11: 23 am ˜µÂâòÙOË˃÷iûŸ·él€Ì–Y¬¯ªºöžªáÉ”assignment #011
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11:01:03 **** 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 --> If you know the length of the string you can determine the wavelengths of the first few harmonics. We know that a complete cycle of a wave goes from node to antinode (positive) to node to antinode (negative). The full cycle consists of 4 nodes. We can reason that the first harmonic has a single node-antinode distance which is the distance of the string and it corresponds to 1/4 wavelength, so the wavelength for the first node must be 4 x the length of the string. This is done for the other three wavelengths using 3/4, 5/4, and 7/4 times the length of the string.
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11:05:19 ** 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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RESPONSE --> okay I understand.
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11:11:10 **** 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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RESPONSE --> To find the frequencies of the first few harmonices we use the equation V = lambda times frequency. Rearranging this equation we see that frequency equals velocity/ lambda
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11:12:33 ** 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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RESPONSE --> okay. I got this one right.
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11:14:21 ** We divide tension by mass per unit length: v = sqrt ( tension / (mass/length) ). **
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RESPONSE --> Enter, as appropriate, an answer to the question, a critique of your answer in response to a given answer, your insights regarding the situation at this point, notes to yourself, or just an OK. Always critique your solutions by describing any insights you had or errors you makde, and by explaining how you can make use of the insight or how you now know how to avoid certain errors. Also pose for the instructor any question or questions that you have related to the problem or series of problems The velocity is equal to the square root of the tension/ mass per unit length.
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11:14:28 **** gen phy explain in your own words the meaning of the principal of superposition
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RESPONSE --> different section
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11:14:33 ** 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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RESPONSE --> okay
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11:14:39 **** 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 --> different section
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11:14:45 ** 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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RESPONSE --> okay
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11:14:53 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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RESPONSE --> another section. didn't do
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11:14:58 ** 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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RESPONSE --> okay
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11:15:03 **** Describe your sketch for t = 0 and state how the shapes differ at t = .0005 and t = .0010.
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RESPONSE --> okay
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11:15:09 ** 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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RESPONSE --> okay
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11:15:25 **** If mass / unit length is .500 kg / m what is the tension?
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RESPONSE --> okay
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11:15:42 ** 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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RESPONSE --> okay
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11:17:19 **** What is the average power?
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RESPONSE --> If you know the velocity and E (elastic modulus) you can solve for inertia factor (power) using the equation: V = sqrt elastic modulus/ intertia factor
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11:22:05 ** 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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RESPONSE --> I double checked the math. i came up with 17.35 watts or approx. 17 watts.
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GÜî֎Ǩ‡¹™åûÖšâ°§Âù¦¤Áè¦ assignment #011 011. `Query 10 Physics II 06-23-2009"
course PHY 122 June 23 @ 11: 23 am ˜µÂâòÙOË˃÷iûŸ·él€Ì–Y¬¯ªºöžªáÉ”assignment #011
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11:01:03 **** Query introductory set six, problems 11-14 **** given the length of a string how do we determine the wavelengths of the first few harmonics?
......!!!!!!!!...................................
RESPONSE --> If you know the length of the string you can determine the wavelengths of the first few harmonics. We know that a complete cycle of a wave goes from node to antinode (positive) to node to antinode (negative). The full cycle consists of 4 nodes. We can reason that the first harmonic has a single node-antinode distance which is the distance of the string and it corresponds to 1/4 wavelength, so the wavelength for the first node must be 4 x the length of the string. This is done for the other three wavelengths using 3/4, 5/4, and 7/4 times the length of the string.
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11:05:19 ** 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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RESPONSE --> okay I understand.
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11:11:10 **** 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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RESPONSE --> To find the frequencies of the first few harmonices we use the equation V = lambda times frequency. Rearranging this equation we see that frequency equals velocity/ lambda
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11:12:33 ** 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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RESPONSE --> okay. I got this one right.
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11:14:21 ** We divide tension by mass per unit length: v = sqrt ( tension / (mass/length) ). **
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RESPONSE --> Enter, as appropriate, an answer to the question, a critique of your answer in response to a given answer, your insights regarding the situation at this point, notes to yourself, or just an OK. Always critique your solutions by describing any insights you had or errors you makde, and by explaining how you can make use of the insight or how you now know how to avoid certain errors. Also pose for the instructor any question or questions that you have related to the problem or series of problems The velocity is equal to the square root of the tension/ mass per unit length.
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11:14:28 **** gen phy explain in your own words the meaning of the principal of superposition
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RESPONSE --> different section
.................................................
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11:14:33 ** 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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RESPONSE --> okay
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11:14:39 **** 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 --> different section
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11:14:45 ** 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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RESPONSE --> okay
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11:14:53 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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RESPONSE --> another section. didn't do
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11:14:58 ** 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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RESPONSE --> okay
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11:15:03 **** Describe your sketch for t = 0 and state how the shapes differ at t = .0005 and t = .0010.
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RESPONSE --> okay
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11:15:09 ** 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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RESPONSE --> okay
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11:15:25 **** If mass / unit length is .500 kg / m what is the tension?
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RESPONSE --> okay
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11:15:42 ** 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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RESPONSE --> okay
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11:17:19 **** What is the average power?
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RESPONSE --> If you know the velocity and E (elastic modulus) you can solve for inertia factor (power) using the equation: V = sqrt elastic modulus/ intertia factor
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11:22:05 ** 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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RESPONSE --> I double checked the math. i came up with 17.35 watts or approx. 17 watts.
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GÜî֎Ǩ‡¹™åûÖšâ°§Âù¦¤Áè¦ assignment #011 011. `Query 10 Physics II 06-23-2009"