course Phy 122 A lab entitled 07-02-2007_____force_vectors was posted to my access page and is not my work. I think it's a Phy 121 lab. Just thought you'd need to know! Thanks. ]ŽåÎØy ‰Üƒ¯s·µˆ°šàÜ»fÑ^ÇE–xassignment #021
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22:32:27 **** Query introductory set six, problems 11-13 **** given the length of a string how do we determine the wavelengths of the first few harmonics?
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RESPONSE --> The length of the string is equal to half the wavelength. L = 1/2 lambda Therefore, for the first harmonic, wavelength is double the string length multiplied by one half-wave. L = 1*1/2 lambda lambda = 2 L For the second harmonic, wavelength is equal to the string length because you multiply by 2 half-waves. L = 2*1/2 lambda lambda = L For the third harmonic, wavelength is 2/3 the string length because you multiply by 3 half-waves. L = 3*1/2 lambda lambda = 2/3 L For the next harmonics, you simply multiply 1/2 lambda by the number of half-waves and solve for lamda, the wavelength.
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22:32:45 ** 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 --> I understand this concept.
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22:39:09 **** 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 --> If we know the wavelengths and velocities, it is simple to solve for the frequency of the harmonics. Frequency is the number of peaks that pass a given point in a second. Frequency is found by dividing velocity by wavelength. f = v / lambda = (m/s) / (m/peak) = peaks/second
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22:39:17 ** 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 --> I understand this concept
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22:39:49 **** 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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RESPONSE --> From Class notes #13 and assignment 20, we are given the relationship: velocity = `sqrt (T/mu) where T is the tension on the string and mu is the mass per unit of length of the string. Therefore, v = `sqrt (T / m/L)
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22:40:04 ** We divide tension by mass per unit length: v = sqrt ( tension / (mass/length) ). **
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RESPONSE --> I answered this question correctly.
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22:40:38 **** gen phy explain in your own words the meaning of the principal of superposition
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RESPONSE --> I am not a general physics student, but in chapter 11 I read that the principal of superposition explains that when two waves pass through a region and meet they cause either destructive interference or constructive interference. Destructive interference is caused by waves that have opposite displacements (one up and one down) and when they meet they ""cancel"" each other out. Constructive interference is cause by waves that have the same displacement and when they meet they combine to form a larger wave that is the product of their displacements.
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22:40:48 ** 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 --> I understand this concept.
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22:41:43 **** 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 --> The questions for this assignment have been the same as the ones for the assignment 20 query. To say that the angle of reflection is equal to the angle of incidence can be explained by the incident ray and reflected ray. The angle formed by an incident ray with a perpendicular is called the angle of incidence. That ray is reflected on the opposite side of the perpendicular as the reflecting ray. Therefore, the reflecting ray would make an equal angle with the perpendicular to the reflecting surface. This angle is called the angle of reflection.
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22:41:51 ** 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 --> I understand.
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22:42:02 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 --> I am not a university physics student.
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22:42:06 ** 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 --> Ok.
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22:42:26 **** Describe your sketch for t = 0 and state how the shapes differ at t = .0005 and t = .0010.
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RESPONSE --> This is a follow-up question for the university physics students.
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22:42:31 ** 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 --> Ok.
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22:43:30 **** If mass / unit length is .500 kg / m what is the tension?
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RESPONSE --> This question is a follow-up on the previous university physics question, but I know the necessary formula. We know from Class notes #13 that velocity = `sqrt(T/mu) = `sqrt(T / m/L) Therefore, T = (v^2)(mu) = (v^2)(m/L) = v^2 * .500kg/m
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22:43:37 ** 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 --> Ok.
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22:43:57 **** What is the average power?
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RESPONSE --> This is a follow-up question for university physics and I do not yet know how to calculate average power.
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22:44:01 ** 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 --> Ok.
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