course phys 202 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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Given Solution: 3 ** 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. ** Your Self-Critique: I didn’t make the note that a string free at one end has different nodes because the first is ¼ the wavelength instead of half. This makes all the others also change their fractions. So the 2nd is 4/3, the 3rd is 5/4 and the 4th is 7/4. Your Self-Critique Rating: 3 ********************************************* 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: We know the wavelengths and can use this to determine the frequency if we now the velocity of wave propagation. V = sqrt(tension/mass per unit length). We can divide velocity by wavelength to get the frequency.
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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. ** Your Self-Critique: OK Your Self-Critique Rating: OK ********************************************* Question: **** Given the tension and mass density of a string how do we determine the velocity of the wave in the string? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: I kind of explained this in the last problem. Velocity is equal to the square root of the tension divided by the mass per unit length.
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Given Solution: ** We divide tension by mass per unit length: v = sqrt ( tension / (mass/length) ). ** Your Self-Critique: OK Your Self-Critique Rating: OK ********************************************* Question: **** gen phy explain in your own words the meaning of the principal of superposition YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: Superposition is what happens when two waves within the same medium add together. If they are in phase they can sum together making larger amplitudes. If they are out of phase they can cancel out.
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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. ** Your Self-Critique: OK Your Self-Critique Rating: OK ********************************************* Question: **** gen phy what does it mean to say that the angle of reflection is equal to the angle of incidence? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: When the angle of reflection is equal to the angle of incidence it means the ray is reflected at an equal angle on the other side of the perpendicular of the surface.
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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 ** Your Self-Critique: OK Your Self-Critique Rating: OK ********************************************* 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? ********************************************* Question: **** Describe your sketch for t = 0 and state how the shapes differ at t = .0005 and t = .0010.