#$&* course Phy 122 2/21, 1830hrs 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: `aSTUDENT ANSWER AND INSTRUCTOR RESPONSE: Energy = 2*pi^2*m*f^2*A^2 INSTRUCTOR RESPONSE: ** You should understand the way we obtain this formula. We assume that every point of the string in in SHM with amplitude A and frequency f. Since the total energy in SHM is the same as the maximum potential or the max kinetic energy, all we need to do is calculate the max potential energy or kinetic energy of each point on the string and add up the results. Since we know mass, frequency and amplitude, we see that we can calulate the max kinetic energy we can get the result we desire. Going back to the circular model, we see that frequency f and amplitude A imply reference point speed = circumference / period = circumference * frequency = 2 `pi A f. The oscillator at its maximum speed will match the speed of the reference point, so the maximum KE is .5 m v^2 = .5 m (2 `pi A f)^2 = 2 `pi^2 m f^2 A^2. ** STUDENT QUESTION I found the equation above in the book and understand, but my answer was based on problem 17. I don’t understand why the solution is different? INSTRUCTOR RESPONSE Your solution was in terms of omega; the equation you quote was an intermediate step in the solution process. That equation is accurate, but it doesn't express the result in terms of the quantities given here. The given information didn't include omega, but rather gave the frequency of the oscillation. So putting everything in terms of f, A and mass m: omega = 2 pi f, so vMax = omega * A = 2 pi f * A and total energy = .5 m vMax^2 = .5 m * ( 2 pi f * A)^2, which when expanded is equal to the expression in the given solution. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):2 ------------------------------------------------ Self-critique Rating:NA ********************************************* Question: `qIf the ends of two strings are driven in phase by a single simple harmonic oscillator, and if the wave velocities in the strings are identical, but the length of one string exceeds that of the other by a known amount, then how do we determine whether a given frequency will cause the 'far ends' of the strings to oscillate in phase? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Even if the wave velocities are =, it stands to reason that they will not be in phase unless the shorter string is even with the size of the wave. Even then, given identical velocities, the whip will crack on the shorter end while the longer length is still in travel. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** the question here is whether the far ends of the strings are at the same phase of motion, which occurs only if their lengths differ by exactly one, two, three, ... wavelengths. So we need to find the wavelength corresponding to the given frequency, which need not be a harmonic frequency. Any frequency will give us a wavelength; any wavelength can be divided into the difference in string lengths to determine whether the extra length is an integer number of wavelengths. Alternatively, the pulse in the longer string will be 'behind' the pulse in the shorter by the time required to travel the extra length. If we know the frequency we can determine whether this 'time difference' corresponds to a whole number of periods; if so the ends will oscillate in phase ** STUDENT QUESTION: The pulse of the longer string will take obviously longer than the shorter string but if the frequency is the same they will be oscillating at the same rate. Im not sure if I truly understand. INSTRUCTOR RESPONSE: If the strings are of the same length then, given the specified conditions, their ends will oscillate in phase. When a peak arrives at the end of one string, a peak will arrive simultaneously at the end of the other. If you trim a little bit off the end of one of the strings, this won't be the case. When a peak arrives at the end of the untrimmed string, the peak will have already passed the end of the trimmed string, which is therefore oscillating ahead of the phase of the end of the untrimmed string. The ends of both strings will therefore be oscillating with the same frequency, but out of phase. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): If both strings were the same length, then they would be in phase, I believe ------------------------------------------------ Self-critique Rating:2
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Given Solution: Crests of the wave travel at 5 m/s and are separated by one wavelength, or 40 m. Thus a crest will pass the boat every 40 m / s (5 s) = 8 seconds. Your Self-Critique: got it. Its simple to figure mathematically but difficult to figure what is being asked sometimes in word problems. This one I saw immediately, being a navy eod diver. Your Self-Critique Rating:3 ********************************************* Question: Openstax: A car has two horns, one emitting a frequency of 199 Hz and the other emitting a frequency of 203 Hz. What beat frequency do they produce? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: 203 Hz - 199Hz = 4 Hz difference. In seconds, this states that every second, they will not be in phase 4 times. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: A wave with frequency 199 Hz will come in and out of phase with a wave having frequency 203 Hz four times every second. So the beat frequency is 4 Hz. Your Self-Critique: Got it. Your Self-Critique Rating:3 ********************************************* Question: Openstax: (a) Seismographs measure the arrival times of earthquakes with a precision of 0.100 s. To get the distance to the epicenter of the quake, they compare the arrival times of S- and P-waves, which travel at different speeds. Figure 16.48) If S- and P-waves travel at 4.00 and 7.20 km/s, respectively, in the region considered, how precisely can the distance to the source of the earthquake be determined? (b) Seismic waves from underground detonations of nuclear bombs can be used to locate the test site and detect violations of test bans. Discuss whether your answer to (a) implies a serious limit to such detection. (Note also that the uncertainty is greater if there is an uncertainty in the propagation speeds of the S- and P-waves.) YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: The ‘d is 3.2 km/s This difference will be divided by the difference in time to get distance. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: One wave travels 3.20 km/s faster than the other. If the distance to the epicenter is `ds, then the time between the arrival of the two waves will be `dt = `ds / (3.20 km/s). If `dt is uncertain by 0.10 second, then `ds will be uncertain by 3.20 km/s * 0.10 s = .32 km, or 320 meters. Your Self-Critique: I believe that I am thinking along the same lines Your Self-Critique Rating:3 " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!