#$&* course Phy 232
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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): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* 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: The two strings have the same velocities but one is longer than the other, than the longer one will take a longer amount of time to oscillate. Thus, the two will not oscillate in phase given the same frequency. 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): So if the time difference corresponds to a whole number of periods, then the two will oscillate in phase. What about if the time difference does not correspond to a whole number of periods?