#$&* course Phy 122 013. `Query 11
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Given Solution: `aSTUDENT ANSWER AND INSTRUCTOR RESPONSE: Energy = 2*pi^2*m*f^2*A^2 INSTRUCTOR RESPONSE: ** You should understand the way we obtain this formula. We assume that every point of the string in in SHM with amplitude A and frequency f. Since the total energy in SHM is the same as the maximum potential or the max kinetic energy, all we need to do is calculate the max potential energy or kinetic energy of each point on the string and add up the results. Since we know mass, frequency and amplitude, we see that we can calulate the max kinetic energy we can get the result we desire. Going back to the circular model, we see that frequency f and amplitude A imply reference point speed = circumference / period = circumference * frequency = 2 `pi A f. The oscillator at its maximum speed will match the speed of the reference point, so the maximum KE is .5 m v^2 = .5 m (2 `pi A f)^2 = 2 `pi^2 m f^2 A^2. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `qIf the ends of two strings are driven in phase by a single simple harmonic oscillator, and if the wave velocities in the strings are identical, but the length of one string exceeds that of the other by a known amount, then how do we determine whether a given frequency will cause the 'far ends' of the strings to oscillate in phase? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: If we have the frequency then we can decide if the time difference between the two strings relates to the number of periods. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `qGeneral College Physics and Principles of Physics 11.38: AM 550-1600 kHz, FM 88-108 mHz. What are the wavelength ranges? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: AM RANGE 3e8/5.5e5=545.5m to 3e8/1.6e6=187.5m FM RANGE 3e8/8.8e7= 3.41m to 3e8/1.08e8=2.78m confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a At 3 * 10^8 m/s: a frequency of 550 kHz = 550 * 10^3 Hz = 5.5 * 10^5 Hz will correspond to a wavelength of 3 * 10^8 m/s / (5.5 * 10^5 cycles / sec) = 545 meters. a frequency of 1600 kHz = 1.6* 10^6 Hz will correspond to a wavelength of 3 * 10^8 m/s / (1.6 * 10^6 cycles / sec) =187 meters. The wavelengths for the FM range are calculated similarly. a frequency of 88.0 mHz= 88.0 * 10^6 Hz = 8.80 * 10^7 Hz will correspond to a wavelength of 3 * 10^8 m/s / (8.80 * 10^7 cycles / sec) = 3.41 meters. The 108 mHz frequency is calculated similarly and corresponds to a wavelength of 2.78 meters. STUDENT QUESTION I don’t understand where the v came from? How did you get the velocity to work the problem? INSTRUCTOR RESPONSE This is electromagnetic radiation. Its propagation velocity is the speed of light. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `qGeneral College Physics and Principles of Physics 11.52: What are the possible frequencies of a violin string whose fundamental mode vibrates at 440 Hz? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 2*440Hz=880Hz 3*440Hz=1320Hz 4*440Hz=1760Hz confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThe fundamental mode for a string fixed at both ends fits half a wavelength onto the string and therefore has a wavelength equal to double its length. The next three harmonics fit 2, 3 and 4 half-wavelengths into the length of the string and so have respectively 2, 3 and 4 times the frequency of the fundamental. So the first 4 harmonics are fundamental frequency = 440 Hz First overtone or second harmonic frequency = 2 * 440 Hz = 880 Hz Second overtone or third harmonic frequency = 3 * 440 Hz = 1320 Hz Third overtone or fourth harmonic frequency = 4 * 440 Hz = 1760 Hz STUDENT QUESTION Where does the 4th come from? Didn’t see that in the notes? INSTRUCTOR RESPONSE The pattern of the frequencies should be clear from the way the wavelengths are determined. The pattern follows the obvious pattern of the first three harmonics. The nth harmonic will contain n half-wavelengths in the distance covered by a single half-wavelength of the fundamental. So its frequency will be n times higher than the fundamental. The frequency of the nth harmonic is therefore n times that of the fundamental. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!