course Phy 122 }w{ɢZH҈Dassignment #022
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21:53:49 Query introductory set six, problems 15-18 how do we determine the energy of a standing wave given the amplitude and frequency of the wave and the mass of the string?
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RESPONSE --> In order to determine the energy of a standing wave with the amplitude and frequency we use the kinetic energy formula. KE = 1/2 m*v^2 The velocity of the wave can be found by the formula which applies to circular orbits: velocity = omega*amplitude. Where omega is the product of 2pi times the frequency. omega = (2pi rad/cycle)(frequency) velocity = omega * amplitude = (2pi*f)(A) After we have found the velocity of the wave, we can calculate the total energy of the wave. KE = 1/2*m*(2pi*f*A)^2
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21:54:28 STUDENT 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. **
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RESPONSE --> I understand this concept and I feel that I explained adequately although not as thoroughly as the instructor's solution.
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22:02:27 If 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?
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RESPONSE --> In order to know if the strings oscillate in phase you must first find the wavelengths of the given frequencies. You then calculate the number of wavelengths in the given difference in distance between the two strings (lambda/`dL). If these calculations come out to be whole numbers, the waves will exhibit constructive interference. If the resulting numbers are a whole number plus half, the waves act in destructive interference. Therefore, in order for the strings to oscillate in phase, the result of the wavelength divided by the difference in distance must be a whole number.
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22:04:24 ** 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 **
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RESPONSE --> I feel that I explained this concept correctly. However, I did not go on to mention that the pulse in the longer string will be ""slower"" than the shorter string, and if that time difference is related to whole number of periods, the ends oscillate in phase.
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22:17:58 General College Physics and Principles of Physics 11.38: AM 550-1600 kHz, FM 88-108 mHz. What are the wavelength ranges?
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RESPONSE --> The book also gives that the velocity is 3.00*10^8m/s. If we know the frequencies and velocities, it is simple to find the corresponding wave velocities. lambda = v/f AM 550-1600 kiloHz = 550*10^3 - 1600*10^3Hz FM 88-106 megaHz = 88*10^6 - 106*10^6Hz AM wavelengths: (3.00*10^8m/s) / (550*10^3Hz) = 545.45m (3.00*10^8m/s) / (1600*10^3Hz) = 187.5m range: 187.5m - 545.45m FM wavelengths: (3.00*10^8m/s) / (88*10^6Hz) = 3.41m (3.00*10^8m/s) / (108*10^3Hz) = 2.77m range: 2.77m - 3.41m
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22:18:22 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.
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RESPONSE --> I understand this concept and I answered the problem correctly.
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22:27:03 General College Physics and Principles of Physics 11.52: What are the possible frequencies of a violin string whose fundamental mode vibrates at 440 Hz?
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RESPONSE --> To find the possible frequencies of the first four harmonics, we first must look at the fundamental mode, 440Hz. In the class notes, I read that the first harmonic wavelength is double the length of the string. The fundamental mode frequency is the frequency of the first harmonic. Therefore, it fits only one half-wave. harmonic 1: f = 1 * 440Hz = 440Hz The second harmonic is equal to the length of the string. Therefore, it fits 2 half-waves onto the length. harmonic 2: f = 2 * 440Hz = 880Hz The third harmonic will fit three half-waves on the string. harmonic 3: f = 3*440Hz = 1320Hz The fourth harmonic fits 4 half-waves on the string. harmonic 4: f = 4*440Hz = 1760Hz
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22:27:36 The 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
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RESPONSE --> I answered this problem correctly. I remember by formulas and so I try to keep this relationship in mind for these problems: f = (# half-waves)(fundamental mode vibration)
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22:28:13 General College Physics Problem: Earthquake intensity is 2.0 * 10^6 J / (m^2 s) at 48 km from the source. What is the intensity at 1 km from the source?
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RESPONSE --> I am not a general physics student and I am not yet familiar with the concept of earthquake intensity.
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22:28:25 The wave is assumed spherical so its surface area increases as the square of its distance and its intensity, which is power / surface area, decreases as the square of the distance. So the intensity at 1 km will be (48 km / 1 km)^2 = 2300 times as great, or 2300 * 2.0 * 10^6 J / (m^2 s) = 4.6 * 10^9 J/(m^2 s).
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RESPONSE --> Ok.
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22:28:51 At what rate did energy pass through a 5.0 m^2 area at the 1 km distance?
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RESPONSE --> This is a follow-up question for the previous general physics question on earthquake intensity.
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22:29:02 Through a 5 m^2 area the rate of energy passage is therefore 4.6 * 10^9 J / (m^2 s) * 5.0 m^2 = 2.3 * 10^10 J / s, or 23 billion watts.
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RESPONSE --> Ok.
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