course Phy 202 7/6 11:30p 015. `Query 13
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Given Solution: `qSTUDENT RESPONSE: The logitudinal waves had a higher velocity. `aThat doesn't provide evidence that the high-pitched wave was longitudinal, since we didn't directly measure the velocity of those waves. The higher-pitches waves were damped out much more rapidly by touching the very end of the rod, along its central axis, than by touching the rod at the end but on the side. The frequency with which pulses arrive at the ear determines the pitch. The amplitude of the wave affects its intensity, or energy per unit area. For a given pitch the energy falling per unit area is proportional to the square of the amplitude. Intensity is also proportional to the square of the frequency. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok
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Given Solution: `aThe intensity at 120 dB is found by solving the equation dB = 10 log(I / I_threshold) for I. We get log(I / I_threshold) = dB / 10, so that I / I_threshold = 10^(120 / 10) = 12and I = I_threshold * 10^12. Since I_threshold = 10^-12 watts / m^2, we have for dB = 120: I = 10^-12 watts / m^2 * 10^12 = 1 watt / m^2. The same process tells us that for dB = 20 watts, I = I_threshold * 10^(20 / 10) = 10^-12 watts / m^2 * 10^2 = 10^-10 watts / m^2. Dividing 1 watt / m^2 by 10^-10 watts / m^2, we find that the 120 dB sound is 10^10 times as intense, or 10 billion times as intense. A more elegant solution uses the fact that dB_1 - dB_2 = 10 log(I_1 / I_threshold) - ( 10 log(I_2 / I_threshold) ) = 10 log(I_1 / I_threshold) - ( 10 log(I_2 / I_threshold) ) = 10 {log(I_1) - log( I_threshold) - [ ( log(I_2) - log(I_threshold) ]} = 10 { log(I_1) - log(I_2)} = 10 log(I_1 / I_2). So we have 120 - 20 = 100 = 10 log(I_1 / I_2) and log(I_1 / I_2) = 100 / 10 = 10 so that I_1 / I_2 = 10^10. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok Self-critique Rating: 3 ********************************************* Question: `qquery gen phy 12.30 length of open pipe, 262 Hz at 21 C? **** gen phy What is the length of the pipe? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: First we find velocity of the sound waves V= (331+0.60*21) V= 343.6 m/s Next we have to find the wavelength of the sound. The equation we have is frequency= velocity/wavelength. This can be manipulated to Wavelength= velocity/frequency Wavelength= (343.6 m/s)/(262 Hz) Wavelength= 1.3 m Because we have an open pipe, there will be antinodes at the ends and therefore the wavelength will be 2 times the length of the pipe. So Length= (1.3m)/2 L= 0.65 m
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Given Solution: `aGOOD STUDENT SOLUTION First we must determine the velocity of the sound waves given the air temperature. We do this using this formula v = (331 + 0.60 * Temp.) m/s So v = (331 + 0.60 * 21) m/s v = 343.6 m/s The wavelength of the sound is wavelength = v / f = 343.6 m/s / (262 Hz) = 0.33 meters. So 262 Hz = 343.6 m/s / 4 * Length Length = 0.33 meters f = v / (wavelength) 262 Hz = [343 m/s] / (wavelength) wavelength = 1.3 m. So the wavelength is 1.3 m. If it's an open pipe then there are antinodes at the ends and the wavelength is 2 times the length, so length of the the pipe is about 1.3 m / 2 = .64 m, approx.. Had the pipe been closed at one end then there would be a node and one end and an antinode at the other and the wavelength of the fundamental would have therefore been 4 times the length; the length of the pipe would then have been 1.3 m / 4 = .32 m. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok Self-critique Rating: 3 "