#$&* course PHY 202 015. `Query 15 was submitted 1 Jul 2011 around 11:50 PM. 015. `Query 15
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Given Solution: `qSTUDENT RESPONSE: The longitudinal waves had a higher velocity. That 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 ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `qquery General College Physics and Principles of Physics 12.08: Compare the intensity of sound at 120 dB with that of a whisper at 20 dB. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: When comparing the intensity of sound at 120 dB with that of a whisper at 20 dB, the intensity at 120 dB can be found by solving the equation dB = 10 log(I / I_threshold) for I. log(I / I_threshold) = dB / 10, so that I / I_threshold = 10^(120 / 10) = 12 and 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. 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 therefore the solution is 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. A whisper therefore has one ten billionth the intensity that sound at the pain threshold has. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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: To determine the length of the pipe, we must first determine the velocity of the sound waves given the air temperature using formula: v = (331 + 0.60 * Temp.) m/s v = (331 + 0.60 * 21) m/s v = 343.6 m/s The wavelength of the sound is wavelength = v / f wavelength = 343.6 m/s / (262 Hz) = 1.33 m The pipe is open, so it has antinodes at both ends. • The fundamental frequency occurs when there is a single node between these antinodes. So the length of the pipe corresponds to two node-antinode distances. • Between a node and an adjacent antinode the distance is 1/4 wavelength. In this case this distance is 1/4 * 1.33 m = .33 m • The two node-antinode distances between the ends of the pipe therefore correspond to a distance of 2 * .33 m = .66 m. We conclude that the pipe is .64 meters long. 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 1.33 m / 4 = .33 m. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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) = 1.33 meters, approx.. The pipe is open, so it has antinodes at both ends. • The fundamental frequency occurs when there is a single node between these antinodes. So the length of the pipe corresponds to two node-antinode distances. • Between a node and an adjacent antinode the distance is 1/4 wavelength. In this case this distance is 1/4 * 1.33 meters = .33 meters, approx.. • The two node-antinode distances between the ends of the pipe therefore correspond to a distance of 2 * .33 meters = .66 meters. We conclude that the pipe is .64 meters long. Had the pipe been closed at one end then there would be a node at one end and an antinode at the other. The wavelength of the fundamental would have therefore been 4 times the length; the length of the pipe would then have been 1.33 m / 4 = .33 m. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: 3 " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!