#$&* course phy122 I had difficulty understanding this assignment. Please provide insight Professor. If your solution to stated problem does not match the given solution, you should self-critique per instructions at http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm.Your solution, attempt at solution. If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.
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Given Solution: `qSTUDENT RESPONSE: The logitudinal 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): ------------------------------------------------ Self-critique Rating: ********************************************* 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: Professor I used the explanation of the solution below to figure out this problem. I do however have some question and I will post them below in the self critique. dB=10log(I/I_threshold) log(I/I_threshold)=dB/10 I/I_threshold=10^(120/10)=12 I=I_threshold *10^12 Since I_threshold = 10^12 watts/m^2, we have for dB=120 I=10^-12 watts/m^2 x 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^-10w/m^2 Dividing 1w/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. 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): I think I now understand where the 10^12 came from. I had to look at it over and over again to figure it out. It seems like the simplest of things can confuse me sometimes. ------------------------------------------------ Self-critique Rating: ********************************************* Question: Openstax: Dolphins make sounds in air and water. What is the ratio of the wavelength of a sound in air to its wavelength in seawater? Assume air temperature is 20.0ºC . YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: I have reviewed the answer for this multiple times and I still don’t think that I understand. I am guessing that the propagation speeds are something I should know off the top of my head. I now know that the speed of sound in water is 5 times more than that in air. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Using propagation speeds 340 m/s and 1600 m/s for sound in water and air, respectively: The frequency of a particular sound is determined by the frequency at which the vocal cords vibrate, and is the same in water as in air. Since the speed of sound in water is roughly 5 times that in air, the same number of peaks will be distributed over about 5 times the distance in water as opposed to in air. So the wavelength in water will be about 5 times that in air. Your Self-Critique: Would you mind to explain this in more detail? Your Self-Critique Rating:
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Given Solution: dB = 10 log( I / I_0), so log( I / I_0) = dB / 10. In this case dB = 91 so log(I / I_0) = 91/10 = 9.1. It follows that I / I_0 = 10^9.1 = 1.3 * 10^9, approx.. so that I = 1.3 * 10^9 * I_0 = 1.3 * 10^9 * (10^-12 watts / m^2) = 1.3 * 10^-3 watts / m^2, or about .0013 watts / m^2. Your Self-Critique: Professor the 10^-12 that is in the problem above, I now know is the hearing threshold intensity)
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Given Solution: A 90-dB sound has intensity I = 10^9 * I_0, where I_0 is the hearing threshold intensity 10^-12 watts / m^2. For a sound of twice that intensity I = 2 * 10^9 * I_0, the dB level is dB = 10 log( I / I_0) = 10 log(2 * 10^9 * I_0 / I_0) = 10 log( 2 * 10^9) = 10 * 9.3 = 93. For a sound 1/5 that intensity dB = 10 log( I / I_0) = 10 log(0.2 * 10^9 * I_0 / I_0) = 10 log( 0.2 * 10^9) = 10 * 8.3 = 83. Note that the first sound is 10 times as intense as the second (double a number is 10 times as great as 1/5 that number), and the dB difference between these sounds is 10, adding validation to the principle that a sound 10 times as intense as another exceeds the former to 10 dB. Your Self-Critique: Your Self-Critique Rating: ********************************************* Question: Openstax: (a) What is the fundamental frequency of a 0.672-m-long tube, open at both ends, on a day when the speed of sound is 344 m/s? (b) What is the frequency of its second harmonic? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: The length of the tube is one half the wavelength. So 2*.672=1.34 m The wavelength of the second harmonic is .67m 1.34*344m/s=460.96m/s .67m*344m/s=230.48m/s Frequency is found by dividing velocity/wavelength: 460.96/1.34=344 Hmmm, how do I find the frequency….this can’t be right. Please help professor. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: If a tube is open at both ends, its fundamental mode of vibration has an antinode at both ends, with no additional antinodes. The configuration of nodes and antinodes is thus ANA, which corresponds to one-half wavelength. Thus the length of the tube is one-half wavelength, and the wavelength in this case is lamba_fundamental = 2 * .672 m = 1.34 m, approx.. (the positions of the antinodes are not exactly at the ends of the tube, so a four-significant-figure result would imply more precision than can be expected). Every harmonic has antinodes at the ends. The second harmonic has an antinode in the middle as well, with a node (as always) between two adjacent antinodes. So the configuration of this harmonic is ANANA, corresponding to a full wavelength. The wavelength of the second harmonic is thus lambda_2d_harmonic = .67 m, approx.. The frequencies are easily calculated from the wavelengths and the 344 m/s propagation velocity. Your Self-Critique: Your Self-Critique Rating: ********************************************* Question: Openstax: (a) Ear trumpets were never very common, but they did aid people with hearing losses by gathering sound over a large area and concentrating it on the smaller area of the eardrum. What decibel increase does an ear trumpet produce if its sound gathering area is 900 cm^2 and the area of the eardrum is 0.500 cm^2 , but the trumpet only has an efficiency of 5.00% in transmitting the sound to the eardrum? (b) Comment on the usefulness of the decibel increase found in part (a). YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: 900cm^2/0.5cm^2=1800cm^2 1800cm^2*.05=90 dB=10log(90)=19.54dB confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The sound is gathered from an area which is 900 cm^2 / (0.5 cm^2) = 1800 times the area, which if transmitted without loss to the eardrum would result in 1800 times the intensity. However transmission is only 5% efficient, so the increase in intensity is by a factor of only 5% of 1800 or 90. Every 10 times the intensity results in a 10 dB increase. 90 is almost 10 times 20, so the increase in intensity would be nearly 20 dB. This could be quite useful for someone with hearing loss, though modern hearing aids do much better. Note: A more precise estimate that 20 dB would be useless in this context, since eardrums vary significantly in size (0.500 cm^2 implies 3-significant-figure accuracy, which is absurd in this context). However more precise calculations are possible. In this case, dB = 10 log(90) would yield 19.5 dB rather than 20, and the .5 dB difference would be insignificant. Your Self-Critique: Your Self-Critique Rating: ********************************************* Question: Openstax: What frequencies will a 1.80-m-long tube produce in the audible range at 20.0ºC if: (a) The tube is closed at one end? (b) It is open at both ends? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: Propagation speed is 340 m/s Audible range is 30hz-15000hz Longest possible length 4*1.8=7.2m (will produce sound around 50 hz) 4/3 x 1.8m=2.4m 4/5 * 1.8m= 1.45m 4/7 *1.8=1.03m 15000 hz would have wavelength of 340m/s / 15000s^-1=2.3*10^-2 I have looked at the solution for this problem and now I am lost. I can’t figure out where the 78 wavelengths came from below. Where did the 311/4 come from? Please explain with more detail. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: We reason out our results by considering possible node-antinode configurations. If the tube is closed at one end then the possible node-antinode configurations are NA, NANA, NANANA, NANANANA, ..., with the length of the tube being 1/4, 3/4, 5/4, ... times the wavelength of the sound. Possible wavelengths will therefore be 4, 4/3, 4/5, 4/7, ... . times the length of the tube. Assuming propagation speed 340 m/s and audible range 30 Hz - 15 000 Hz, we easily see that the longest possible wavelength, which is 4 * 1.8 m = 7.2 m, will produce a sound around 50 Hz, clearly in the audible range. The next few wavelengths are 4/3 * 1.8 m = 2.4 m, 4/5 * 1.8 m = 1.45 m, 4/7 * 1.8 m = 1.03 m (approx.), etc. A 15 000 Hz sound would have wavelength 340 m/s / (15 000 s^-1) = 2.3 * 10^-2 m, or .023 m. At this wavelength the tube spans a little more than 78 wavelengths. 78 = 312 / 4, so if the tube is 311 / 4 times the wavelength the wavelength will be at least .023 m and have a frequency less than about 15 000 Hz. If the tube length is 313/4 times the wavelength the frequency will exceed 15 000 Hz. Thus if, and only if, tube length is equal to 1/4, 3/4, 5/4, ..., 311/4 times the wavelength, the wavelength will produce audible sound. Audible wavelengths will therefore be 4 * 1.8 m, 4/3 * 1.8 m, 4/5 * 1.8 m, ..., 4/311 * 1.8 meters. The corresponding frequencies can be obtained by dividing the speed of sound by each wavelength. For example, wavelength 4/311 * 1.8 meters implies frequency 340 m/s / (4/311 * 1.8 meters) = 14700 Hz, approx.. If the tube is open at both ends then the node-antinode configurations are ANA, ANANA, ANANANA, ... and the length of the tube will be equal to 2, 1, 2/3, 1/2, ... times the wavelength of the sound (these numbers come from the fact that the respective node-antinode configurations comprise 2/4, 4/4, 6/4, 8/4, ... of the length of the tube). From the above reasoning we see that the tube will produce audible sounds up to the harmonic where tube length is 312/4 times the wavelength of the sound. Your Self-Critique: Your Self-Critique Rating: