#$&* course PHY 202 3/1/15 at 7:40PM 015. `Query 13
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Given Solution: `qSTUDENT RESPONSE: The logitudinal waves had a higher velocity. That doesn't provide evidence that the high-pitched wave was longitudinal, since we didn't directly measure the velocity of those waves. The higher-pitches waves were damped out much more rapidly by touching the very end of the rod, along its central axis, than by touching the rod at the end but on the side. The frequency with which pulses arrive at the ear determines the pitch. The amplitude of the wave affects its intensity, or energy per unit area. For a given pitch the energy falling per unit area is proportional to the square of the amplitude. Intensity is also proportional to the square of the frequency. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): So, I understand that pitch is not determined by whether or not the wave is transverse or longitudinal; that is dependent on the frequency, as explained in the above solution. I think I made my error in determining what exactly the question was asking for, as I was still thinking about the longitudinal vs. transverse wave conversation in the video as opposed to realizing that the question above specifically asked about the differences in pitch. ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: `qquery General College Physics and Principles of Physics 12.08: Compare the intensity of sound at 120 dB with that of a whisper at 20 dB. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 120dB = 10*log(I / 1.0*10^-12W/m^2) I = 1W/m^2 20dB = 10*log(I / 1.0*10^-12W/m^2) I = 1.0*10^-10W/m^2 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The 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: ********************************************* 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: Wavelength = velocity/frequency If frequency stays the same, then the ratio of velocity of a sound in air to its velocity in seawater should be equivalent to the ratio of the wavelength of a sound in air to its wavelength in seawater. Velocity of sound in air = 343m/s Velocity of sound in seawater = 1560m/s Velocity ratio = 343/1560 = 0.22 So, the ratio of a wavelength of a sound in air to its wavelength in seawater should be 0.22, which suggests that the wavelength in air is 0.22 times that of the wavelength in seawater. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Using propagation speeds 340 m/s and 1600 m/s for sound in water and air, respectively: The frequency of a particular sound is determined by the frequency at which the vocal cords vibrate, and is the same in water as in air. Since the speed of sound in water is roughly 5 times that in air, the same number of peaks will be distributed over about 5 times the distance in water as opposed to in air. So the wavelength in water will be about 5 times that in air. Your Self-Critique: I took a slightly different approach, but since 0.22 is approximately 1/5, my answer was correct and had only flipped the ratio. Your Self-Critique Rating: ********************************************* Question: Openstax: The warning tag on a lawn mower states that it produces noise at a level of 91.0 dB. What is this in watts per meter squared? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: 91dB = 10*log( I / 1.0*10^-12W/m^2) I = 1.26*10^-3 W/m^2 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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: OK Your Self-Critique Rating: ********************************************* Question: Openstax: (a) What is the decibel level of a sound that is twice as intense as a 90.0-dB sound? (b) What is the decibel level of a sound that is one-fifth as intense as a 90.0-dB sound? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: a) 90dB = 10*log(I / 1.0*10^-12W/m^2) I = 1*10^-3 W/m^2 2I = 2*10^-3 W/m^2 B = 10*log(2*10-3 / 1.0*10^-12) B = 93dB b) 1/5 * I = 2*10^-4 W/m^2 B = 10*log(2*10-4 / 1.0*10^-12) B = 83dB confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: A 90-dB sound has intensity I = 10^9 * I_0, where I_0 is the hearing threshold intensity 10^-12 watts / m^2. For a sound of twice that intensity I = 2 * 10^9 * I_0, the dB level is dB = 10 log( I / I_0) = 10 log(2 * 10^9 * I_0 / I_0) = 10 log( 2 * 10^9) = 10 * 9.3 = 93. For a sound 1/5 that intensity dB = 10 log( I / I_0) = 10 log(0.2 * 10^9 * I_0 / I_0) = 10 log( 0.2 * 10^9) = 10 * 8.3 = 83. Note that the first sound is 10 times as intense as the second (double a number is 10 times as great as 1/5 that number), and the dB difference between these sounds is 10, adding validation to the principle that a sound 10 times as intense as another exceeds the former to 10 dB. Your Self-Critique: OK Your Self-Critique Rating: ********************************************* Question: Openstax: (a) What is the fundamental frequency of a 0.672-m-long tube, open at both ends, on a day when the speed of sound is 344 m/s? (b) What is the frequency of its second harmonic? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: a) f1 = v/2L f1 = 344m/s / (2*0.672m) f1 = 256 Hz b) f2 = 2f1 f2 = 2(256Hz) f2 = 512 Hz confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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: OK 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: a) The area of the eardrum is 1800 times smaller than the area of the ear trumpet, so this increases the intensity of incoming sound by a factor of 1800 (intensity = power/area). The log of such an increase is 3.3, which is then multiplied by 10 to give the decibels, so this result would lead to a 33dB increase via the ear trumpet. But since the trumpet only has an efficiency of 5% in transmitting sound, the decibel increase is only 1.65dB. b) A decibel increase of 1.65dB seems too small to make a noticeable difference. 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: It appears that I applied the 5% efficiency factor too early in my calculations, which definitely makes a big difference since the equation involved is a logarithm. Your Self-Critique Rating:3 ********************************************* 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: a) Fundamental frequency, f1 = v/4L f1 = 343m/s / (4*1.8m) f1 = 47.6 Hz First overtone, third harmonic, f3 = 3v/4L f3 = (3*343) / (4*1.8) f3 = 142.9 Hz Second overtone, fifth harmonic, f5 = 5v/4L f5 = (5*343) / (4*1.8) f5 = 238.2 Hz Third overtone, seventh harmonic, f7 = 7v/4L f7 = (7*343) / (4*1.8) f7 = 333.5 Hz b) Fundamental frequency, f1 = v/2L f1 = 343m/s / (2*1.8m) f1 = 95.3 Hz First overtone, second harmonic, f2 = v/L f2 = 343/1.8 f2 = 190.6 Hz Second overtone, third harmonic, f3 = 3v/2L f3 = (3*343) / (2*1.8) f3 = 285.8 Hz Third overtone, fourth harmonic, f4 = 4v/2L f4 = (4*343) / (2*1.8) f4 = 381.1 Hz 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: I neglected the “audible frequency” part of the question, so while I found a number of frequencies that are audible, I did not find the maximum audible frequency possible in either tube. Wavelength = 343m/s / 15,000Hz Wavelength = 0.023 m 1.8m / 0.023m = 78.3 wavelengths to fit the tube 78.3 = L / 4 L = 313.2m So, if the length of the tube is less than 313.2/4 times the wavelength, the resulting frequency will be at the upper-but-still-audible limits of human hearing.
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Given Solution: GOOD 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 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK **** I do have a question though. In the textbook, parts b and c of this question ask for the fundamental wavelength and frequency of the fundamental harmonic, and for the wavelength and frequency of the wave in the outside air. It seems to me that the wavelength and frequency would be the same in both b and c as they are in my answer above. For b, the fundamental frequency, this is what I calculated above here, so wavelength would be 2L = 1.31m and the resulting frequency would be the original given 262Hz. For part c, since the tube is open at both ends, the wavelength and frequency in the tube would be the same as if they traveled into the outside air. Is my reasoning correct here???? **** ------------------------------------------------ Self-critique Rating:
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Given Solution: ** Since fR = fS ( 1 - v/c) we have v = (fR / fS - 1) * c = 3 * 10^8 m/s * ( 4.586 * 10^14 Hz / (4.568 * 10^14 Hz) - 1) = 1.182 * 10^6 m/s, approx. In the 949 years since the explosion the radius of the nebula would therefore be about 949 years * 365 days / year * 24 hours / day * 3600 seconds / hour * 1.182 * 10^6 m/s = 3.5 * 10^16 meters, the diameter about 7 * 10^16 meters. 5 minutes of arc is 5/60 degrees or 5/60 * pi/180 radians = 1.4 * 10^-3 radians. The diameter is equal to the product of the distance and this angle so the distance is distance = diameter / angle = 7 * 10^16 m / (1.4 * 10^-3) = 2.4 * 10^19 m. Dividing by the distance light travels in a year we get the distance in light years, about 6500 light years. CHECK AGAINST INSTRUCTOR SOLUTION: ** There are about 10^5 seconds in a day, about 3 * 10^7 seconds in a year and about 3 * 10^10 seconds in 1000 years. It's been about 1000 years. So those streamers have had time to move about 1.177 * 10^6 m/s * 3 * 10^10 sec = 3 * 10^16 meters. That would be the distance of the closest streamers from the center of the nebula. The other side of the nebula would be an equal distance on the other side of the center. So the diameter would be about 6 * 10^16 meters. A light year is about 300,000 km/sec * 3 * 10^7 sec/year = 9 * 10^12 km = 9 * 10^15 meters. So the nebula is about 3 * 10^16 meters / (9 * 10^15 m / light yr) = 3 light years in diameter, approx. 5 seconds of arc is 5/60 of a degree or 5 / (60 * 360) = 1 / 4300 of the circumference of a full circle, approx. If 1/4300 of the circumference is 6 * 10^16 meters then the circumference is about 4300 times this distance or about 2.6 * 10^20 meters. The circumference is 1 / (2 pi) times the radius. We're at the center of this circle since it is from here than the angular diameter is observed, so the distance is about 1 / (2 pi) * 2.6 * 10^20 meters = 4 * 10^19 meters. This is about 4 * 10^19 meters / (9 * 10^15 meters / light year) = 4400 light years distant. Check my arithmetic. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q **** query univ phy 16.74 / 16.66 (21.26 10th edition). 200 mHz refl from fetal heart wall moving toward sound; refl sound mixed with transmitted sound, 85 beats / sec. Speed of sound 1500 m/s. What is the speed of the fetal heart at the instant the measurement is made? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: . ** 200 MHz is 200 * 10^6 Hz = 2 * 10^8 Hz or 200,000,000 Hz. The frequency of the wave reflected from the heart will be greater, according to the Doppler shift. The number of beats is equal to the difference in the frequencies of the two sounds. So the frequency of the reflected sound is 200,000,085 Hz. The frequency of the sound as experienced by the heart (which is in effect a moving 'listener') is fL = (1 + vL / v) * fs = (1 + vHeart / v) * 2.00 MHz, where v is 1500 m/s. This sound is then 'bounced back', with the heart now in the role of the source emitting sounds at frequency fs = (1 + vHeart / v) * 2.00 MHz, the 'old' fL. The 'new' fL is fL = v / (v - vs) * fs = v / (v - vHeart) * (1 + vHeart / v) * 2.00 MHz. This fL is the 200,000,085 Hz frequency. So we have 200,000,085 Hz = 1500 m/s / (v - vHeart) * (1 + vHeart / v) * 2.00 MHz and v / (v - vHeart) * (1 + vHeart / v) = 200,000,085 Hz / (200,000,000 Hz) = 1.000000475. A slight rearrangement gives us (v + vHeart) / (v - vHeart) = 1.000000475 so that v + vHeart = 1.000000475 v - 1.000000475 vHeart and 2.000000475 vHeart = .000000475 v, with solution vHeart = .000000475 v / (2.000000475), very close to vHeart = .000000475 v / 2 = .000000475 * 1500 m/s / 2 = .00032 m/s, about .3 millimeters / sec. ** STUDENT COMMENT My final answer was twice the answer in the given solution. I thought that I used the Doppler effect equation correctly; however, I may have solved for the unknown incorrectly. INSTRUCTOR RESPONSE The equations tell you the frequency that would be perceived by a hypothetical detector on the heart. Suppose that each time the detector records a 'peak', it sends out a pulse. The pulses are sent out at the frequency of the detected wave. The source of these pulses is the detector, which is moving toward the 'listener', and as a result they are detected at an even higher frequency. Thus the doubled number of beats. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!