course Phy 202 assignment #013ñCèËêBy×ßïçb¾„˜‚´zþ˜ž”ª
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18:14:51 query experiment to be viewed and read but not performed: transverse and longitudinal waves in aluminum rod what is the evidence that the higher-pitched waves are longitudinal while the lower-pitched waves are transverse?
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RESPONSE --> Lower pitched waves are longer waves that travel further while the high pitched are shorter waves. Longitudinal waves are travel shorter distances and therefore are more costumed to being high pitched.
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18:15:20 STUDENT 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. **
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RESPONSE --> I was completely off. i get that the velocity is faster on longitudinal waves. What I still don't understand how that differs from transverse and how the 2 types of waves are different
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18:15:42 Query 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.
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RESPONSE --> THe intensity of the sound 120dB is 1.0*10^10 times greater then that of a sound at 20dB. According to the chart in the book or by working it out using hte equation beta=10log (I/I0) 120dB=10 log (I/1.0*10W/m^2)=1 W/m^2 20dB=10 log (I/1.0*10W/m^2)=1*10^-10
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18:15:59 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.
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RESPONSE --> I got the same answer by the first way. I had not thought about using dB2-dB1.
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18:57:35 Query gen phy 12.30 length of open pipe, 262 Hz at 21 C? **** gen phy What is the length of the pipe?
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RESPONSE --> L=v/2f in order to find the v, 331m/s +.60T is used. 331+60*21=343.6m/s L=343.6m/s / 2*262Hz=.656m THe frequancy of the standing wave is 262Hz and the wavelength is lambda=v/f 343.6m/s/262=1.31
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18:58:57 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) = 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. **
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RESPONSE --> I don't know where I got the 2*v from but all the rest of my math was correct. The rest of my reasoning was correct also.
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18:58:59 **** Univ phy 16.72 (10th edition 21.32): Crab nebula 1054 A.D.;, H gas, 4.568 * 10^14 Hz in lab, 4.586 from Crab streamers coming toward Earth. Velocity? Assuming const vel diameter? Ang diameter 5 arc minutes; how far is it?
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18:59:01 ** 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) = 1.182 * 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. **
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18:59:04 **** query univ phy 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?
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18:59:06 . ** 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. **
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