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course Phy 242
3/15 6
110307 PhysicsFrom 110302:
`q006. When I threw the 'singing rod' at you, at the instant it left my hand it was about 15 meters away. The first 'peak' of the sound wave that left the end of the rod after I released it was 15 meters from you and hadn't yet reached your ear.
Moving at 340 m/s, how long did it take that first peak to get to your ear?
approx. .044sec
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The rod didn't hit you in the ear. I threw it so that it would land a few meters in front of you, so as to avoid damaging the rod (or you). Suppose that there was a microphone at the point where the rod hit, so that the rod actually hit the microphone at a distance 12 meters from where I released it, and suppose that you were listening to the signal from the microphone
You will agree that in traveling that 12 meters the rod emitted a number of peaks. How long did it take the rod to get to that point, and how many peaks were produced, if we assume that the frequency of the sound produced is 2500 Hz?
Freq of steel bar is approx. 2960Hz and speed of sound is 340m/s and Listener at rest hears freq of 2500Hz we can use equation
f_L = [v/(v + v_rod)]*(f_rod) 2960Hz
2500Hz = [360m/s/(360m/s + v_rod)]* 2960Hz
approx .8446 = 360m/s/(360m/s+v_rod)
.8446*(360m/s+v_rod) =360m/s
304.056m/s + .8446v_rod = 360m/s/v_rod
.8446v_rod = approx. 56m/s
v_rod = approx. 66m/s
At 66m/s it takes
12m = 66m/s*`dt
`dt = 12m/66m/s = .1818sec
So it would take rod approx. .1818sec to get to microphone
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@& You're using an equation here, but the question is structured so as to lead you to the equation.*@
@& However I didn't give you a speed for the rod, so you didn't have much alternative.
66 m/s is roughly 50% faster than a good fastball. Javelin throw over 1000 feet. I'll work on it.
Bottom line, of course, is that the frequency didn't change that much.
Good calculation.*@
How long did it take first peak to reach the microphone?
Moving at 340m/s and traveling 12m time required is
`dt = 340m/s/12m = approx. .035sec
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Assuming that the rod traveled at 20 meters / second, how long did it take the rod to reach the microphone?
`dt = 12m/20m/s = .6sec
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From the time the first peak reached the microphone, until the rod hit the microphone, how much time elapsed?
time elapsed is
.6sec - .035sec = .565sec
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How many peaks were therefore detected by the microphone, and during what time interval were they detected?
Peaks detected are
F_mic = (360m/s+20m/s)/(360m/s/F_rod)
= (1+ (20m/s/360m/s)) *2960Hz
= approx. 3124Hz or 3124cycles/sec
Time elapsed was .565sec so that
3124cycles/sec *.565sec =approx. 1765 cycles or peaks
This is within .565sec
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What therefore was the frequency detected by the microphone?
That would be the 3124Hz I believe
?????I found all answers using different formulas from book if there was a more logical way to work through the questions could you give me an outline on what that approach would be????????
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@& I think you just did it. Had you used 20 m/s with the given formula you should have reached the same conclusion.*@
`q007. Suppose that rather than me throwing the rod at you, you threw the microphone at the rod at 20 m/s, while I held it stationary. Use an analysis similar to the one above to find the frequency that would be detected by the microphone. Your answer will be close to, but not identical with, the answer you got above.
F_mic = (360m/ - 20m/s)/(360m/s/F_rod)
= (1- (20m/s/360m/s)) *2960Hz
=approx. .9444*2960Hz
= approx. 2795.5Hz
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@& Again you should think this through using the reasoning process outlined above. However if you know the formulas and use them correctly, you'll be OK.*@
`q008. If there was a microphone stationary at the rod, and its signal was mixed with the signal from the microphone you threw, how many beats per second would be observed?
beat freq = difference in freq
F_beat = 2960Hz - 2795.5Hz = 164.5Hz = 164.5cycles/sec
Or 164.5 beats per second
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`q009. From the measurements you made in Monday's class, what do you get for the wavelength of laser light?
****
I cant recall what this pertains to
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@& We ran the laser light through the transparency and measured the distance between peaks, and the distance from the laser to the target.*@
For 110307
`q001. The log of a number is the power to which you need to raise 10 to get that number. What are the logs of the following numbers?
100
10000
.0001
10 000 000 000 000
10^5 / 10^14
100 / 10^7
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2, 4, -4, 13, -9, -5
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`q002. Quickly sketch the following graphs
log(x) vs. x from x = 1 to x = 100
log(x) vs. x from x = 10 to x = 1 000
log(x) vs. x from x = 100 000 to 10 000 000
log(x) from x = 0.1 to x = 10.
What do your graphs have in common?
Same basic curve for all graphs
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`q003. Using your graphs from before, estimate the following:
log(3)
log(5)
log(300)
log(500)
log(2 000 000)
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.3, .7, 2.4, 2.6, 6.2
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`q004. Relabel one or more of your graphs to estimate each of the following:
log(300 000 000)
log(5 000 000 000)
log(.00003)
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9.4, 10.6, ???I had a difficult time figuring out how to set this one up so finally used common sense knowing 10^-4 = .0001, I approx. it to be close to -4.3?????
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@& .0003 is between 10^-4 and 10^-3, so the log would be between -4 and -3.*@
@& The graph would have vertical coordinates -5, -4 and -3. .0003 would be three tenths of the way between 10^-4 and 10^-3. When you project up and over you'll end up about halfway between -4 and -3, which would put you at about -3.5.*@
`q005. The hearing threshold is defined to be 10^-12 watts / m^2. We can stand sounds up to intensity 1 watt / m^2 without pain. 1 watt / m^2 is called the pain threshold.
How many times more intense than the hearing threshold is the pain threshold?
What is the log of this ratio?
****
10^(-12) is 1,000,000,000,000 smaller than 1 so that Pain threshold is 1,000,000,000,000 times greater than the hearing threshold.
log ratio = log(10^12) = 12
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`q006. Based on your graph(s), what is the log of the ratio of each of the following intensities to the hearing threshold?
.0001 watts / m^2
10^-8 watts / m^2
.03 watts / m^2
5 * 10^-7 watts / m^2
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10^8/10^12, 10^4/10^12, 10^10/10^12, 5*10^5/10^12
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`q007. If the log of the ratio of intensity to hearing threshold intensity is the given number, then what is the intensity?
5
9
3.5
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5*10^12, 9*10^12, 3.5*10^12
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@& If the log of a number is 5, the number is 100 000.
So 5 would imply an intensity ratio of 100 000, or 10^5, and an intensity of 10^5 * 10^-12 = 10^-7.*@
`q008. The decibel level of a sound is 10 times the log of the ratio of its intensity to the hearing threshold intensity. What is the decibel level of each of the following, which you have seen in a previous question:
.0001 watts / m^2
10^-8 watts / m^2
.03 watts / m^2
5 * 10^-7 watts / m^2
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10dB*( 10^8/10^12) = (10^-3)dB,
10dB*10^-8 = 10^-7dB,
10dB*( 10^10/10^12) = (10^-1)dB,
10dB*( 5 * 10^-7) = 5 * (10^-6)dB,
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@& .0001 watts / m^2is 10^8 times the hearing threshold, so the log of the ratio is 8.
db = 10 * log( intensity ratio) = 10 * 8 = 80 db*@
`q009. What is the intensity of a sound with each of the following decibel levels?
80 dB
25 dB
53 dB
110 dB
`q010. What is the ratio of the intensities of two sounds whose decibel levels differ by 33 dB?
3.3
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@& That's the ratio of the logs.*@
University Physics:
`q011. If the waveforms y_1 = A cos(omega * t) and y_2 = A cos( (omega + `dOmega) t are mixed, what is the equation of the combined wave function, in terms of sines and cosines of omega * t and `dOmega * t?
What is the maximum amplitude of the resulting beats?
Acos(omega * t) + A cos( (omega + `dOmega) t)
Acos(omega * t) + [A( cos( omega*t)cos( `dOmega* t) + sin(omega*t)sin( `dOmega* t)
Acos(omega * t) + A( cos( omega*t)cos( `dOmega* t) + A sin(omega*t)sin( `dOmega* t)
Acos(omega * t)(1 + cos( `dOmega* t)), cos fn runs from -1 to 1 so inside fn 1 + cos(
`dOmega* t) goes from 0 to 2
Assuming cos(omega * t) has value of 1 when cos( `dOmega* t) has the value of 1, which is not necessarily true max value will be 2 and min value will be 0.
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@& A is the same for both, so it would be 2 A rather than 2.*@
`q012. A speaker suspended by its power cord oscillates back and forth with an amplitude of 10 cm and a frequency of 40 cycles / minute. It emits sound at 1000 Hz. A microphone is mounted in front of the speaker, and another behind it, so that when the speaker moving toward one microphone it is moving away from the other. Both speakers are at some distance from the speaker.
The sound collected by the microphones is mixed into a single sound.
What is the maximum frequency of the resulting beats?
Im not sure Im clear on how to approach this one.
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@& The speaker oscillates as a pendulum, so you can find its maximum velocity.
There will be a doppler shift in the sound detected at each microphone.
You would calculate the frequencies detected by the two microphones.*@
... energy?
... run a log, a parabola, a power, a series of such ... run at a parabola and bounce off ...
"
@& Good work on many, but not all, of these questions.*@
@& See my notes and give some of these another shot.
&#Please see my notes and, unless my notes indicate that revision is optional, submit a copy of this document with revisions and/or questions, and mark your insertions with &&&& (please mark each insertion at the beginning and at the end).
Be sure to include the entire document, including my notes.
If my notes indicate that revision is optional, use your own judgement as to whether a revision will benefit you. &#
*@