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.
016. `Query 14
Question:
`qquery Principles of Physics and General College Physics
12.40: Beat frequency at 262 and 277 Hz; beat frequency two octaves lower.
Your solution:
Confidence Rating:
Given Solution:
`aThe beat frequency is the difference in the frequencies,
in this case 277 Hz - 262 Hz = 15 Hz.
One ocatave reduces frequency by half, so two octaves lower
would give frequencies 1/4 as great. The
difference in the frequencies would therefore also be 1/4 as great, resulting
in a beat frequency of 1/4 * 15 Hz = 3.75 Hz.
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Question:
What frequency is received by a person watching an oncoming ambulance moving at 110 km/h and emitting a steady 800-Hz sound from its siren? The speed of sound on this day is 345 m/s.
What frequency does she receive after the ambulance has passed?
Your solution:
Confidence Rating:
Given Solution:
The general formula for the Doppler shift is
f ' = (c + v_
where v_r is the velocity of the receiver, v_s the velocity of the source, and c the speed of sound. f is the frequency of the source, f ' the frequency observed by the receiver. v_r is considered positive if the receiver is moving in the direction of the source, and v_s is considered positive if the source is moving in the direction of the receiver.
In this case, the receiver is presumably stationary, while the source (the ambulance) is moving at 110 km/h = 110 * 1000 m / (3600 s) = 30 m/s. The received frequency is thus
f ' = (345 m/s + 0) / (345 m/s - 30 m/s) * 800 Hz = 876 Hz.
After the ambulance has passed, it is no longer approaching the source and its velocity is negative. In this case we have
f ' = (345 m/s + 0) / (345 m/s - (- 30 m/s)) * 800 Hz = 736 Hz.
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36. Two eagles fly directly toward one another, the first at 15.0 m/s and the
second at 20.0 m/s (both relative to the air). Both screech, the first one
emitting a frequency of 3200 Hz and the second one emitting a frequency of 3800
Hz. What frequencies do they receive if the speed of sound is 330 m/s?
Question:
Your solution:
Confidence Rating:
Given Solution:
The general formula for the Doppler shift is
f ' = (c + v_
where v_r is the velocity of the receiver, v_s the velocity of the source, and c the speed of sound. f is the frequency of the source, f ' the frequency observed by the receiver. v_r is considered positive if the receiver is moving in the direction of the source, and v_s is considered positive if the source is moving in the direction of the receiver.
Let's assume that the first eagle is the source and the second the receiver. In this case, the receiver is approaching the source at 20 m/s, while the source is approaching the receiver at 15 m/s, and the source is emitting frequency 3200 Hz. The received frequency is thus
f ' = (330 m/s + 20 m/s) / (330 m/s - 15 m/s) * 3200 Hz = 3556 Hz, approx.
In the second case the first eagle is the receiver, the second the source. The source is approaching the receiver at 20 m/s, the receiver approaching the source at 15 m/s, so
f ' = (330 m/s + 15 m/s) / (330 m/s - 20 m/s) * 3800 Hz = 4230 Hz, approx.
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Question:
`qquery gen phy problem 12.46 speakers 1.8 meters apart,
listener three meters from one and 3.5 m from the other ****
gen phy what is the lowest
frequency that will permit destructive interference at the location of the
listener?
Your solution:
Confidence Rating:
Given Solution:
`aSTUDENT SOLUTION:
To solve this problem, I first realize that for destructive interference
to occur, the path difference is an odd multiple of half of the wavelength(ex.
1/2, 3/2, 5/2). I used Fig.12-17 in the
text to help visualize this problem.
Part (A) of the problem asks to calculate the lowest frequency at which
destructive interference will occur at the point where two loudspeakers are
2.5m apart and one person stands 3.0m from one speaker and 3.5m from the other. The text states that 'destructive
interference occurs at any point whose distance from one speaker is greater
than its distance from the other speaker by exactly one-half wavelength.' The path difference in the problem is fixed,
therefore the lowest frequency at which destructive interference will occur is
directly related to the longest wavelength.
To calculate the lowest frequency, I first have to calculate the longest
wavelength using the equation
'dL ='lambda/2, where `dL is the path difference.
'lambda=2*'dL
=2(3.5m-3.0m)=1m
Now I can calculate the frequency using
f=v/'lambda. The
velocity is 343m/s which is the speed of sound.
f=343m/s/1m=343 Hz.
Thus, the lowest frequency at which destructive interference
can occur is at 343Hz.
Keeping in mind that destructive interference occurs if the
distance equals an odd multiple of the wavelength, I can calculate (B) part
of the
problem.
To determine the next wavelength, I use the equation
'dL=3'lambda/2
wavelength=2/3(3.5m-3.0m)
=0.33m
Now I calculate the next highest frequency using the
equation f=v/wavelength.
f^2=343m/s/0.33m=1030Hz.
I finally calculate the next highest frequency.
'del L=5/2 'lambda
wavelength=0.20m
f^3=343m/s/0.2m=1715 Hz.
INSTRUCTOR EXPLANATION:
The listener is .5 meters further from one speaker than from
the other. If this .5 meter difference
results in a half-wavelength lag in the sound from the further speaker, the
peaks from the first speaker will meet the troughs from the second. If a half-wavelength corresponds to .5
meters, then the wavelength must be 1 meter.
The frequency of a sound with a 1-meter wavelength moving at 343 m/s
will be 343 cycles/sec, or 343 Hz.
The next two wavelengths that would result in destructive
interference would have 1.5 and 2.5 wavelengths corresponding to the .5 m path
difference. The wavelengths would
therefore be .5 m / (1.5) = .33 m and .5 m / (2.5) = .2 m, with corresponding
frequencies 343 m/s / (.33 m) = 1030 Hz and 343 m/s / (.2 m) = 1720 Hz, approx.
****
**** gen phy
why is there no highest frequency that will permit destructive
interference?
** You can get any number of half-wavelengths into that .5
meter path difference. **
STUDENT COMMENT:
After reading the solution I understand the formula I
am supposed to use a bit better, but I am still kind of confused about
the concept of destructive interference.
INSTRUCTOR RESPONSE:
As you change position the relative alignment of 'peaks' and
'valleys' change.
Sometimes peaks from one path arrive at the same time as peaks from the other
(in which case valleys will arrive with valleys), and the interference is
constructive.
Sometimes peaks from one path arrive at the same time as valleys from the other,
and the interference is destructive.
When one path is a whole number of wavelengths longer than the other, peaks meet
peaks and the waves reinforce.
When one path is a half a wavelength longer than the other, peaks meet valleys
and the waves cancel; the same happens when one path is half a wavelength plus a
whole number of wavelengths longer than the other.
STUDENT QUESTION
I got the lowest frequency fine. And I was on the right track with my reasoning.
But which way do you suggest to solve problems like this: your way, or the
student’s solution.
INSTRUCTOR RESPONSE
The two solutions are completely equivalent. If you really
understand it one way, you'll understand it the other.
However in the interest of time, you should pick one. Whichever way makes more sense to you, that's the way you should think of it.
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Question:
`qgen phy what must
happen in order for the sounds from the two speakers to interfere
destructively, assuming that the sources are in phase?
Your solution:
Confidence Rating:
Given Solution:
`a** The path difference has to be and integer number of wavelengths plus a half wavelength. **
STUDENT QUESTION
The book tells me that for these two speakers to
interfere destructively, the distance from one speaker has to be greater than
its distance from the other speaker by one-half wavelength. Destructive
interference would occur if the distance would equal 1/2, 3/2, 5/2,…
wavelengths.
INSTRUCTOR RESPONSE
That is correct.
The given solution to the original problem says this as well.
The book's explanation of course gives you a third option for the most appropriate way to think of the problem.
In any case you need to understand why those path differences result in destructive interference. Once you're clear on that, a wide variety of interference problems become pretty straightforward.
CRAB NEBULA PROBLEM?
This Query will exit.
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