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Speakers & sound waves |
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| Nov10-07, 08:15 PM | #1 |
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Speakers & sound waves
Two loudspeakers are placed side by side a distance d apart. A listener observes constructive interference while standing in front of the loudspeakers, equidistant from both of them. The distance from the listener to the point half-way between the speakers is l.
One of the loudspeakers is then moved directly away from the other. Once the speaker is moved a distance r from its original position, the listener, who is not moving, observes destructive interference for the first time. Find the speed of sound v in the air if both speakers emit a tone of the same frequency f. http://session.masteringphysics.com/...11/104507C.jpg ______________________________ I know that velocity = (wavelength)(frequency), and the path length difference for the case of destructive interference is =0.5(wavelength). And the distance between the observer and the speaker that has been moved is sqrt((0.5d+r)^2 + l^2). How do I put everything together to get the speed of sound in the air using only those variables introduced in the problem???? |
| Nov10-07, 10:26 PM | #2 |
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What is the distance between the observer and the speaker that has not been moved?
What is the difference between the two distances? |
| Nov10-07, 11:09 PM | #3 |
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now what???? I already had these in my head, but I didn't know how to follow through and put it together. Anyone can help me? |
| Nov10-07, 11:28 PM | #4 |
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Speakers & sound waves
So far so good. That's the difference expressed in terms of d, l and r. What is it expressed in terms of wavelength?
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| Nov10-07, 11:33 PM | #5 |
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I don't know. Every wavelength cycle measures 2pi so (wavelength)(sqrt((0.5d+r)^2 + l^2)-sqrt(0.5d^2+l^2)) / 2pi . Am I right?
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| Nov11-07, 12:39 AM | #6 |
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You do know! "I know that ... the path length difference for the case of destructive interference is =0.5(wavelength). "
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| Nov11-07, 08:44 AM | #7 |
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You know, the simplest way to do this question is to make use of the young's double slit experiment equations? This is effectively the same thing, only the waves are sound waves and the screen is the person.
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