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TrueTears

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Re: TrueTears question thread
« Reply #285 on: June 29, 2009, 05:03:54 pm »
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When compression occurs this means the sound is very loud (ie it is produces loudest sound)

But when rarefaction occurs does this also produce the loudest sound?
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NE2000

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Re: TrueTears question thread
« Reply #286 on: June 29, 2009, 06:33:28 pm »
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When compression occurs this means the sound is very loud (ie it is produces loudest sound)

But when rarefaction occurs does this also produce the loudest sound?

That's the way I saw it. Because if you graph the pressure, those are both points of maximum amplitude.
« Last Edit: June 29, 2009, 06:35:01 pm by NE2000 »
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kamil9876

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Re: TrueTears question thread
« Reply #287 on: June 29, 2009, 06:37:43 pm »
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Also, if you look at the particles in some space. And see how they expand and compress etc. The motion at max compression and max expansion is the same, velocity is 0 since they turn around. Therefore kinetic energy is zero at these points. At normal air pressure however the kinetic energy is a max. (this is analogous to harmonic motion in springs). Hence the kinetic energy being zero implies that that max kinetic energy must have been changed to another energy - sound :)
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TrueTears

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Re: TrueTears question thread
« Reply #288 on: June 29, 2009, 06:41:48 pm »
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Yeah thanks, I got my head around it after posting it, I thought about it as springs like kamil said, when compression occurs, the sound wave can be thought of as "compressed spring" hence storing potential energy. When it is in rarefaction it undergoes "stretching" also like a spring hence storing potential energy. Both cases produces the loudest sound because the energy stored is the maximum energy.
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NE2000

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Re: TrueTears question thread
« Reply #289 on: June 29, 2009, 06:43:20 pm »
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Yep kamil's explanation made it clearer for sure :)
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TrueTears

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Re: TrueTears question thread
« Reply #290 on: June 29, 2009, 08:18:32 pm »
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A material absorbs 90% of the sound energy that falls upon it, what is the equivalent number of dB ?
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Re: TrueTears question thread
« Reply #291 on: June 29, 2009, 08:57:31 pm »
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let be the decibels of the sound of 100% of the sound energy, and be the decibels of 10% of the sound energy:




However



The A and t cancel out and so:




and therefore:




Hence the difference in decibels is 10dB
« Last Edit: June 29, 2009, 11:56:37 pm by kamil9876 »
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NE2000

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Re: TrueTears question thread
« Reply #292 on: June 29, 2009, 09:11:19 pm »
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Or you could just consider it in terms of the attenuation or loss formula from Unit Three. As intensity is proportional to power, it just comes out to be 0.1 in the log and you get -10 dB. Which equates to a loss of 10 dB.

Which now that I actually fully read kamil's post is basically just what kamil did. :laugh:
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TrueTears

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Re: TrueTears question thread
« Reply #293 on: June 29, 2009, 09:36:34 pm »
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Thanks.

2 students pat and morgan are discussing an experiment testing the nature of sound. They imagine a speaker with a dust particle motionless in front of it and discuss what will happen to the particle when the speaker is turned on. pat says that since there is energy transferred by the wave the particle will gain energy and a succession of little impulses will push the particle continuously away from the speaker. Morgan agrees that energy is carried by the wave however the result of pressure variations will cause the dust particle to move back and forth about its original position.

Which one or more of the follow statements is true
A. Pat is correct
B. morgan is correct
C. A longitudinal pressure wave interacts with the dust particle
D. The particle will move from high pressure areas into low pressure areas

I know the answer is B C and D, but I'm wondering how exactly does compression and rarefaction in a sound wave cause the particle to oscillate?
« Last Edit: June 29, 2009, 09:38:06 pm by TrueTears »
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TrueTears

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Re: TrueTears question thread
« Reply #294 on: June 30, 2009, 01:44:10 am »
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The threshold of hearing varies with frequency. To calculate intensity in decibels, I and Io must be at the same frequency. [the question will provide you with enough information to solve it]

Thanks Mao, but for sound intensity level, you know how if the frequency is lower then a higher sound intensity level is needed in order to have the same loudness with if the frequency was higher then a lower sound intensity is needed.

So

If the frequency decreases then the sound wave carries less energy so you will need a higher "I", but this I is at a different frequency to so how do you work out L?

Furthermore if the frequency increases then the sound wave carries more energy so a lower "I" is needed to produce same loudness, but again this "I" is at a different frequency to so how do you work out L?
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Mao

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Re: TrueTears question thread
« Reply #295 on: June 30, 2009, 01:52:23 am »
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The threshold of hearing varies with frequency. To calculate intensity in decibels, I and Io must be at the same frequency. [the question will provide you with enough information to solve it]

Thanks Mao, but for sound intensity level, you know how if the frequency is lower then a higher sound intensity level is needed in order to have the same loudness with if the frequency was higher then a lower sound intensity is needed.

So

If the frequency decreases then the sound wave carries less energy so you will need a higher "I", but this I is at a different frequency to so how do you work out L?

Furthermore if the frequency increases then the sound wave carries more energy so a lower "I" is needed to produce same loudness, but again this "I" is at a different frequency to so how do you work out L?

Io changes with frequency. To calculate intensity at a frequency other than 1000hz, you will be given the respective threshold of hearing. (Or a graph to extrapolate the data from)

Another formula to remember for intensity is , this one is usually more useful.
« Last Edit: June 30, 2009, 01:54:55 am by Mao »
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TrueTears

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Re: TrueTears question thread
« Reply #296 on: June 30, 2009, 02:05:53 am »
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Okay, thanks Mao.

Also is the fundamental frequency always the one that resonates the strongest?
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NE2000

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Re: TrueTears question thread
« Reply #297 on: June 30, 2009, 09:30:32 am »
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Thanks.

2 students pat and morgan are discussing an experiment testing the nature of sound. They imagine a speaker with a dust particle motionless in front of it and discuss what will happen to the particle when the speaker is turned on. pat says that since there is energy transferred by the wave the particle will gain energy and a succession of little impulses will push the particle continuously away from the speaker. Morgan agrees that energy is carried by the wave however the result of pressure variations will cause the dust particle to move back and forth about its original position.

Which one or more of the follow statements is true
A. Pat is correct
B. morgan is correct
C. A longitudinal pressure wave interacts with the dust particle
D. The particle will move from high pressure areas into low pressure areas

I know the answer is B C and D, but I'm wondering how exactly does compression and rarefaction in a sound wave cause the particle to oscillate?

Just imagine those diagrams of compression and rarefaction. The whole idea of waves is a transfer of energy without 'net' movement of particles. But there is still particle movement. So if you have an animation or a diagram or something you can see all those small dots moving out (rarefaction) and in (compression). That's the oscillation. The dust particle just represents one of those small dots.
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TrueTears

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Re: TrueTears question thread
« Reply #298 on: June 30, 2009, 06:09:19 pm »
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Thanks.

If you have a sound source of one frequency placed at the open end of a pipe that is closed the other end. Then only ONE standing wave will form right? [by ONE standing wave I mean a standing wave of only a certain frequency, not like heaps of frequencies mixed together]
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NE2000

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Re: TrueTears question thread
« Reply #299 on: June 30, 2009, 06:53:08 pm »
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Thanks.

If you have a sound source of one frequency placed at the open end of a pipe that is closed the other end. Then only ONE standing wave will form right? [by ONE standing wave I mean a standing wave of only a certain frequency, not like heaps of frequencies mixed together]

Based on what you said on the thread on multiple harmonics I would think yes. A sound wave with a node at the open end and an antinode at the closed end will form and the frequency will depend on the frequency of sound you put into the pipe.
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