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October 22, 2025, 01:27:43 pm
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Author Topic: Dekoyl's Questions  (Read 24822 times)  Share 

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kamil9876

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Re: Dekoyl's Questions
« Reply #105 on: August 16, 2009, 08:22:49 pm »
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drops to a lower energy level. So change is negative.

It's simply the conservation of energy:








is positive since it went from 0(wasn't there) to something.
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Voltaire: "There is an astonishing imagination even in the science of mathematics ... We repeat, there is far more imagination in the head of Archimedes than in that of Homer."

dekoyl

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Re: Dekoyl's Questions
« Reply #106 on: August 26, 2009, 08:51:31 pm »
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Brain fart: Is it possible to calculate the frequency of electrons?

I'm doing my English language analysis at the moment and this popped into my head. :| Not cool
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/0

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Re: Dekoyl's Questions
« Reply #107 on: August 26, 2009, 08:56:57 pm »
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Not sure but I'm guessing that you can get it from the de Broglie wavelength



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dekoyl

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Re: Dekoyl's Questions
« Reply #108 on: August 26, 2009, 09:00:55 pm »
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Oh okay thanks /0. I just flipped through the Jacaranda book and there's a question that just leaves the frequencies of electrons blank (p351)
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Mao

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Re: Dekoyl's Questions
« Reply #109 on: August 26, 2009, 10:06:48 pm »
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Mao weighs 70kg, he is walking at 1m/s. Hence, he has a wavelength of 9.47E-36 m, hence a frequency of 1.06E35 Hz.

TAKE THAT GAMMA RAY, I BEAT YOU BY 10^16 = 10 MILLION BILLION TIMES.
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dekoyl

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Re: Dekoyl's Questions
« Reply #110 on: November 10, 2009, 01:23:58 am »
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A tube of length 0.765m closed at one end, with the speaker at the open end and a microphone is placed 0.085m from the closed end:

[_O___________ <|
   ^mic                ^speaker

The frequency of the sound from the speaker is increased from 0Hz until the microphone detects a maximum intensity of sound.
Q: What number harmonic is presen when the microphone detects maximum amplitude for the first time?

Thanks  :(
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/0

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Re: Dekoyl's Questions
« Reply #111 on: November 10, 2009, 01:29:45 am »
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Wouldn't it be the first harmonic? If you're increasing frequency from 0Hz the first resonance you'll come across is the fundamental.
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dekoyl

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Re: Dekoyl's Questions
« Reply #112 on: November 10, 2009, 01:44:00 am »
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Quote from: /0 on November 10, 2009, 01:29:45 am
Wouldn't it be the first harmonic? If you're increasing frequency from 0Hz the first resonance you'll come across is the fundamental.
The answers says 5th harmonic.

I can't offer any reasons as to why because I'm learning as I do trial exams (ha.)
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TrueTears

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Re: Dekoyl's Questions
« Reply #113 on: November 10, 2009, 02:20:29 am »
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Tricky Q this one:

There must be an antinode at the mic.

Thus the wavelength is (Since there's an antinode at the closed end)

So now





Now we have n = 1,3,5,7, 9

Which is the 5th resonant freq AKA 9th harmonic. I think answer is wrong if it says 5th harmonic.
« Last Edit: November 10, 2009, 02:23:07 am by TrueTears »
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dekoyl

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Re: Dekoyl's Questions
« Reply #114 on: November 10, 2009, 02:31:33 am »
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Thanks TT!

This is the exact solution:

Quote
Therefore there must be wavelengths present in the tube at this time. This is the 5th harmonic.

That's the exact working out they have (which I don't understand). :(
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TrueTears

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Re: Dekoyl's Questions
« Reply #115 on: November 10, 2009, 02:38:28 am »
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Yeah that's just like my solution, I'm not sure what its talking about 2.25 wavelength present, but I'm pretty sure that is meant to be the 5th resonant freq. Probably a mistake (this was a neap exam right?). So it should be 9th harmonic.
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kamil9876

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Re: Dekoyl's Questions
« Reply #116 on: November 10, 2009, 02:45:31 am »
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What they did is count how many 1/4 wavelengths are in the pipe for the largest wavelength that satisfies the condition pointed out by TT. ie coz 9*(1/4)=2 and a quarter.

if u look at the patterns formed for this kind of tube you can find the formula:
(prove this lol, you can do so by induction which I know you have learnt from umep :P)

Knowing that 9 quarters of a wavelength fit inside, we solve this as: 2n-1=9 which gives n=5
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naved_s9994

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Re: Dekoyl's Questions
« Reply #117 on: November 10, 2009, 07:48:31 am »
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Quote from: /0 on August 26, 2009, 08:56:57 pm
Not sure but I'm guessing that you can get it from the de Broglie wavelength





What H would you use?
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crappy

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Re: Dekoyl's Questions
« Reply #118 on: November 10, 2009, 01:26:17 pm »
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Quote from: naved_s9994 on November 10, 2009, 07:48:31 am
Quote from: /0 on August 26, 2009, 08:56:57 pm
Not sure but I'm guessing that you can get it from the de Broglie wavelength





What H would you use?

You would use H in joules, it would be illogical to use in in electron volts.
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ElectricalEng@Monash (2nd year)

dekoyl

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Re: Dekoyl's Questions
« Reply #119 on: November 10, 2009, 03:47:50 pm »
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We have a 2 open-ended tube and a 1 open-ended tube (with the other end closed). For a scenario (not necessarily a tube) that has TWO closed ends, are both ends an antinode?
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