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September 21, 2025, 01:20:21 pm
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Author Topic: resonant frequency question 13 vcaa 2009  (Read 1139 times)  Share 

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TyErd

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resonant frequency question 13 vcaa 2009
« on: November 10, 2010, 09:22:40 am »
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help? lol
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Whatlol

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Re: resonant frequency question 13 vcaa 2009
« Reply #1 on: November 10, 2010, 09:25:09 am »
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length stays the same but now you do lambda = 4L
so v/4l = f
333 /(4 x 0.432) = 193
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SixWinged

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Re: resonant frequency question 13 vcaa 2009
« Reply #2 on: November 10, 2010, 09:27:41 am »
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Alternatively, think of the diagrammatic model for closed/open pipes. The closed end pipe shows 1 quarter of a wavelength at the fundamental while the open end pipe shows half a wavelength. From that it can be said that, for the same length pipe, the wavelength is double in the closed end pipe (as only 1/4 of it can fit in the pipe), and hence the frequency is halved. 385/2 = 193
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TyErd

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Re: resonant frequency question 13 vcaa 2009
« Reply #3 on: November 10, 2010, 09:31:44 am »
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so we can disregard the n in the formula?
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Whatlol

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Re: resonant frequency question 13 vcaa 2009
« Reply #4 on: November 10, 2010, 09:35:06 am »
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Quote from: TyErd on November 10, 2010, 09:31:44 am
so we can disregard the n in the formula?

its for fundemental frequency so n = 1
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TyErd

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Re: resonant frequency question 13 vcaa 2009
« Reply #5 on: November 10, 2010, 09:53:07 am »
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oh yeah thanks, also can you help me with question 11 in vcaa 2008
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TyErd

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Re: resonant frequency question 13 vcaa 2009
« Reply #6 on: November 10, 2010, 09:54:35 am »
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for sound btw
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Re: resonant frequency question 13 vcaa 2009
« Reply #7 on: November 10, 2010, 09:57:32 am »
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k so just do lambda = 4L
0.325 /4 = 0.81 m
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TyErd

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Re: resonant frequency question 13 vcaa 2009
« Reply #8 on: November 10, 2010, 10:00:39 am »
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oh yeah thanks, umm i dont really get question 10 either
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Re: resonant frequency question 13 vcaa 2009
« Reply #9 on: November 10, 2010, 10:10:22 am »
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when you have a resonance frequency, a standing wave is established and  constructive interference occurs which will increase the intensity of the sound , so you will expect the sound to be louder.
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TyErd

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Re: resonant frequency question 13 vcaa 2009
« Reply #10 on: November 10, 2010, 10:13:04 am »
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okay thanks, also is the position - time graphs the opposite of pressure -time graphs??
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TyErd

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Re: resonant frequency question 13 vcaa 2009
« Reply #11 on: November 10, 2010, 10:15:38 am »
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like say if you're given the pressure time graph, how would you draw the displacement time graph
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Re: resonant frequency question 13 vcaa 2009
« Reply #12 on: November 10, 2010, 10:22:48 am »
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i think the point of 0 pressure is max displacement
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TyErd

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Re: resonant frequency question 13 vcaa 2009
« Reply #13 on: November 10, 2010, 10:25:06 am »
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ok thanks
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