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metsjetsfan
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So my teacher is horrible, and I pretty much have to teach myself everything. With this question, I am getting different answers than my teacher, and my teacher has been unable to explain to me satisfactorily why and how she got her answers.
So please help me out, and even give some kind of concept explanation if you can.
Thanks
A sound wave resonates inside an open pipe filled with air at room temperature at 2nd harmonic and 1st overtone. The length of the pipe is 33 cm.
a-determine the wavelength of the resonating sound wave
b-determine the frequency of the tuning fork
c-determine the next higher frequency that will resonate in a pipe of this length
d-if the open pipe is replaced with a pipe which is closed at one end, what would have to be the length of the closed pipe for the original tuning fork to resonate at its fundamental frequency?
I actually found this equation on my own and my teacher didn't use it, but here it is:
f=nv/2L
(a)-pretty simple i thought. wavelength=2L/n so wavelength= 33 cm
(b)-this was ok, though not completely sure.
f=nv/2L=2*343/(2*0.33)=1039 hz
(c)-this is where the problems arrive:
since i believe it's asking for the third harmonic (the next higher frequency that will resonate) i thought i should just do:
f=nv/2L so f= 3*343/ (2*0.33) = 1559 hz
but my teacher said that L= 2*wavelength, and wavelength = L/2 = 0.33/2= .166
and f = 343/.166 = 2078 hz
but i think that is the fourth harmonic, which is not what we're looking for. This is where my problem comes in terms of concept.
(d) ok this was the hardest part. the diagram shows what looks like a fourth of a wavelength in the pipe with a node at the closed end. so i thought that I should set the fundamental frequency of the first tuning fork, which i found to be 520, equal to v/4L.
I then got 4L = 0.66, L= 16.5 cm
But my teacher just did L=wavelength / 4 = .33/4 = 8.25 cm
PLEASE HELP ME. My teacher is killing me, and I could really use guidance on this stuff.
Thanks to anyone who responds.
So please help me out, and even give some kind of concept explanation if you can.
Thanks
Homework Statement
A sound wave resonates inside an open pipe filled with air at room temperature at 2nd harmonic and 1st overtone. The length of the pipe is 33 cm.
a-determine the wavelength of the resonating sound wave
b-determine the frequency of the tuning fork
c-determine the next higher frequency that will resonate in a pipe of this length
d-if the open pipe is replaced with a pipe which is closed at one end, what would have to be the length of the closed pipe for the original tuning fork to resonate at its fundamental frequency?
Homework Equations
I actually found this equation on my own and my teacher didn't use it, but here it is:
f=nv/2L
The Attempt at a Solution
(a)-pretty simple i thought. wavelength=2L/n so wavelength= 33 cm
(b)-this was ok, though not completely sure.
f=nv/2L=2*343/(2*0.33)=1039 hz
(c)-this is where the problems arrive:
since i believe it's asking for the third harmonic (the next higher frequency that will resonate) i thought i should just do:
f=nv/2L so f= 3*343/ (2*0.33) = 1559 hz
but my teacher said that L= 2*wavelength, and wavelength = L/2 = 0.33/2= .166
and f = 343/.166 = 2078 hz
but i think that is the fourth harmonic, which is not what we're looking for. This is where my problem comes in terms of concept.
(d) ok this was the hardest part. the diagram shows what looks like a fourth of a wavelength in the pipe with a node at the closed end. so i thought that I should set the fundamental frequency of the first tuning fork, which i found to be 520, equal to v/4L.
I then got 4L = 0.66, L= 16.5 cm
But my teacher just did L=wavelength / 4 = .33/4 = 8.25 cm
PLEASE HELP ME. My teacher is killing me, and I could really use guidance on this stuff.
Thanks to anyone who responds.