# Homework Help: Harmonic Waves and pipes

1. Apr 6, 2009

### metsjetsfan

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

1. The problem statement, all variables and given/known data

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?

2. Relevant equations

I actually found this equation on my own and my teacher didn't use it, but here it is:

f=nv/2L

3. 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.

2. Apr 6, 2009

### LowlyPion

For an open ended column then you have the wavelength is the length of the column.

http://www.glenbrook.k12.il.us/GBSSCI/PHYS/Class/sound/u11l5c.html [Broken]

You get a frequency of 1039hz looks like to me.

The third harmonic for an open ended pipe is one more 1/2 wavelength. (3 times the fundamental).

For the closed end pipe:
http://www.glenbrook.k12.il.us/GBSSCI/PHYS/Class/sound/u11l5d.html [Broken]

Recall that the wavelength of the tuning fork before was 33cm, so 1/4 of that would be 33/4 wouldn't it?

Last edited by a moderator: May 4, 2017
3. Apr 6, 2009

### metsjetsfan

so is the answer to part c 1559 then????

Last edited: Apr 6, 2009
4. Apr 6, 2009

### metsjetsfan

and for part d, my problem is that the fundamental frequency for the original tuning fork would be 520 hz because that is half of the 1039 hz from part b.

but then that means the wavelength= 340 m/s / 520 hz = .65 m. And then 1/4 of .65= .1625.

I understand the shortcut for why it is 33/4 but why does it not work the long way?