# Resonance in air column

1. May 4, 2006

### jimmy_neutron

Here's the situation: you have a pipe in a bucket of water and can vary the length of the air column in the pipe by raising or lowering it.

Here's my question, hope someone can help.
The speed of sound in helium is three times greater than the speed of sound in air (340m/s). Under these conditions will there be any resonances as you change the pipe length. If so what lengths of the pipe will give you resonances (first three resonant lengths).

Where do I get started? thanks!!

2. May 4, 2006

### hellraiser

The frequencies possible in a pipe closed one end are
f = nv / (4l)
n is an odd number (1,3,5,...)
So for first three resonant freqs. you have to put the value of n as 1,3,5 respectively and solve.

3. May 4, 2006

### andrevdh

Using the basic wave equation
$$v=\lambda f$$
The wavelength in air will be
$$\lambda_a=\frac{v_a}{f}$$
The wavelength in helium gas will be
$$\lambda_h\frac{v_h}f}$$
which comes to
$$\lambda_h=\frac{3v_a}{f}$$
therefore the wavelength in helium will be three times of that in air
$$\lambda_h=3\lambda_a$$
Now as you raise the pipe resonance will form when its length is equal to one quarter of a wavelength $l=\frac{\lambda}{4}$, the next resonance will form as the pipe length can hold what fraction of the wavelength?

Last edited: May 4, 2006
4. May 4, 2006

### jimmy_neutron

Ok, so for the first resonance it will be wavelength/4
For the second resonance (the third harmonic) it will be 3x wavelength/4
And for the third resonance it will be 5x wavelength/4

How do I find the wavelength in order to calculate the length of the pipe?

thanks.

5. May 4, 2006

### nrqed

If you are looking for numerical values of lengths, you are stuck. It seems to me that the question is incomplete. Are you sure they don't provide the frequency of the source of sound?? Without that, you can't answer the question! (if they give the speed of sound, this suggests that they must have given the frequency somewhere)

6. May 5, 2006

### andrevdh

One can conclude from the wave equation
$$v=\lambda f$$
that if the frequency changes the wavelength will change and therefore the length at which standing waves will form, no matter in which medium you perform the experiment. All one can say from such an analysis is how the resonance lengths in air will relate to those in helium. I assume that since the lengths in air is known according to the formulas to be
$$l_1=\frac{\lambda}{4}$$
the corresponding length for the same frequency in helium will be
$$3l_1$$
see if you can find the other two in terms of $l_1$

Last edited: May 5, 2006