What are the resonance heights for a tuning fork and resonance tube experiment?

[endeavor]

"A tuning fork with a frequency of 440 Hz is held above a resonance tube partially filled with water. Assuming that the speed of sound in air is 342 m/s, for what heights of the air column will resonances occur?"

f₁ = 440 Hz
v = 342m/s
I'm not sure if this is referring to an open pipe, or a closed pipe.
For an open pipe,
f_n = (nv)/(2L) = nf₁
v/(2L) = f₁
L = v/(2f₁)
L = 0.3886 m
But this is the length for the sound traveling up and down... i think. So the length is 0.194m. However, the answer gives 3 values, 0.194m, 0.583m, 0.972m. Even for a closed pipe, I there is only 1 value...
I must be doing something wrong. I don't think the speed of sound in water is a factor, because my first answer seems to be correct...

 

[Andrew Mason]

"A tuning fork with a frequency of 440 Hz is held above a resonance tube partially filled with water. Assuming that the speed of sound in air is 342 m/s, for what heights of the air column will resonances occur?"

f₁ = 440 Hz
v = 342m/s
I'm not sure if this is referring to an open pipe, or a closed pipe.

This is a closed pipe. There is a node at the bottom and an anti-node at the open end. So resonances occur where the length of the pipe is L = \lambda/4, 3\lambda/4, 5\lambda/4,... (2n+1)\lambda/4

Using the universal wave equation:

\lambda = v/f

Substituting the resonance criterion: \lambda = 4L/(2n+1) = v/f

resonance occurs at:

L = (2n+1)v/4f = (2n+1)*342/4*440 = (2n+1).194 m.

AM

 

[endeavor]

This is a closed pipe. There is a node at the bottom and an anti-node at the open end. So resonances occur where the length of the pipe is L = \lambda/4, 3\lambda/4, 5\lambda/4,... (2n+1)\lambda/4

Using the universal wave equation:

\lambda = v/f

Substituting the resonance criterion: \lambda = 4L/(2n+1) = v/f

resonance occurs at:

L = (2n+1)v/4f = (2n+1)*342/4*440 = (2n+1).194 m.

AM

That makes sense, except for why f is kept as the initial frequency. The formula I have is:
L = (mv)/(4f_m) where m = 1,3,5,...
and
f_m = mf₁
thus
L = v/(4f₁)
why do I use only the initial frequency here?

 

[Andrew Mason]

That makes sense, except for why f is kept as the initial frequency. The formula I have is:
L = (mv)/(4f_m) where m = 1,3,5,...
and
f_m = mf₁
thus
L = v/(4f₁)
why do I use only the initial frequency here?

In this case, there is a tuning fork which provides only the fundamental frequency - a pure sine wave. There are no higher frequencies to resonate in the air column. So you are dealing with the fundamental frequency only.

AM

 

[endeavor]

In this case, there is a tuning fork which provides only the fundamental frequency - a pure sine wave. There are no higher frequencies to resonate in the air column. So you are dealing with the fundamental frequency only.

AM

Oh, Ok. Thanks!