# Resonant frequencies in a tube with adjustable length

• Von Neumann
In summary: If you only hit the fork at the fundamental frequency, it will make only that frequency. So the question is meaningless.
Von Neumann
Problem:

Determine the quantity of resonant frequencies that a 1024 Hz tuning fork will have in a 1m long tube with an adjustable water level. Find the length of the air column for each frequency. Assume the speed of sound is 344m/s.

Solution:

The resonant frequencies are odd integer multiples of the fundamental so,

f=(2n-1)f'

(where f' is the 1st resonant, n the number of nodes, and f is the nth resonant frequency)

f=(2n-1)v/$\lambda$
=(2n-1)v/(4l)
= [(n-1/2)v]/(2l)

Solving for l,

l=[(n-1/2)v]/(2f)

Substituting values of n into the equation until l > 1m,

n=1
l=8.4cm

n=2
l=25.2cm

n=3
l=42.0cm

n=4
l=58.8cm

n=5
l=75.6cm

n=6
l=92.4cm

Plugging in n=7 yields a length of 1.08 m, so the wave is no longer within the tube. Therefore there are 6 resonant frequencies.

Is this correct?

Last edited:
I think unfortunately this question as stated is meaningless. Tuning forks only make one frequency so why is the question asking about the quantity of resonant frequencies? This has misdirected your thinking.

If the question actually is asking " at what lengths of tube up to 1m does resonance occur for this single frequency?" then I agree with your answers but not your reasoning.

The condition for resonance in a half closed pipe is:
$$L=\frac{λ}{4}(2n-1)$$
Where λ has one value given by 344/1024.
Your final equation is equivalent to this if you make f=1024 only ie only one frequency, which is what you must have done.
If we can have any frequency as long as it is an odd multiple of 1024Hz then there are infinite answers since we have an infinite number of possible λ values.
ie
if f=1024Hz we get 6 lengths
if f=3*1024 we get 18 lengths
if f=5*1024 we get 30 lengths
if f=(2n-1)*1024 we get 6*(2n-1) lengths

Which clearly leads to an infinite sum.

Von Neumann said:
(where f' is the 1st resonant, n the number of nodes, and f is the nth resonant frequency)

f=(2n-1)v/$\lambda$
What exactly is λ here? I think you want the wavelength that goes with f, not f'. So the equation becomes: fn=v/λn=(2n-1)f'. The harmonics have higher frequencies, so shorter wavelengths.
This leads to the infinite sequence apelling demonstrates.

haruspex said:
What exactly is λ here? I think you want the wavelength that goes with f, not f'. So the equation becomes: fn=v/λn=(2n-1)f'. The harmonics have higher frequencies, so shorter wavelengths.
This leads to the infinite sequence apelling demonstrates.

As already stated, that is precisely the equation I used; namely f=(2n-1)f'. λ is the wavelength associated with the specific tuning fork used in the experiment. Essentially, a tuning fork is an instrument that resonates at a certain constant pitch when struck. In this case the frequency is 1024 Hz. Since f = v/λ => λ = v/f = 344/1024.

As apelling stated, since the condition for resonance in an open-end pipe only supports odd multiples of quarter wavelengths, we get l=λ/4(2n-1). Since the maximum length l of the adjustable air column is 1m, there is only a finite number permitted within the tube for the frequency at hand. Hope that clears it up.

Thanks guys.

Von Neumann said:
since the condition for resonance in an open-end pipe only supports odd multiples of quarter wavelengths, we get l=λ/4(2n-1).
What you have calculated is the number of lengths of air column that will resonate to a given frequency, 1024Hz. The question clearly asks for a number of frequencies. This led both apelling and me to conclude you needed to consider harmonics of 1024Hz, and we both arrived at the answer that there would be an infinite number of them.
But I now see that the question is quite misleading, though you seem to have read it as intended. It's not the "resonant frequencies that a 1024 Hz tuning fork will have", it's the fundamental frequencies of the pipe that will have 1024Hz as a harmonic.

## What are resonant frequencies in a tube with adjustable length?

Resonant frequencies in a tube with adjustable length refer to the specific frequencies at which a tube will vibrate and produce a standing wave when air is blown into it. These frequencies are dependent on the length of the tube and can be adjusted by changing the length of the tube.

## How do resonant frequencies in a tube with adjustable length affect sound production?

The resonant frequencies in a tube with adjustable length determine the pitch of the sound produced. When air is blown into the tube, it will vibrate at its resonant frequencies, producing distinct pitches. By adjusting the length of the tube, the resonant frequencies and therefore the pitch of the sound can be changed.

## What is the formula for calculating resonant frequencies in a tube with adjustable length?

The formula for calculating resonant frequencies in a tube with adjustable length is f = nv/4L, where f is the frequency, n is the harmonic number (1, 2, 3, etc.), v is the speed of sound, and L is the length of the tube.

## What is the relationship between tube length and resonant frequencies in a tube with adjustable length?

The length of the tube and the resonant frequencies have an inverse relationship. This means that as the length of the tube increases, the resonant frequencies decrease and vice versa. This relationship follows the formula f ∝ 1/L.

## How can resonant frequencies in a tube with adjustable length be used in practical applications?

Resonant frequencies in a tube with adjustable length have various practical applications, such as in musical instruments like flutes and organ pipes. They can also be used in industrial settings for tuning and testing of equipment. Additionally, they can be used in scientific experiments to study the properties of sound and acoustics.

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