# Speed of sound from resonant length of tube vs tuning fork wavelengths

• ChiralSuperfields
In summary: I agree, I am just waiting for the prof to get back to me. I agree that if we did do a graph of frequency of turning fork vs wavelength of sound measured from the tube then the area under the graph of would...
ChiralSuperfields
Homework Statement
I am trying to find the speed of sound given that gradient of resonant length of the tube (for the fundamental) vs tuning fork wavelength
Relevant Equations
Theoretically ##L_{res}= λ/4##

However, for the graph ##L_{res} =0.2381λ##
The graph is,

I do not understand why how it is possible to find the speed of sound from the gradient for this graph. Can someone please help?

Many thanks!

Callumnc1 said:
Homework Statement:: I am trying to find the speed of sound given that gradient of resonant length of the tube (for the fundamental) vs tuning fork wavelength
Relevant Equations:: Theoretically ##L_{res}= λ/4##

However, for the graph ##L_{res} =0.2381λ##

The graph is,
View attachment 323524
I do not understand why how it is possible to find the speed of sound from the gradient for this graph. Can someone please help?

Many thanks!
How do you know the tuning fork wavelength? The mechanics of a fork determine its frequency.

ChiralSuperfields
haruspex said:
How do you know the tuning fork wavelength? The mechanics of a fork determine its frequency.

We calculated found it by dividing the speed of sound (343 m/s) by the frequency read on each turning fork. But for some reason we are meant to graph resonant length of the tube vs tuning fork wavelength instead of resonant length vs 1/4f.

Many thanks!

Callumnc1 said:
We calculated found it by dividing the speed of sound (343 m/s) by the frequency read on each turning fork.
Why would you do that if the object of the exercise is to determine the speed of sound?

ChiralSuperfields
haruspex said:
Why would you do that if the object of the exercise is to determine the speed of sound?

I have no idea. It does not make sense to me. But that is what we were meant to do.

Many thanks!

Callumnc1 said:

I have no idea. It does not make sense to me. But that is what we were meant to do.

Many thanks!
I feel sure you have misunderstood. Were you given the task in writing or verbally? If in writing, can you post it? If verbally, are there any other students you can confirm it with?

ChiralSuperfields
haruspex said:
I feel sure you have misunderstood. Were you given the task in writing or verbally? If in writing, can you post it? If verbally, are there any other students you can confirm it with?

Callumnc1 said:
I saw your posted extract, then it vanished. But I saw enough to explain as one misunderstanding on your part and one typo.

You were not supposed to use the speed of sound to find the wavelength of the tuning fork. Indeed, a tuning fork does not have a specific wavelength. What it has is a frequency, and the resulting wavelength depends on the speed of sound in the air, which can vary from day to day.
You were supposed to deduce the wavelength produced by the tuning fork from the resonant length of the tube.

The typo is that it should have said to plot the frequency against the wavelength.

ChiralSuperfields
haruspex said:
I saw your posted extract, then it vanished. But I saw enough to explain as one misunderstanding on your part and one typo.

You were not supposed to use the speed of sound to find the wavelength of the tuning fork. Indeed, a tuning does not have a specific wave. What it has is a frequency, and the resulting wavelength depends on the speed of sound in the air, which can vary from day to day.
You were supposed to deduce the wavelength prod by the tuning fork from the resonant length of the tube.

The typo is that it should have said to plot the frequency against the wavelength.

Sorry, the extract have be found in post #9.

Many thanks!

My group did actually do a graph of resonant length vs frequency

This gives the correct gradient to find the speed of sound. However, that is not the specific question in bold at the bottom of post #9.

Many thanks!

haruspex said:
I saw your posted extract, then it vanished. But I saw enough to explain as one misunderstanding on your part and one typo.

You were not supposed to use the speed of sound to find the wavelength of the tuning fork. Indeed, a tuning fork does not have a specific wavelength. What it has is a frequency, and the resulting wavelength depends on the speed of sound in the air, which can vary from day to day.
You were supposed to deduce the wavelength produced by the tuning fork from the resonant length of the tube.

The typo is that it should have said to plot the frequency against the wavelength.

I agree, I am just waiting for the prof to get back to me. I agree that if we did do a graph of frequency of turning fork vs wavelength of sound measured from the tube then the area under the graph of would be equal to the speed of sound.

Many thanks!

Last edited:
Callumnc1 said:
if we did do a graph of frequency of turning fork vs wavelength of sound measured from the tube then the area under the graph of would be equal to the speed of sound.
Umm, no it wouldn’t. The product of x and y at each point would be an estimate of the speed of sound.
I should have written that the graph has to be the inverse of one against the other, e.g. 1/f against λ.

ChiralSuperfields
Callumnc1 said:
My group did actually do a graph of resonant length vs frequency

View attachment 323537
This gives the correct gradient to find the speed of sound. However, that is not the specific question in bold at the bottom of post #9.

Many thanks!
I'm guessing that is 1/(4f), not (1/4)f.

ChiralSuperfields
haruspex said:
Umm, no it wouldn’t. The product of x and y at each point would be an estimate of the speed of sound.
I should have written that the graph has to be the inverse of one against the other, e.g. 1/f against λ.

Why would the product of f and λ be an estimate? I'm guessing it is because the area under the f and λ would have to be approximated by using rectangles and triangles. But actually, we could integrate to find the area?

I guess we could have 1 as the inverse of the other as you mention.

Many thanks!

Callumnc1 said:
Why would the product of f and λ be an estimate?
Because ##v=fλ##.
If you integrate ##\int f.d\lambda=\int\frac v{\lambda}.d\lambda=v\ln(\frac{\lambda_f}{\lambda_i})##.

ChiralSuperfields
haruspex said:
Because ##v=fλ##.
If you integrate ##\int f.d\lambda=\int\frac v{\lambda}.d\lambda=v\ln(\frac{\lambda_f}{\lambda_i})##.

Thats integral it not an estimate is it? By the way you were correct!! It should be resonant length of tube against inverse frequency (1/f).

Many thanks!

haruspex said:
I'm guessing that is 1/(4f), not (1/4)f.
Yes, thank you @haruspex!

• Introductory Physics Homework Help
Replies
5
Views
1K
• Introductory Physics Homework Help
Replies
4
Views
2K
• Introductory Physics Homework Help
Replies
10
Views
2K
• Introductory Physics Homework Help
Replies
9
Views
3K
• Introductory Physics Homework Help
Replies
3
Views
1K
• Introductory Physics Homework Help
Replies
15
Views
7K
• Introductory Physics Homework Help
Replies
4
Views
2K
• Introductory Physics Homework Help
Replies
12
Views
3K
• Introductory Physics Homework Help
Replies
4
Views
6K
• Introductory Physics Homework Help
Replies
10
Views
570