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

[member 731016]

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!

 

[haruspex]

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.

 

[member 731016]

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

Thank you for your reply @haruspex!

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!

 

[haruspex]

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?

 

[member 731016]

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

Thank you for your reply @haruspex!

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

Many thanks!

 

[haruspex]

Thank you for your reply @haruspex!

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?

 

[member 731016]

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?

Thanks your for your reply @haruspex! I will just find it.

 

[haruspex]

Thanks your for your reply @haruspex! I will just find it.

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.

 

[member 731016]

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.

Thank you for your reply @haruspex!

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

Many thanks!

 

[member 731016]

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!

 

[member 731016]

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.

Thank you for your reply @haruspex!

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!

 

[haruspex]

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.

 

[haruspex]

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.

 

[member 731016]

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.

Thank you for your reply @haruspex!

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!

 

[haruspex]

Why would the product of f and λ be an estimate?

Because $v=fλ$.
If you integrate $\int f.d\lambda=\int\frac v{\lambda}.d\lambda=v\ln(\frac{\lambda_f}{\lambda_i})$.

 

[member 731016]

Because $v=fλ$.
If you integrate $\int f.d\lambda=\int\frac v{\lambda}.d\lambda=v\ln(\frac{\lambda_f}{\lambda_i})$.

Thank you for your reply @haruspex!

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!

 

[member 731016]

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

Yes, thank you @haruspex!