How to find the wavelength of the sound of the tuning fork?

  • #1
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Homework Statement


Ko_Fis_2014.png

1) find the wavelength of the sound of the tuning fork (cm)
2) what's the vertical distance x from the top of the pipe to the antinode above the pipe (cm)
3) find the frequency of the tuning fork (Hz)

Homework Equations


velocity = wavelength . frequency

for pipe closed at one end
n = 1,3,5
wavelength for n times harmonic = 4.lenght/n

The Attempt at a Solution


I don't know for which harmonic the tuning fork exciting at 16.7 cm and 50.7 cm
I assume for 16.7 is 1st harmonic
so the wavelength = 4*16.7 = 66.8 cm
since the 50.7 approximately 3 times 16.7, I assume that the n value is 3, which is the 2nd harmonic
so the wavelength = 4*50.7/3 = 67.6 cm

so the wavelength is 67 cm??

2) can you give me clue, where to start?

3) f = v / wavelength
f = 340 / 0.67 m = 507 Hz
 

Answers and Replies

  • #2
epenguin
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You seem to have done the hard parts OK. For (2) I think you only need to know the meaning of 'antinode'.
https://en.m.wikipedia.org/wiki/Node_(physics)

Fix stuff in your head - don't leave this problem without sketching a picture of the amplitude against height on top of these pictures, which would also help answer.

I guess I answered this question because I did that same experiment in English 'A' Level exam many years ago mnyah mnyah.
(I got it a bit wrong because I forgot the /4 , but I did write my answer was unreasonable so probably got nearly full marks.)
 
  • #3
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You seem to have done the hard parts OK. For (2) I think you only need to know the meaning of 'antinode'.
https://en.m.wikipedia.org/wiki/Node_(physics)

Fix stuff in your head - don't leave this problem without sketching a picture of the amplitude against height on top of these pictures, which would also help answer.

I guess I answered this question because I did that same experiment in English 'A' Level exam many years ago mnyah mnyah.
(I got it a bit wrong because I forgot the /4 , but I did write my answer was unreasonable so probably got nearly full marks.)
but my answer slightly different than the right answer. For example the frequency is 500Hz..
English A level... is there physics on that?
 
  • #4
epenguin
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I meant an English examining board for Physics - not British as there were/are different examining boards/syllabuses for Scotland. Even if it was Physics they might still have docked me a mark if I had wiritten 'wavelenght'. o0)

It looks like it was along time ago, as I need to be reminded 'read the question'. They do tell you that the tuning fork has an antinode above the top of the column, so the fairly small discrepancy between your two result values is no doubt due to your taking into account just the column length, not the whole distance from antinode to node at water surface.
 
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  • #5
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I meant an English examining board for Physics - not British as there were/are different examining boards/syllabuses for Scotland. Even if it was Physics they might still have docked me a mark if I had wiritten 'wavelenght'. o0)

It looks like it was along time ago, as I need to be reminded 'read the question'. They do tell you that the tuning fork has an antinode above the top of the column, so the fairly small discrepancy between your two result values is no doubt due to your taking into account just the column length, not the whole distance from antinode to node at water surface.
hmm... from the websites I found that the x distance is 0.6 times radius... but I still don't understand about number 2?
for number 1 the right answer is 68 cm.. why? which column length to be taken?
 
  • #6
haruspex
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I assume for 16.7 is 1st harmonic
No need to make that assumption. Let it be the nth harmonic.
so the wavelength = 4*16.7
Even if it is the first harmonic, that is not right. Where is the known antinode in relation to the known node?
 
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  • #7
epenguin
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The way I think of it, resonance happens like this: Systems like this air column have their own natural frequency at which they will vibrate if erxcited. Your classic SHM is the basic model. Resonance happens when you push something the same way it's going anyway, and when like in SHM it changes its direction you push it that way too - at any rate force applied at the same frequency and direction as the natural frequency. So it is reasonable to think that you get resonance in your problem this way: Think of the tuning fork that is sending the wave; it's maximum velocity is at zero displacement going say towards the tube and you will get the maximum kick going the other way half a period later. That half a period is the time you want the compression wave to go down the tube and bounceback and assist the fork just when it's going maximum velocity the other way. So it looks to me that the position of the tuning fork is where you have to measure from.

On this assumption having two measurements, not given x, you could calculate x by algebra. Since apparently you have been given the answers you could also do a calculation without algebra and see if this reasoning corresponds.

As you have found out, what happens at the end of the tube is complicating. I don't think you would have been set the problem that depended on complicated calculations for which you don't have any theory, or the theory is beyond you. But hopefully these complications are the same for both your measurements so can be cancelled out. So if you haver/had sketched diagrams like I suggested in my first post, what do they tell you about the difference in the measured lengths?

Very faint thing in the back of my mind is telling me that I knew about this and used it when I did this experiment myself all those years ago, (Quite likely it is in your textbook too - I found it in my son's).

It looks like this experiment isolates some simple feature out of a complicated situation. As sometimes in teaching, particularly exam oriented teaching, the complications are not acknowledged, students feel uncomfortable till they are shown how to get the right answer. In this bit of the course people talk glibly of tuning forks, taking them for granted. It is not all that evident why a tuning fork works, why it has to be the shape it is. They are used in the abovementioned textbook, but no explanation - they not even in the index. So realise now I had never been told and didn't really know. Wikipedia article tells you something. Made me also realise the only reason experiment this worked for me is that you hear the resonating column actually louder than the tuning fork. It would have expected most of the energy to be on the fork. But maybe somebody who has properly studied and taught these things can explain better than somebody who is just realising them!
 
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  • #8
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hmm... from the websites I found that the x distance is 0.6 times radius... but I still don't understand about number 2?
for number 1 the right answer is 68 cm.. why? which column length to be taken?
The 0.6 times pipe radius you refer to is usually referred to as the end correction.....but you don't need to know its value for this problem. You do, however, need to know that an antinode appears at a total height equal to length of tube plus length of end correction. All should become clear if you draw sketches showing the wave profiles for two succesive resonances and marking in the positions of the nodes and antinodes.
 
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  • #9
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Thanks for all replies. Sorry, i have problem opening this forum before..
I think i get it from internet that
L1 + end-correction = lambda/4
While
L2 - L1 = lambda/2

Using those equations i can find the answer

The open end side of pipe start with antinode
While the closed end side ended with node
 
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  • #10
haruspex
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Thanks for all replies. Sorry, i have problem opening this forum before..
I think i get it from internet that
L1 + end-correction = lambda/4
While
L2 - L1 = lambda/2

Using those equations i can find the answer

The open end side of pipe start with antinode
While the closed end side ended with node
Right. And presumably you also got L2 + end-correction =3 lambda/4.
 
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  • #11
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Right. And presumably you also got L2 + end-correction =3 lambda/4.
Ok i get it, thanks
 
  • #12
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Ok i get it, thanks
Please i didn't get question 1 pleasee
 
  • #13
haruspex
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Please i didn't get question 1 pleasee
Do you understand the concept of an end correction at an open end of a pipe?
 
  • #14
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Do you understand the concept of an end correction at an open end of a pipe?
yeah.. End correction = 0.6 * pipe raduis as stated above ...
 
  • #15
haruspex
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yeah.. End correction = 0.6 * pipe raduis as stated above ...
That's a rough guide but need not be exact in all cases. In the given question it can be calculated from the data.
But I was asking if you understood the concept, not how to calculate it.
 
  • #16
Merlin3189
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The advice of penguin & others to sketch diagrams is EXCELLENT. I do it for anything but the most trivial questions.

But the answer here is in the wording of the question, "Successive resonances of ...."
Nodes are always ##\frac {λ}{2}## apart, as are antinodes. (As you would see in sketches.)
So the distance between successive resonances are also ##\frac {λ}{2}## apart
And that would be equally true for pipes open at both ends.

End correction is an interesting issue and you should be aware of its existence, but is not IMO essential to parts 1 and 3, nor even to answer part 2, though you might be puzzled by this if you were unaware.
End effect also happens on vibrating strings, but I've never seen this level of question acknowledge that.
It also generalises to other situations in physics. For example, you often don't know the exact length of a pendulum or spring, because you can't easily know where the effective ends are.
 

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