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Frequency of a tuning fork

  • Thread starter K3nt70
  • Start date
  • #1
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Homework Statement



A tuning fork is set into vibration above a vertical open tube filled with water, as seen in the figure below.

http://img207.imageshack.us/img207/5817/dgian1686mm6.gif [Broken]

The water level is allowed to drop slowly. As it does so, the air in the tube above the water level is heard to resonate with the tuning fork when the distance from the tube opening to the water level is L1 = 0.100 m and L2 = 0.450 m. What is the frequency of the tuning fork? Assume that the speed of sound in air is 343 m/s.



Homework Equations



f1 = [tex]\frac{\upsilon}{4L}[/tex]

where v = 343 m/s



The Attempt at a Solution




I know that when the distance between the tuning fork and surface of the water is 0.1m. This means it is the first harmonic. Since it it a closed tube, i use the formula for the first harmonic of a closed tube, but my frequency comes out to 858 Hz which is incorrect.


I'm sure that i have to use the second length they give to us, but i'm not sure what its for..


cheers
 
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Answers and Replies

  • #2
Errr, I think the question is wrong or something, because the resonant lengths are supposed to be 1/4 lambda (lambda is wavelength), 3/4 lambda, 5/4 lambda, etc...

So I think either your L1 or L2 is incorrectly given. But otherwise, I think the formula you're using should be fine...I don't think you have to use L2. Maybe they put it in to trick you? Haha, hope that helped a little?
 
  • #3
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well. i'm fairly sure the question is correct; when there are problems with assignment questions, they can change them asap (since its over the internet). And seeing as this assignment is somewhat old, (im reviewing for a test) they've had a lot of time to fix it ;P
 
  • #4

Homework Statement



A tuning fork is set into vibration above a vertical open tube filled with water, as seen in the figure below.

http://img207.imageshack.us/img207/5817/dgian1686mm6.gif [Broken]

The water level is allowed to drop slowly. As it does so, the air in the tube above the water level is heard to resonate with the tuning fork when the distance from the tube opening to the water level is L1 = 0.100 m and L2 = 0.450 m. What is the frequency of the tuning fork? Assume that the speed of sound in air is 343 m/s.

Oh, I just did this
Well, v=fLambda
Lamda=2(L2-L1)
so Lambda=.7
so now 343=fLambda
f=343*.7
f=247.1

Hope that helps
 
Last edited by a moderator:
  • #5
oh. yeah. warmfire540's answer looks right. gee...this is a good sign... i have a physics exam on monday and i can't get these questions right!!!
 
  • #6
82
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Its actually f = v/lambda, so 343/0.7 = 490 which is correct. why did you set lambda = 2(L2-L1)? it works to get the correct final answer, but i don't know the logic behind it.
 
Last edited:
  • #7
Because L2-L1 = 1/2 lambda because the difference between any 2 resonant lengths in a closed tube is 1/2lambda. So knowing that:

1/2lambda = L2-L1
lambda = 2(L2-L1)
 
  • #8
82
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ooohhhh ok. wow that's a pretty important rule - can't believe i didn't know that. Thanks for your help!
 
  • #9
Its actually f = v/lambda, so 343/0.7 = 490 which is correct. why did you set lambda = 2(L2-L1)? it works to get the correct final answer, but i don't know the logic behind it.
Yeah, you're right goodjob
Just didn't see that while typing it out! Hope it helped though
:tongue2:
 
  • #10
1
0
Its actually f = v/lambda, so 343/0.7 = 490 which is correct. why did you set lambda = 2(L2-L1)? it works to get the correct final answer, but i don't know the logic behind it.
actually knowing [tex]\lambda[/tex]/4=L1
and 3[tex]\lambda[/tex]/4=L2
we can say (L2-L1)= 3[tex]\lambda[/tex]/4 - [tex]\lambda[/tex]/4
= 2[tex]\lambda[/tex]/4
=[tex]\lambda[/tex]/2
(L2-L1)=[tex]\lambda[/tex]/2
2(L2-L1)=[tex]\lambda[/tex] <--- shown
 

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