Wavelength of Tuning Fork

  • Thread starter HelgaMan
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  • #1
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Homework Statement


The problem asks to find the wavelength of a tuning fork with 1000Hz frequency. Then they give you two consecutive resonant lengths which are 25 cm and 76 cm.

The way they measured the resonant lengths is by placing the tuning fork over a tube that has 1 end open and 1 end closed so that the waves come back or whatever and the lengths are the lengths of the tubes that had resonance.


Homework Equations



The equation i tried to use was:

Resonant length = n/4 * wavelength + end correction.


The Attempt at a Solution


So, first of all, i thought that the two consecutive harmonics were the 2nd and 3rd ones.. i cant seem to find where i justified that, so maybe thats not right, but my work assumes it is.. i guess. :p

anywho, i subtracted the 2 equations to eliminate end correction. (n's are odd because one end is open and one end is closed on the tube)

76 = 5/4 * wavelength + e
- 25 = 3/4 * wavelength + e
51 = 2/4 * wavelength

so wavelength equals 102 cm, however.. thats about 3 times as much as what it should be, which is 33-ish centimeters, and i think its because my method only works for maybe the first and 2nd harmonics or something, because it works when its just the 1st and 2nd harmonics... or i forgot to take something into consideration, i dunno, thats why im asking :D

anywho, any help would be greatly appreciated, thx. :biggrin:

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
LowlyPion
Homework Helper
3,090
5
So, first of all, i thought that the two consecutive harmonics were the 2nd and 3rd ones.. i cant seem to find where i justified that, so maybe thats not right, but my work assumes it is.. i guess. :p

... (n's are odd because one end is open and one end is closed on the tube)

Maybe you should resolve this conflict?

Is 2nd a permitted harmonic?
 
  • #3
5
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uhmm, well the harmonics jsut mean that

n would 3 and 5 which are the 2nd and 3rd when there is a standing wave and one end is a node and the other is an anti node
 
  • #4
LowlyPion
Homework Helper
3,090
5
uhmm, well the harmonics jsut mean that

n would 3 and 5 which are the 2nd and 3rd when there is a standing wave and one end is a node and the other is an anti node

And you're certain that it's not the 1st and 3rd harmonic?
 
  • #5
5
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hmm, im not sure what the actual harmonics are,

but, i googled the wave length of a 1000 Hz wave and it should be approximately 33 cm,

and if you do it as if it were the 1st and 3rd, then the answer would be 51 cm, which i htink is a bit too far off
 

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