What is the Wavelength of a Tuning Fork with 1000Hz Frequency?

AI Thread Summary
The discussion focuses on calculating the wavelength of a tuning fork with a frequency of 1000 Hz, using resonant lengths of 25 cm and 76 cm from a tube with one end open and one closed. The method employed involves the resonant length equation, but there is confusion regarding which harmonics are being used, with assumptions made about the 2nd and 3rd harmonics. The calculated wavelength of 102 cm appears incorrect, as it significantly deviates from the expected value of approximately 33 cm for a 1000 Hz wave. Participants suggest that the harmonics might actually be the 1st and 3rd rather than the assumed 2nd and 3rd, leading to a need for clarification on the harmonic series in this context. The conversation emphasizes the importance of correctly identifying harmonics to achieve accurate wavelength calculations.
HelgaMan
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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 can't seem to find where i justified that, so maybe that's 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.. that's 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, that's why I am asking :D

anywho, any help would be greatly appreciated, thx. :biggrin:
 
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HelgaMan said:
So, first of all, i thought that the two consecutive harmonics were the 2nd and 3rd ones.. i can't seem to find where i justified that, so maybe that's 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?
 
uhmm, well the harmonics just 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
 
HelgaMan said:
uhmm, well the harmonics just 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?
 
hmm, I am 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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