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

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Homework Help Overview

The problem involves determining the wavelength of a tuning fork with a frequency of 1000Hz, using measurements from resonant lengths of a tube with one end open and one end closed. The resonant lengths provided are 25 cm and 76 cm.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the relationship between resonant lengths and harmonics, questioning whether the measured lengths correspond to the 2nd and 3rd harmonics or possibly the 1st and 3rd harmonics. There is also an exploration of the end correction factor in the context of the problem.

Discussion Status

The discussion is ongoing, with participants offering insights into harmonic definitions and questioning the assumptions made about which harmonics are involved. Some guidance has been provided regarding the nature of standing waves in the tube, but no consensus has been reached on the correct interpretation of the harmonics.

Contextual Notes

There is uncertainty regarding the correct identification of harmonics, and participants are considering the implications of using different harmonic assumptions on the calculated wavelength. Additionally, the expected wavelength for a 1000Hz wave is noted, but the relevance of this information to the problem is still under discussion.

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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