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

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SUMMARY

The wavelength of a tuning fork with a frequency of 1000 Hz is calculated using the resonant lengths of 25 cm and 76 cm. The correct formula applied is the resonant length equation, which incorporates the end correction. By analyzing the resonant lengths, the wavelength is determined to be 102 cm, but this value is incorrect as the expected wavelength is approximately 33 cm. The confusion arises from the identification of harmonics, where the correct harmonics for a tube with one end closed are the 3rd and 5th, not the 2nd and 3rd.

PREREQUISITES
  • Understanding of wave mechanics and harmonic frequencies
  • Familiarity with the resonant length equation: Resonant length = n/4 * wavelength + end correction
  • Knowledge of standing waves in closed-end tubes
  • Basic skills in algebra for manipulating equations
NEXT STEPS
  • Research the properties of standing waves in closed-end tubes
  • Learn about the relationship between frequency and wavelength in sound waves
  • Study the concept of harmonics and their identification in wave phenomena
  • Explore the application of end corrections in acoustic measurements
USEFUL FOR

Students studying physics, particularly those focusing on wave mechanics, acoustics, and harmonic analysis. This discussion is also beneficial for educators teaching sound wave properties and resonance in tubes.

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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