Finding length with two given frequencies

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SUMMARY

The discussion centers on calculating the depth of a mine shaft using resonant frequencies of 83.72 Hz and 107.64 Hz, with the speed of sound in air at 20 °C set at 343 m/s. The equation used is f = n*Vs/(4L), where L represents the depth of the shaft. Participants highlight that the assumption of the lower frequency being the fundamental mode is incorrect, as the next resonant frequency does not align with the expected harmonic series. This leads to the conclusion that both frequencies represent higher harmonics, necessitating a different approach to determine the shaft's depth.

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leinad0213
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Homework Statement



You find an abandoned mine shaft and wish to measure its depth. Using an audio oscillator of variable frequency, you note that you can produce successive resonances at frequencies of 83.72 Hz and 107.64 Hz. What is the depth of the shaft? Assume the temperature in the shaft is 20 °C.

Vs = 343 m/s

Homework Equations



f = n*Vs/(4L)

The Attempt at a Solution



83.72 = n1*343/(4L) 107.64 = n2*343/(4L)

When I solve for L on one of the equations to plug into the other. The L's end up canceling each other out, so I don't get L which is what I am trying to find.
 
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leinad0213 said:

Homework Statement



You find an abandoned mine shaft and wish to measure its depth. Using an audio oscillator of variable frequency, you note that you can produce successive resonances at frequencies of 83.72 Hz and 107.64 Hz. What is the depth of the shaft? Assume the temperature in the shaft is 20 °C.

Vs = 343 m/s

Homework Equations



f = n*Vs/(4L)

The Attempt at a Solution



83.72 = n1*343/(4L) 107.64 = n2*343/(4L)

When I solve for L on one of the equations to plug into the other. The L's end up canceling each other out, so I don't get L which is what I am trying to find.

Looks like you have assumed the lower frequency is the fundamental. If it was, the next frequency to resonate would be 3 times. clearly 107 is not 3 x 83, so neither is the fundamental.
 

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