Finding length with two given frequencies

[leinad0213]

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.

 

[PeterO]

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.