Open Tube Resonance: Fundamental Frequency

[physics123]

Homework Statement

A long tube that is open at both ends is used to construct a musical instrument. The sound waves that enter the tube are generated by a taut wire with a tension of 600 N and a linear mass density of 0.031 kg/m.
If the length of the tube is 2.4 m and a hole is cut in the side of the tube at 0.8 m, what is the fundamental resonant frequency of the new system? A) 214 Hz B) 107 Hz C) 71 Hz D) 143 Hz E) 321 Hz

Homework Equations

λ= v/2L

The Attempt at a Solution

I broke the pipe up into two pieces and found that the frequencies of the 1.6 m tube and the 0.8m tube are 107Hz and 214Hz, respectively. These are both answers available. If I add these together, I also get a third answer. Which is correct?

 

[BvU]

What is the efffect of the hole on the standing wave pattern in the tube (cf a flute) ? So what's the fundamental wavelength ?

 

[physics123]

The fundamental wavelength of which part of the tube? The hole effectively breaks the pipe into two different pieces does it not?

 

[BvU]

Correct. But which wavelength 'survives' ?

 

[physics123]

Not sure I catch what you mean by 'survives'. The way I see this, you now how two different pipes of different lengths and I'm unsure which to focus on.

 

[BvU]

The one that 'survives' is the one that fits in both sections ... !

 

[physics123]

So that would mean the smaller section of pipe would be the 'surviving' one in these questions?

 

[BvU]

Right. The biggest wavelength that fits in there also fits in the other section (as a second harmonic). The other way around not.