**a. **For open-open tube, frequency is:

$\overline{){\mathbf{f}}{\mathbf{=}}\frac{\mathbf{n}\mathbf{v}}{\mathbf{2}\mathbf{L}}}$

** **f_{1} = (1)(v)/2L = v/2L

Solving for L:

L = **v/2f _{1} **

The lowest frequency in most humans' audible range is 20 Hz.

a. What is the length of the shortest open-open tube needed to produce this frequency?

L = ___ m

b. What is the length of the shortest open-closed tube needed to produce this frequency?

L = ___ m

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