Question #117c1

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Question #117c1

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Answer 1 · 2017-02-16T12:34:18

It would be less than that of air.

Explanation:

The wavelength of each harmonic is a function of the tube length so independent of the gas used. The speed of a wave is given by:

$c = f λ$ $c=flamda$

Since $λ$ $lamda$ for the harmonic is the same as air (independent of the gas), then since the speed ( $c$ $c$ ), is lower in $C O_{2}$ $CO_2$ , the frequency ( $f$ $f$ ) must also be lower for each harmonic.

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Question #117c1

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