Question

A certain organ pipe, open at both ends, produces a fundamental frequency of 300 Hz in air. If the pipe is filled with helium at the same temperature, what fundamental frequency fHe will it produce? Take the molar mass of air to be 28.8 g/mol and the molar mass of helium to be 4.00 g/mol. Now consider a pipe that is stopped (i.e., closed at one end) but still has a fundamental frequency of 300 Hz in air. How does your answer to first question, fHe, change?

Answer 1

Answer:

Explanation:

Let L be the length of organ pipe . For fundamental note

wave length λ = 2L

Frequency = velocity / λ

300 = velocity / λ

velocity in air= 300 x 2L

= 600 L

In helium velocity changes as follows

velocity ∝ 1 / √molar mass

velocity in helium =√ ( 28.8 / 4 ) x velocity in air

velocity in helium = 2.68  x velocity in air

velocity in helium = 2.68  x 600 L

Frequency = velocity in helium / 2L

= 2.68 X 600L / 2L

= 804 Hz.

For open organ pipe also increase will be 2.68 times only because the increase is due to change in the medium from air to helium .