Question

Now consider a pipe that is stopped (i.e., closed at one end) but still has a fundamental frequency of 285 Hz in air. How does your answer to Part A, fHe, change? View Available Hint(s) Now consider a pipe that is stopped (i.e., closed at one end) but still has a fundamental frequency of 285 in air. How does your answer to Part A, , change? a. fHe increases. b. fHe decreases. c. fHe stays the same.

Answer 1

835.2267 Hz

Step-by-step explanation:

M = Molar mass

`\gamma` = Specific heat ratio

L = Length of tube

For open ends the frequency is

`f=(v)/(2L)`

For closed ends the frequency is

`f=(v)/(4L)`

So,
`f\propto v`

Speed of sound is given by

`v=\sqrt{(\gamma RT)/(M)}\`

So,
`v\propto \sqrt{(\gamma)/(M)}`

From the above relations we get

`(f_h)/(f_a)=\sqrt{(\gamma_hM_a)/(\gamma_aM_h)}\\\Rightarrow (f_h)/(f_a)=\sqrt{(1.67* 28.8)/(1.4* 4)}\\\Rightarrow (f_h)/(f_a)=2.93062`

`f_h=f_a2.93062\\\Rightarrow f_h=285* 2.93062\\\Rightarrow f_h=835.2267\ Hz`

The frequency of helium is 835.2267 Hz

It does not matter if the tube is closed or open as
`f\propto v`