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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.

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boffeemadrid

Answer:

835.2267 Hz

Explanation:

M = Molar mass

[tex]\gamma[/tex] = Specific heat ratio

L = Length of tube

For open ends the frequency is

[tex]f=\dfrac{v}{2L}[/tex]

For closed ends the frequency is

[tex]f=\dfrac{v}{4L}[/tex]

So, [tex]f\propto v[/tex]

Speed of sound is given by

[tex]v=\sqrt{\dfrac{\gamma RT}{M}}\[/tex]

So, [tex]v\propto \sqrt{\dfrac{\gamma}{M}}[/tex]

From the above relations we get

[tex]\dfrac{f_h}{f_a}=\sqrt{\dfrac{\gamma_hM_a}{\gamma_aM_h}}\\\Rightarrow \dfrac{f_h}{f_a}=\sqrt{\dfrac{1.67\times 28.8}{1.4\times 4}}\\\Rightarrow \dfrac{f_h}{f_a}=2.93062[/tex]

[tex]f_h=f_a2.93062\\\Rightarrow f_h=285\times 2.93062\\\Rightarrow f_h=835.2267\ Hz[/tex]

The frequency of helium is 835.2267 Hz

It does not matter if the tube is closed or open as [tex]f\propto v[/tex]

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