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

Answer:

835.2267 Hz

Explanation:

M = Molar mass

$\gamma$ = Specific heat ratio

L = Length of tube

For open ends the frequency is

$f=\dfrac{v}{2L}$

For closed ends the frequency is

$f=\dfrac{v}{4L}$

So, $f\propto v$

Speed of sound is given by

$v=\sqrt{\dfrac{\gamma RT}{M}}$

So, $v\propto \sqrt{\dfrac{\gamma}{M}}$

From the above relations we get

$\dfrac{f_h}{f_a}=\sqrt{\dfrac{\gamma_hM_a}{\gamma_aM_h}}\\\Rightarrow \dfrac{f_h}{f_a}=\sqrt{\dfrac{1.67\times 28.8}{1.4\times 4}}\\\Rightarrow \dfrac{f_h}{f_a}=2.93062$

$f_h=f_a2.93062\\\Rightarrow f_h=285\times 2.93062\\\Rightarrow f_h=835.2267\ Hz$

The frequency of helium is 835.2267 Hz

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