Question

Unless indicated otherwise, assume the speed of sound in air to be v = 344 m/s. A pipe closed at both ends can have standing waves inside of it, but you normally don’t hear them because little of the sound can get out. But you can hear them if you are inside the pipe, such as someone singing in the shower. (a) Show that the wavelengths of standing waves in a pipe of length L that is closed at both ends are λn=2L/n and the frequencies are given by fn=nv/2L=nf1, where n=1,2,3,… (b) Modeling it as a pipe, find the frequency of the fundamental and the first two overtones for a shower 2.50 m tall. Are these frequencies audible?

Answer 1

Answer:

68.8 Hz

137.6 Hz, 206.4 Hz

Explanation:

L = Length of tube = 2.5 m

v = Velocity of sound in air = 344 m/s

Distance between nodes is given by

$L=\dfrac{\lambda}{2}+m\dfrac{\lambda}{2}\\\Rightarrow \dfrac{\lambda(n+1)}{2}=L\\\Rightarrow \lambda=\dfrac{2L}{n+1}$

Where n = 0, 1, 2, 3, ...

Making n+1 = n

$\lambda=\dfrac{2L}{n}$

where n = 1, 2, 3 .....

For fundamental frequency n = 1

$\lambda=\dfrac{2\times 2.5}{1}\\\Rightarrow \lambda=5\ m$

Frequency is given by

$f=\dfrac{v}{\lambda}\\\Rightarrow f=\dfrac{344}{5}\\\Rightarrow f=68.8\ Hz$

The fundamental frequency is 68.8 Hz

First overtone

$2f=2\times 68.8=137.6\ Hz$

Second overtone

$3f=3\times 68.8=206.4\ Hz$

The overtones are 137.6 Hz, 206.4 Hz