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

The wavelengths and frequencies of standing waves in a closed pipe are determined by the pipe's length and the harmonic number. For a 2.50 m closed pipe, the fundamental frequency is 68.8 Hz, the first overtone is 137.6 Hz, and the second overtone is 206.4 Hz, all of which are audible to humans.

Step-by-step explanation:

To find the wavelengths and frequencies of standing waves in a closed pipe, we can use the relationships for resonant wavelengths λn and frequencies fn as stated in the question:

λn = 2L/n fn = nv/2L For a pipe that is closed at both ends, such as a shower, the wavelengths of standing waves formed are harmonics of the fundamental wavelength, which is twice the length of the shower.

(a) To calculate the wavelengths, we use λn = 2L/n where L is the length of the shower and n is the harmonic number (1, 2, 3, ...), which yields:

• λ1 (Fundamental) = 2L
• λ2 (First overtone) = 2L/2 = L
• λ3 (Second overtone) = 2L/3
• (b) To find the frequencies, we take fn = nv/2L with v = 344 m/s and L = 2.50 m:

• f1 (Fundamental frequency) = 1 * 344 m/s / (2 * 2.50 m) = 68.8 Hz
• f2 (First overtone) = 2 * 344 m/s / (2 * 2.50 m) = 137.6 Hz
• f3 (Second overtone) = 3 * 344 m/s / (2 * 2.50 m) = 206.4 Hz

All of these frequencies are within the audible range for humans, which is typically from about 20 Hz to 20 kHz.

Answer 2

68.8 Hz

137.6 Hz, 206.4 Hz

Step-by-step explanation:

L = Length of tube = 2.5 m

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

Distance between nodes is given by

`L=(\lambda)/(2)+m(\lambda)/(2)\\\Rightarrow (\lambda(n+1))/(2)=L\\\Rightarrow \lambda=(2L)/(n+1)`

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

Making n+1 = n

`\lambda=(2L)/(n)`

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

For fundamental frequency n = 1

`\lambda=(2* 2.5)/(1)\\\Rightarrow \lambda=5\ m`

Frequency is given by

`f=(v)/(\lambda)\\\Rightarrow f=(344)/(5)\\\Rightarrow f=68.8\ Hz`

The fundamental frequency is 68.8 Hz

First overtone

`2f=2* 68.8=137.6\ Hz`

Second overtone

`3f=3* 68.8=206.4\ Hz`

The overtones are 137.6 Hz, 206.4 Hz