A plumber is trying to fix a clog in a vertical pipe, but does not know how far down in the pipe the clog and the water level is. He is a good whistler, so he whistles into the open end of the pipe and finds that a frequency of 197hz (cycles per second) returns a loud resonant sound. He knows the speed of sound is 340 m/s. Calculate the depth of the air in the pipe (in meters) above the clog.

Question

Dylan Rivas

Physics

12 August, 17:35

A plumber is trying to fix a clog in a vertical pipe, but does not know how far down in the pipe the clog and the water level is. He is a good whistler, so he whistles into the open end of the pipe and finds that a frequency of 197hz (cycles per second) returns a loud resonant sound. He knows the speed of sound is 340 m/s. Calculate the depth of the air in the pipe (in meters) above the clog.

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Answers (1)

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Spears · Answer 1

the shortest distance to the obstruction is 0.431 m

Explanation:

We can see this system as an air column, where the plumber is open and where the water is closed, in the case when he hears the sound there is a phenomenon of resonance and superposition of waves with constructive interference.

For the lowest resonance we must have a node where the water is and a maximum where the plumber is a quarter of the wavelength

λ = ¼ L

If we are in a major resonance specifically the following resonance. We have a full wavelength plus a quarter of the wavelength

λ = 4L / 3

The general formula is

λ = 4L / n n = 1, 3, 5, 7, ...

In addition the wave speed is the product of the frequency by the wavelength

v = λ f

Let's replace

v = (4L / n) f

L = v n / (4 f)

Now we can calculate the depth or length of the air column

If we have the first standing wave n = 1

L = 340 1 / (4 197)

L = 0.431 m

If it is the second resonance n = 3

L = 340 3 / (4 197)

L = 1.29 m

We can see the shortest distance to the obstruction is 0.431 m