Beat frequency in organ pipe

[dimpledur]

Homework Statement

Chapter 12, problem 34. An auditorium has organ pipes at the front and at the rear of the
hall. Two identical pipes, one at the front and one at the back, have fundamental frequencies
of 264.00 Hz at 20.0°C. During a performance, the organ pipes at the back of the hall are at
25.0°C, while those at the front are still at 20.0°C. What is the beat frequency when the two
pipes sound simultaneously? Use 3 significant figures.

The Attempt at a Solution

Basically, I am not sure whether to assume the organ pipes are closed at one end, and open at the other, or open at both ends. My textbook says it can be both...

Anyways, what I was going to do was just take a ratio of the frequencies.

f1 / f2 = (n1*v1/4L) / (n2*v2/4L)

The only problem here is I have 2 unknowns. I don't know n2 or f2..

 

[Delphi51]

I think you would use n1 = n2 = 1 since it mentions fundamental frequency.
Do you understand why the pipe at the back would have a different frequency due to the higher temperature there?

The temperature affects the speed of sound in the air inside the pipe, so v2 will be different from v1. You will need a formula that tells how the speed of sound varies with temperature.

 

[nlightner]

I would first clarify with the instructor or textbook which type of organ pipe is being referred to in the problem. If it is not specified, I would consider both possibilities and provide solutions for each case.

For a closed-open pipe, I would use the formula f = (n*v)/(4L), where n is the harmonic number (1 for fundamental frequency), v is the speed of sound, and L is the length of the pipe. Using the given information, I would calculate the fundamental frequency for the pipe at the back of the hall:

f2 = (1*343 m/s)/(4*L2)

where L2 is the length of the pipe at the back of the hall. Since the pipe at the front of the hall is at the same temperature, its length would also be the same, and therefore its fundamental frequency would also be 264 Hz.

Now, using the formula for beat frequency, fbeat = |f1-f2|, I would find the beat frequency as:

fbeat = |264 Hz - f2|

For an open-open pipe, I would use the formula f = (n*v)/(2L), where n is the harmonic number (1 for fundamental frequency), v is the speed of sound, and L is the length of the pipe. Using the given information, I would calculate the fundamental frequency for the pipe at the back of the hall:

f2 = (1*343 m/s)/(2*L2)

where L2 is the length of the pipe at the back of the hall. Since the pipe at the front of the hall is at the same temperature, its length would also be the same, and therefore its fundamental frequency would also be 264 Hz.

Now, using the formula for beat frequency, fbeat = |f1-f2|, I would find the beat frequency as:

fbeat = |264 Hz - f2|

In both cases, the beat frequency would be the same, and would depend on the length of the pipe at the back of the hall, which is unknown. Therefore, I would suggest finding the length of the pipe at the back of the hall or providing more information to solve for the beat frequency accurately.