Investigating Beat Frequency in Pipes with Different Temperatures

[BooGTS]

Two pipes, equal in length, are each open at one end. Each has a fundamental frequency of 473 Hz at 297 K. In one pipe the air temperature is increased to 302 K. If the two pipes are sounded together, what beat frequency results?

I think I missed this day entirely due to another project. I know that when you have two frequencies, you cannot simply add them together. It also says "one pipe" but not which one? Apparently, it doesn't matter, but I don't see how you;d know when to use the formula Vair = 331 m/s * sq. rt. of (T/273).

thanks for any help!

 

[BooGTS]

anyone? *bump*

 

[NateTG]

What the problem most likely means is that the temperature change will change the fundamental of one of the pipes. As a result, you'll get two different tones if you strike both pipes simultaneously. What is the frequency of the beat caused by the interfearance between the two?

 

[Andrew Mason]

Two pipes, equal in length, are each open at one end. Each has a fundamental frequency of 473 Hz at 297 K. In one pipe the air temperature is increased to 302 K. If the two pipes are sounded together, what beat frequency results?

I think I missed this day entirely due to another project. I know that when you have two frequencies, you cannot simply add them together. It also says "one pipe" but not which one? Apparently, it doesn't matter, but I don't see how you;d know when to use the formula Vair = 331 m/s * sq. rt. of (T/273).

thanks for any help!

The wavelength of the sound produced by the pipe is determined by its length. The frequency is determined by the speed of the sound wave in the pipe divided by the wavelength (this follows from the universal wave equation: $\lambda f = v$).

So how is the speed, hence frequency, affected by this $5 ^oK$ temperature increase?

AM