Solving Closed Pipe Question: Length Difference of 2 Pipes

[pebbles]

Homework Statement

Once closed organ pipe has a length of 3.2 m. Frequency of the note played by this pipe is 27 Hz. When a second pipe is played at the same time, a 1.50 Hz beat note is heard. By how much is the second pipe too long?

Homework Equations

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The Attempt at a Solution

I'm really lost on this problem. I was thinking Lb-La=1/2 of wavelength, but I don't think that's right.

Any direction is appreciated. Thanks in advance.

 

[Kurdt]

Perhaps this page could get you started.

http://hyperphysics.phy-astr.gsu.edu/hbase/sound/beat.html

 

[Moetasim]

As a scientist, the first step in solving this problem would be to gather all the necessary information and identify any relevant equations. In this case, we know the length of one pipe (La = 3.2 m) and the frequency of the note it produces (fa = 27 Hz). We are also given the frequency of the beat note (fbe = 1.50 Hz).

The next step would be to use the equation for the beat note frequency (fbe = |fa-fb|) to find the frequency of the second pipe (fb). Rearranging the equation, we get fb = fa-fbe. Plugging in the known values, we get fb = 27 Hz - 1.50 Hz = 25.50 Hz.

Now, we can use the formula for the wavelength of a standing wave in a closed pipe (λ = 4L/n) to find the length of the second pipe (Lb). Rearranging the equation, we get Lb = nλ/4, where n is the harmonic number. Since we are dealing with a beat note, the harmonic number for both pipes must be the same. Therefore, we can set n = 1 for both pipes.

Plugging in the values for the wavelength (using the frequency of the second pipe, fb) and the length of the first pipe (La), we get:

Lb = (1)(λ)/4 = (1)(v/fb)/4 = (1)(343 m/s)/(25.50 Hz)/4 = 3.38 m

Finally, to find the difference in length between the two pipes, we can subtract the length of the first pipe (La) from the length of the second pipe (Lb). This gives us:

Lb-La = 3.38 m - 3.2 m = 0.18 m

Therefore, the second pipe is 0.18 m too long compared to the first pipe.