One closed pipe organ has a length of 2.5 m.
The frequency of the note is 34.3 Hz. When a second pipe is played at the same time, a 1.7 Hz beat note is heard. By how much is the second pipe too long?
Would you have to use the frequency of the second pipe to find the length of that pipe?
Yes; you need the frequency of the second pipe.
If you are trying to match pipe fundamental frequencies, and the second pipe is too long (rather than too short), then the natural frequency of the second pipe is 34.3 - 1.7 = 32.6 Hz.
Frequency is inversely proporional to length. If the frequency is low by a factor 34.3/32.6 = 1.052, the length is too long by the same factor.
To find the length of the second pipe, we need to use the frequency of the beat note and some knowledge about beat frequencies.
A beat note is the result of two sound waves with slightly different frequencies interfering with each other. The beat frequency can be calculated by subtracting the frequency of one wave from the frequency of the other.
In this case, the beat frequency is given as 1.7 Hz. Let's call the frequency of the second pipe f2. Since the beat frequency is the difference between the two frequencies, we can write the equation:
f2 - 34.3 Hz = 1.7 Hz
Solving for f2:
f2 = 34.3 Hz + 1.7 Hz
f2 = 36 Hz
Now that we have the frequency of the second pipe, we can use the formula for the fundamental frequency of a closed pipe to find its length.
The fundamental frequency (f1) for a closed pipe is given by:
f1 = (2v) / λ
where:
v is the speed of sound in air (approximately 343 m/s at room temperature)
λ is the wavelength of the sound wave produced by the pipe
We rearrange the equation to solve for the wavelength:
λ = (2v) / f1
Plugging in the known values:
λ = (2 * 343 m/s) / 34.3 Hz
= 20 m
Since the wavelength of the fundamental frequency for the first pipe is 20 m, the length of the pipe must be half the wavelength, so the length of the first pipe is 10 m.
Now, let's find the length of the second pipe.
Using the same formula for the second pipe:
λ = (2 * 343 m/s) / 36 Hz
= 19.06 m
Since the wavelength of the fundamental frequency for the second pipe is 19.06 m, the length of the pipe must also be half the wavelength, so the length of the second pipe is approximately 9.53 m.
To find by how much the second pipe is too long, subtract the length of the first pipe from the length of the second pipe:
9.53 m - 10 m = -0.47 m
Therefore, the second pipe is 0.47 meters shorter than the first pipe.