Sound of Music - Closed Organ Pipe

[whitehorsey]

1. One closed organ pipe has a length of 2.40m.
a. What is the frequency of the note played by this pipe?
b. When a second pipe is played at the same time, a 1.40 Hz beat note is heard. By how much is the second pipe too long?



2. f = nv/2L (n = 1)
change of f = absolute value of f₂ - f₁



3. a. f = nv/2L (n = 1)
= 1(343)/2(2.4)
= 71.46 Hz

b. change of f = absolute value of f₂ - f₁
-1.4 = f₂ - 71.46 1.4 = f₂ - 71.46
f₂ = 70.06, 72.86

f = nv/2L
70.06 = 343/2L 72.86 = 343/2L
140.12 L = 343 145.72 L = 343
L = 2.4479 L = 2.3538
2.4 - 2.4479 = 0.048 m

 

[anathema]

2.3538 - 2.4 = 0.0462 m

Therefore, the second pipe is too long by 0.048 m or 0.0462 m, depending on whether the frequency is 70.06 Hz or 72.86 Hz. This difference in length is causing the beat note of 1.40 Hz to be heard. To eliminate the beat note, the length of the second pipe would need to be adjusted accordingly.

 

[nlightner]

2.3538 - 2.4 = -0.0462 m

Therefore, the second pipe is either 0.048 m too long or 0.0462 m too short. This difference in length causes the beat frequency of 1.4 Hz to be heard. This can be explained by the interference of sound waves from the two pipes, resulting in constructive and destructive interference at different points in time, creating a beat or fluctuation in the overall sound.