SOLUTION SET UP (Figure 1) shows our sketch. Note that the frequency of the n =
ID: 776353 • Letter: S
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SOLUTION SET UP (Figure 1) shows our sketch. Note that the frequency of the n = 2 mode of the open pipe is twice its fundamental frequency, but the n = 5 mode of the stopped pipe has a frequency five times that of the fundamental Let's now apply our equations for the normal modes of open and closed pipes. On a day when the speed of sound is 345 m/s, the fundamental frequency of an open organ pipe is 690 Hz. If the n 2 mode of this pipe has the same wavelength as the n = 5 mode of a stopped pipe, what is the length of each pipe? SOLVE For an open pipe, u/2L, so the length of the open pipe is 345 m/s 2(690 s = 0.250 m The second harmonic of the open pipe has a frequency of f,-2/,-2(690 Hz) 1380 Hz If the wavelengths are the same, then the frequencies are the same, so the frequency of the n = 5 mode of the stopped pipe is also 1380 Hz. The n 5 mode frequency is 5f1-5(v/4L). If this equals 1380 Hz, then 5 ( 345 m/a 4L 1380 Hz and Lstopped 0.313 m stopped REFLECT A final possibility is a pipe that is closed at both ends and therefore has nodes at both ends. This pipe wouldn't be of much use as a musical instrument because there would be no way for the vibrations to get out of it. Part A-Practice Problem: On another day, if the difference between the frequency of the n- 4 mode of the open pipe and the frequency of the n = 9 mode of the stopped pipe is 4.45 Hz what is the speed of sound? Express your answer in meters per second to three significant figures m/s Submit Previous Answers Request Answer Incorrect; Try Again; 2 attempts remaining; no points deducted Figure 1 of 1 o 0.250 m Provide Feedback Next > Open pipe -690 Hz Open pipe: f-2f, Lclesed Stopped pipeExplanation / Answer
Given,
n = 4 ; n = 9 ; f = 4.45 Hz
L = v/2f = 345/2 x 690 = 0.25 m
f' = v/L = 345/0.25 = 1380 Hz
L' = 5v/4L = 5 x345/(4 x 1380) = 0.313 m
We know that, the frequency of nth mode of pipe is:
fn = n v/2L ; fodd = n v/4 L
f2 = 4 v/2L = 4v/2 x 0.25 = 8 v
f9 = 9 v/2L = 9 v/4 x 0.313 = 7.2 v
As per given condition
f2 - f9 = 4.45
8 v - 7.2 v = 4.45
v = 4.45/(8 - 7.2) = 5.563 m/s
Hence, v = 5.563 m/s
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