consider 2 cylindrical pipes of equal length .one of these acts as a closed organ pipes and the other as open organ pipes .the frequency of the third harmonic in the closed pipes is 200 hz higher than the harmonic of the open pipes .calculated the fundamental frequency of the closed pipes .
Answers
It is not clear for which harmonic of the open pipe is to be taken.... I take 2nd...
For the closed pipe (closed on one end):
L = (2n+1) λc /4 , n = 0, 1, 2 , 3 .
For 3rd Harmonic, L = 7λc /4 => λc =
4L/7
frequency = f_c = v/λc = 7 v/4L
For the open pipe (open on both ends)
L = (n+1) λo /2 , n = 0, 1 ,2 ,3
For the 2nd harmonic: λo = 2*L/3
Frequency = f_o = 3v/2L
Given 7v/4L - 3v/2L = 200 Hz
v/4L = 200 Hz
v/L = 800 Hz
fundamental frequency of closed pipe = v/4L = 200 Hz
fundamental frequency of open pipe: = v/2L = 400 Hz
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If the note for the open pipe is the 1st harmonic then :
For the closed pipe (closed on one end):
L = (2n+1) λc /4 , n = 0, 1, 2 , 3 .
For 3rd Harmonic, L = 7λc /4 => λc =
4L/7
frequency = f_c = v/λc = 7 v/4L
For the open pipe (open on both ends)
L = (n+1) λo /2 , n = 0, 1 ,2 ,3
For the 1st harmonic note: λo = L
Frequency = f_o = v/L
Given 7v/4L - v/L = 200 Hz
3v/4L = 200 Hz
v/L = 800/3 Hz
fundamental frequency of closed pipe = v/4L = 200/3 Hz
fundamental frequency of open pipe: = v/2L = 400/3 Hz
