consider two cylindrical pipes of equal length . One of these acts as a closed organ pipe and the other as open organ pipe . The frequency of the third harmonic in the closed pipe is 200 Hz higher than the first harmonic of the open pipe. calculate the fundamental frequency of the closed pipe.
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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 fundamental 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
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 fundamental 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
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1
Li = Lii = L
v3i = 200 + v1ii
v1i - ?
For the closed pipe
L = nλ/4 = nc / 4vi n , vin = nc/4L
where c = 340 m/s for n = 1 we have first harmonic, n = 2 second harmonic, n = 3 third harmonic, etc
For the open pipe:
L = nλ/2 = nc/2vii n , v ii n = nc/2L
3.340/ 4L = 200 + 340/2 L => L = 0.425 (m)
v1i = 340/ 4L = 200 (Hz)
Answer : 200 Hz
v3i = 200 + v1ii
v1i - ?
For the closed pipe
L = nλ/4 = nc / 4vi n , vin = nc/4L
where c = 340 m/s for n = 1 we have first harmonic, n = 2 second harmonic, n = 3 third harmonic, etc
For the open pipe:
L = nλ/2 = nc/2vii n , v ii n = nc/2L
3.340/ 4L = 200 + 340/2 L => L = 0.425 (m)
v1i = 340/ 4L = 200 (Hz)
Answer : 200 Hz
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