Question

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.

Answer 1

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

Answer 2

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