Question

A friend in another city tells you that she has two organ pipes of different lengths, one open at both ends, the other open at one end only. In addition, she has determined that the beat frequency caused by the second-lowest frequency of each pipe is equal to the beat frequency caused by the third-lowest frequency of each pipe. Her challenge to you is to calculate the length of the organ pipe that is open at both ends, given that the length of the other pipe is 1.50 m

L=?

Answer 1

Let the lengths of the organ pipes be L(open at both ends) and L' (open at one end only)

So, the fundamental frequency of the pipe which is open at both ends , f = v/2L

So, second lowest frequency, f2 = v/L

Similarly, the second lowest frequency of the closed pipe :

f2' = 4v/3L'

So, beat frequency , B = f2 - f2' = (v)*( 1/L - 4/3L' )

Similarly, the beat frequency of the third lowest frequency of both :

B' = f3 - f3' = v/(2L/3) - v/(4L'/5)

Now, as per the question :

B = B'

So, (v)*( 1/L - 4/3L' ) = v/(2L/3) - v/(4L/5)

So, (1/L)*(1 - 2/3) = (1/L')*(4/3 - 4/5)

Now, L' = 1.5 m

So, (1/L)*(1 - 2/3) = (1/1.5)*(4/3 - 4/5)

So, L = 0.938 m <-------- answer