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 1.20 m .

I found the answer to be 2.4, but that is incorrect so I need help

Answer 1

lam = wavelength

second lowest with open ends:
lam = Lo
second lowest with closed end
lam = 4/3 Lc

f proportional to 1/Lam
so
f open = k/Lo and f closed = 3k/4Lc
difference frequency = k (1/Lo -3/4Lc)


third lowest open
Lam = (2/3) Lo
third lowest closed
Lam = (4/5) Lc

f open = k (3/2Lo)
fclosed = k(5/4Lc)
difference frequency = k(3/2Lo -4/5Lc)

so
3/2Lo - 4/5Lc = 1/Lo - 3/4Lc
but Lc = 1.2
3/2Lo - 4/6 = 1/Lo - 3/4.8
3/2Lo - 2/2Lo = .6667-.625
1/2Lo = .0417
2 Lo = 24
Lo = 12

Answer 2

It says that is still wrong

Answer 3

Correct answer is 1.5 m