Given : Open pipe, no = 600 Hz, nc 1 = no 1 (first overtones)
For an open pipe, the fundamental frequency,
nO = v/2Lo
The length of the open pipe is
Lo = v /2nO = \(\frac{330}{2×600}\) = 0.275 m
For the open pipe, the frequency of the first overtone is
2nO = 2 x 600 = 1200 Hz
For the pipe closed at one end, the frequency of the first overtone is 3v/Lo
By the data 3v/4L = 1200
∴ Lc = \(\frac{3×330}{4×1200}\) = 0.206 m
The length of the pipe open at both ends is 27.5 cm and the length of the pipe closed at one end is 20.6 cm.