The first overtone frequency of an open organ pipe is equal to the fundamental frequency of a closed organ pipe. If the length of the closed organ pipe is $$20$$ cm. The length of the open organ pipe is ______ cm

The length of the closed organ pipe is $$L_c = 20$$ cm. The fundamental frequency of a closed organ pipe is $$f_c = \frac{v}{4L_c}$$, and we require the first overtone of an open organ pipe to equal this fundamental frequency.

The first overtone (second harmonic) of an open pipe is $$f_o = \frac{2v}{2L_o} = \frac{v}{L_o}$$. Equating this to the fundamental of the closed pipe gives $$\frac{v}{L_o} = \frac{v}{4L_c}$$. Solving for the open pipe length yields $$L_o = 4L_c = 4 \times 20 = 80 \text{ cm}$$. Thus, the length of the open organ pipe is 80 cm.