Frequencies of organ pipes

• For an open organ pipe of length $L_o$, the fundamental frequency is $f_1^{(o)} = \frac{v}{2L_o}$ The first overtone is the second harmonic: $f_{\text{first overtone}}^{(o)} = 2f_1^{(o)} = \frac{v}{L_o}$

• For a closed organ pipe of length $L_c$, the fundamental frequency is $f_1^{(c)} = \frac{v}{4L_c}$

Given condition

The first overtone frequency of the open pipe equals the fundamental frequency of the closed pipe: $\frac{v}{L_o} = \frac{v}{4L_c}$

Substitute $L_c = 20\text{ cm}$

$\frac{1}{L_o} = \frac{1}{4 \times 20} = \frac{1}{80}$

Hence, $L_o = 80\text{ cm}$

Final answer

The length of the open organ pipe is: $\boxed{80\text{ cm}}$

Comparison with stored answer

Stored correct answer = $80$

This matches the derived answer.