Question

An organ pipe closed at one end and another open at both the ends will resonate with each other, if their lengths are in the ratio

Answer 1

1:2

Answer 2

For a pipe closed at one end, the fundamental frequency is

$$f_{\text{closed}} = \frac{v}{4L_c}$$

For an open pipe, the fundamental frequency is

$$f_{\text{open}} = \frac{v}{2L_o}$$

For resonance, $f_{\text{closed}} = f_{\text{open}}$, therefore:

$$\frac{v}{4L_c} = \frac{v}{2L_o}$$

Cancelling $v$ and cross-multiplying gives:

$$2L_o = 4L_c \quad\Longrightarrow\quad L_o = 2L_c$$

Thus, the pipe lengths must be in the ratio

$$L_c : L_o = 1:2$$