1. Fundamental frequency of a closed organ pipe

For a closed organ pipe of length $L_c$, $f_c = \frac{v}{4L_c}$

2. First overtone frequency of an open organ pipe

For an open organ pipe of length $L_o$, the harmonics are: $f_n = \frac{nv}{2L_o}, \quad n=1,2,3,\dots$

So the first overtone is the second harmonic ($n=2$): $f_o = \frac{2v}{2L_o} = \frac{v}{L_o}$

Given: $L_o = 60\text{ cm}$

Hence, $f_o = \frac{v}{60}$

3. Equating the given frequencies

According to the question, $\frac{v}{4L_c} = \frac{v}{L_o}$

Cancel $v$: $\frac{1}{4L_c} = \frac{1}{L_o}$

So, $L_o = 4L_c$

Therefore, $L_c = \frac{L_o}{4} = \frac{60}{4} = 15\text{ cm}$

4. Match with options

The correct option is:

• A: 15 cm