The second overtone of an open pipe has the same frequency as the first overtone of a closed pipe $2 \mathrm{~m}$ long. The length of the open pipe is

Let $L$ be the length of the open pipe. The fundamental frequency of the open pipe is given by $v=\frac{v}{2 L}, v=$ velocity of sound in air The second overtone of the open pipe has a frequency $3 v_O=\frac{3 v}{2 L_O} \mathrm{~Hz}$ The length of the closed pipe $L_C=2 \mathrm{~m}$. The fundamental frequency of the closed pipe is given by $v=\frac{v}{4 L}$ The first overtone of the closed pipe has a frequency $$ 3 v_C=\frac{3 v}{4 L_C}=\frac{3 v}{4 \times 2}=\frac{3 v}{8} \mathrm{~Hz} $$ Now $3 v_O=3 v_C$ or $\frac{3 v}{2 L_O}=\frac{3 v}{8}$ or $2 L_O=8$ or $L_O=4 \mathrm{~m}$