1}}\] and \[{{l}_{2}}\] be the length?s of closed and open pipes respectively. (Neglecting end correction) \[{{l}_{1}}=\frac{{{\lambda }_{1}}}{4}\Rightarrow {{\lambda }_{1}}=4{{l}_{1}}\] and \[{{l}_{2}}=\frac{{{\lambda }_{2}}}{2}\Rightarrow {{\lambda }_{2}}=2{{l}_{2}}\] GIVEN \[{{N}_{1}}={{n}_{2}}\] so \[\frac{v}{{{\lambda }_{1}}}=\frac{v}{{{\lambda }_{2}}}\Rightarrow \frac{v}{4{{l}_{1}}}=\frac{v}{2{{l}_{2}}}=\frac{{{l}_{1}}}{{{l}_{2}}}=\frac{1}{2}\]