A closed organ pipe and an open organ pipe of same length produce 2 beats/second while vibrating in their fundamental modes. The length of the open organ pipe is halved and that of closed pipe is doubled. Then the number of beats produced per second while vibrating in the fundamental mode is

Given, \(\quad \mathrm{f}_{\mathrm{o}}-\mathrm{f}_{\mathrm{c}}=2 \quad \text{...(i)}\)
Frequency of fundamental mode for a closed Frequency of fundar organ pipe, \(f_{c}=\frac{v}{4 L_{c}}\)
Similarly frequency of fundamental mode for an open organ pipe, \(\mathrm{f}_{\mathrm{o}}=\frac{\mathrm{V}}{2 \mathrm{~L}_{\mathrm{o}}}\)
Given \(\quad \mathrm{L}_{\mathrm{c}}=\mathrm{L}_{\mathrm{o}}\)
\(\Rightarrow\) \(f_{0}=2 f_{c} \quad \text{...(ii)}\)
From Eqs. (i) and (ii), we get
\(\mathrm{f}_{\mathrm{o}}=2 \mathrm{f}_{\mathrm{c}}\) \(\mathrm{f}_{\mathrm{o}}=4 \mathrm{~Hz}\) and \(\mathrm{f}_{\mathrm{c}}=2 \mathrm{~Hz}\)
When the length of the open pipe is halved, frequency of fundamental mode is \( \mathrm{f}_{\mathrm{o}}^{\prime}=\frac{\mathrm{v}}{2\left[\frac{\mathrm{L}_{\mathrm{o}}}{2}\right]}=2 \mathrm{f}_{\mathrm{o}}=2 \times 4 \mathrm{~Hz}=8 \mathrm{~Hz} \)
When the length of the closed pipe is doubled, its frequency of fundamental mode is
\(\mathrm{f}_{\mathrm{o}}^{\prime}=\frac{\mathrm{v}}{4\left(2 \mathrm{~L}_{\mathrm{c}}\right)}=\frac{1}{2} \mathrm{f}_{\mathrm{c}}=\frac{1}{2} \times 2=1 \mathrm{~Hz}\)
Hence, number of beats produced per second is
\(\mathrm{f}_{\mathrm{o}}^{\prime}-\mathrm{f}_{\mathrm{c}}^{\prime}=8-1=7\)