Question

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

• A. $8$
  
• B. $7$
  
• C. $2$
  
• D. $6$
  

Answer 1

Answer: (B)

$7$

Let the length of both pipe initially be $L$.

Fundamental frequency of open pipe initially ${\nu }_i=\frac{v}{2L}$

Fundamental frequency of closed pipe initially ${\nu }_i^{\prime }=\frac{v}{4L}$

Thus beat frequency $b_i={\nu }_i-{\nu }_i^{\prime }=\frac{v}{4L}$

$\therefore$ $\frac{v}{4L}=2$

$⟹$ $\frac{v}{8L}=1$

Now the length of open pipe becomes $\frac{L}{2}$ and that of closed pipe becomes $2L$.

So, fundamental frequency of open pipe now ${\nu }_f=\frac{v}{2\left(L/2\right)}=\frac{v}{L}$

Fundamental frequency of closed pipe now ${\nu }_f^{\prime }=\frac{v}{4\left(2L\right)}=\frac{v}{8L}$

Thus new beat frequency $b_f={\nu }_f-{\nu }_f^{\prime }=\frac{7v}{8L}$

$⟹$ $b_f=7\times 1=7$