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 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

Options

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

Answer

D

Explanation

For a closed organ pipe, the frequency of fundamental mode is
`f_(C)=(v)/(4L_(C))`
where v is the velocity of sound in air and `L_(C)` is the length of the closed pipe
Fore an open organ pipe, the frequency of fundamental mode is
`f_(O)=(v)/(2L_(O))`
where `L_(O)` is the length of the oipen pipe
`becauseL_(C)=L_(O)` (Given)
`:.f_(O)=2f_(C)` . . . . (i)
`f_(O)-f_(C)=2` (Given) . . . (ii)
Solving (i) and (ii), we get
`f_(O)=4` Hz, `f_(C)=2` Hz
When the length of the open pipe is halved, its frequency of fundamental mode is
`f_(O)^(.)=(v)/(2((L_(O))/(2)))=2f_(O)=2xx4Hz=8Hz`
When the length of the closed pipe is doubled its frequency of fundamental mode is
`f_(C)^(.)=(v)/(4(2L_(C)))(1)/(2)f_(C)=(1)/(2)xx2Hz=1Hz`
Hence, number of beats produced per second
`=f_(O)^(.)=f_(C)^(.)=8-1=7`