Consider two organ pipes of same length in which are organ is closed and another organ pipe is open. If the fundamental frequency of closed pipe is 250 Hz. Calculate the fundamental frquency of the open pipe.

Explanation

Closed `L=(lambda)/(4) rArr lambda=4`
Open `L=(lambda)/(2) rArr lambda=2L`
`{:("Frequency or"),("closed pipe" (f_(c))):}}=250 Hz`
Formula
`f_(0)=(V)/(lambda)`
`=(V)/(2L)`
`f_(0)=2((V)/(4L))`
`=2f_(c)`
`=2xx250`
`{:("Frequency or"),("the open pipe" (f_(c))):}}=500 Hz`