The funadamental frequency of a closed organ pipe is equal to the first overtone frequency of an open organ pipe. If the length of the open pipe is 60 cm, what is the length closed pipe?

Explanation

In a closed organ pipe,
fundamental freq. `v_(1)=(upsilon)/(4l_(1))`
In an open organ pipe,
fundamental freq. `=(upsilon)/(2l_(2))`
and frq. Of first overtone `v_(20)=(2xx upsilon)/(2l_(2))=(upsilon)/(l_(2))`
As `v_(2)=v_(2)`
`:. (upsilon0)/(4l_(1))=(upsilon)/(l_(2)) or 4l_(1)=l_(2)=60`
`l_(1)=(60)/(4)=15cm`