A tuning fork vibrates at `264 Hz`. Find the length of the shortest closed organ pipe that will resonate with the tuning fork. Speed of sound in air is `350ms^-1`.

Explanation

The resoN/Ant frquency of a closed organ pipe of length `l is (nv)/(4l)` where n is a positive odd integer and v is the speed of sound in air. To resoN/Ate with the given tuning fork
`(nv)/(4l)=264s^-1`
or `l=(nxx350ms^-1)/(4xx264s^-1)`
for l to be minimum `n=1 so that
l_(min)=350/(4xx264)m=33cm.