In a Quinck's experiment , the sound intensity being detected at an appropriate point , changes from minimum for the second time , when the slidable tube is drawn apart by 9.0 cm. If the speed of sound in air be 336 m//s, then what is the frequency of this sounding source ?

In a Quinck's experiment , the sound intensity being detected at an ap

1.	In a Quinck's experiment , the sound intensity being detected at an appropriate point , changes from minimum for the second time , when the slidable tube is drawn apart by 9.0 cm. If the speed of sound in air be 336 m//s, then what is the frequency of this sounding source ?
Answer» Solution :As the SLIDING tube is pulled out by a distance of `9.0 cm`, the path DIFFERNCE EFFECTED is ` 2 xx 9.0 cm = 18.0 cm`. INITIALLY , the position of the tube corresponds to the minimum intensity . When maximum intensity is heard for the first time , tjhe path differnce increased by `lambda//2` and when heard for the second time , there is an additional path differnce of `lambda`. THEREFORE , the total path difference ` = ( lambda)/(2) + lambda = ( 3 lambda)/(2)` `( 3 lambda)/(2) = 18.0 cm` `lambda = 12.0 cm` The frequency of sound emitted by the source will be ` f = (v)/( lambda) = ( 336 m//s) /( 12.0 xx 10^(-2) m) = 2.8 kHz`