In a Quincke's experiment ,the sound intensity has a minimum value I at a particular position.As the sliding tube is pulled out by a distance of 16.5 mm ,the intensioty increases to a maximum of 9I.Take the speed of sound,in air to be 330m/s.(a) Find the frequency of the sound source.
Path difference in the two cases = 2 x 16.5 mm
Path difference for constructive interference = \(n\lambda\)
Path difference for destructive interference = \(\(n+\frac{1}{2}\)\lambda\)
Since maxima is observed during sliding it may be assumed that it should be the nearest one.
Nearest gap between \(n\lambda\) and \(\(n+\frac{1}{2}\)\lambda\) is \(\frac{\lambda}{2}\)
\(\frac{\lambda}{2}=33 mm\)
Now, frequency can be found.
Thank you very much,Sir.