A tube of certain diameter and length 48 cm is open at both ends. Its fundamental frequency of resonance is found to be 320 Hz. The velocity of sound in air is 320 m/s. Estimate the diameter of the tube. One end of the tube is now closed. Calculate the fundamental resonance for the tube.

Explanation

`f_(0)=(v)/(2[L+2e])=(v)/(2[L+2xx0.6r])[as=e0.6r]`
So subsituting the given data,
`320=(320xx100)/(2[48+1.2r]) or r=(10)/(6) cm`
So `D=2r=xx(10//6)=3.33 cm`
Now when one end is closed,
`f_(c)=(v)/(4(L+0.6r))=(340xx100)/(4[48+0.6xx(10//6)])=163.3 Hz`