There is a square membrane of area S. Find the number of natural vibrations perpendicular ot its plane in the frequency interval from omega t o omega+domega if the propogation velocity of vibrations is equal to v.

Solution :Let `xi(x,y,t)` be the displacement of the element at `(x,y)` at time `t`. Then it obeys the equation
`(del^(2)xi)/(at^(2))=V^(2)((del^(2)xi)/(delx^(2))+(del^(2)xi)/(dely^(2)))`
where `xi=0 at x=0,x=l,y=0` and `y=l`
We look for a solution in the FORM
`xi=A sin k_(1) sink_(2) y sin (omegat+delta)`
Then `omega^(2)=V^(2)(k_(1)^(2)+K_(2)^(2))`
`K_(1)=(npi)/(l),k_(2)=(mpi)/(l)`
we write this as
`n^(2)+m^(2)=((t omega)/(piV))^(2)`
Here `n,mgt0`. Each paor `(n,m)` determines a mode. The TOTAL number of MODES whose frequency is `le omega` is the AREA of the quadrant of a circle of radius `(lomega)/(piV)` i.e.,
`N=(pi)/(4)(( omega)/(piV))^(2)`
Then `dN=(l^(2))/(2piV^(2))omega d omega=(S)/(2piv^(2)) omega d omega`.
where `S=l^(2)` is the area of the membrane.