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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. |
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Answer» 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. |
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