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Find the number of natural transverse vibration of a rigtht-angled parallelipoped of volume `V` in the frquency interval from `omega to omega +d omega` if the propogation velocity of vibrations is equal to `v`

Answer» For transverse vibrations of a 3-diamensional continuum (in the from of a cube say). We have the equation
`(del vec(xi))/(delt^(2))=V^(2)vecgrad^(2)vec(xi), di v vec(xi)=0`
Here `vec(xi)=vec(xi)(x,y,z,t)`. We look for solutions in the form
`vec(xi)=vec(A) sink_(1)x,sink_(2)y,sink_(3Ƶ), sin (omega t+delta)`
This requires `omega^(2)=v^(2)(k_(1)^(2)+k_(2)^(2)+k_(3)^(2))`
From the boundary condition that `vec(xi)=0` for `x=0, x=l,y=0,y=l,Ƶ=l`, we get
`k_(1)=(n_(1)pi)/(l),k_(2)=(n_(2)pi)/(l)=(n_(2)pi)/(l),k_(3)=(n_(3)pi)/(l)`
where `n_(1):n_(2),n_(3)` are nonzero positive intergers.
We then get `n_(1)^(2)+n_(2)^(2)+n_(3)^(2)=((lomega)/(pi v))^(2)`
Each triplet `(n_(1),n_(2),n_(3))` determines a possible mode and the number of such modes whose frequency `le omega` is the volume of the all positive octant of a sphere of radius `(l omega)/(pi v)`. Considering also the fact that subsidiary conditions div `vec(xi)=0` implies two independent values of `vec(A)` for we find
`N(omega)=(1)/(8).(4 pi)/(3)((l omega)/(pi v))^(3).2=(V omega^(3))/(3 pi^(2)v^(3))`
Thus `dN=(Vomega^(2))/(pi^(2)v^(3))domega`


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