T 553 solve the following initial value problem for a

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Unformatted text preview: of the membrane located at (x, y ) with sides of length dx and dy is given by the expression ∂2u F = ρ 2 (x, y )dxdy k. ∂t Suppose the force displaces the membrane from a given position u(x, y ) to a new position u(x, y ) + η (x, y ), where η (x, y ) and its derivatives are very small. Then the total work performed by the force F will be F · η (x, y )k = η (x, y )ρ ∂2u (x, y )dxdy. ∂t2 Integrating over the membrane yields an expression for the total work performed when the membrane moves through the displacement η : Work = η (x, y )ρ D ∂2u (x, y )dxdy. ∂t2 (5.14) On the other hand, the potential energy stored in the membrane is proportional to the extent to which the membrane is stretched. Just as in the case of the vibrating string, this stretching is approximated by the integral Potential energy = T 2 D 2 ∂u ∂x + 2 ∂u ∂y dxdy, where T is a constant, called the tension in the membrane. Replacing u by u + η in this integral yields New potential energy = T 2 D 129 ∂u ∂x 2 + ∂u ∂y 2 dxdy ∂u ∂x +T D + T 2 ∂η...
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