SolSec 6.6 - 6- 45 Problems and Solutions Section 6.6 (6.48...

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6- 45 Problems and Solutions Section 6.6 (6.48 through 6.52) 6.48 Calculate the natural frequencies of the membrane of Example 6.6.1 for the case that one edge x = 1 is free. Solution: The equation for a square membrane is ww w tt yy tt += ρ τ with boundary condition given by w (0, y ) = 0, w x ( l,y ) = 0, w ( x ,0) = 0, w ( x,l ) = 0. Assume separation of variables w = X ( x ) Y ( y ) q ( t ) which yields ′′ + == = X X Y Yc q q c 1 2 2 ˙˙ / ωρ where Then qc q 22 0 ω is the temporal equation and =− X X Y Y ωα yields XX YY α γ 2 2 0 0 as the spatial equation where 2 = 2 2 and 2 = 2 + 2 . The separated boundary conditions are X (0) = 0, = Xl () 0 and Y (0) = Y ( l ) = 0. These yield XA xB x B Al l n n l n n =+ = = = = sin cos cos () αα π 0 0 21 2 2
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6- 46 Next Y = C sin γ y + D cos y with boundary conditions which yield D = 0 and C sin l = 0. Thus γπ m ml = and for l = 1 we get a n = () , 21 2 n π for m = m π n,m = 1, 2, 3,… ωα ππ ωπ nm n m nm n m nm cc 22 2 2 2 2 4 4 4 4 4 =+= += −+ = So that ω nm c =− + 4 2 are the natural frequencies.
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6- 47 6.49 Repeat Example 6.6.1 for a rectangular membrane of size
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SolSec 6.6 - 6- 45 Problems and Solutions Section 6.6 (6.48...

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