ch+10+sol+CLASS (1)

ch+10+sol+CLASS (1) - 3 2. Solve for the string response,...

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3 2. Solve for the string response, Equation 10.27, given the following initial conditions: (a) y ( x; 0) = 0 , _ y ( x; 0) = g ( x ) (b) y ( x; 0) = sin x , _ y ( x; 0) = 0 (c) y ( x; 0) = 0 , _ y ( x; 0) = sin x (d) y ( x; 0) = sin x , _ y ( x; 0) = sin x . In all cases, sketch the initial conditions and plot the response y ( x;t ) . Solution: Equation 10.27 is y ( x;t ) = 1 X r =1 r 2 mL sin L A r sin L t + B r cos L t ± : Given initial conditions, y ( x; 0) and _ y ( x; 0) ; the constant coe¢ cients are obtained using the or- thogonality condition, R L 0 mY r Y p dx = ± rp ; where Y r ( x ) is the normalized r th modal function, Y r ( x ) = r 2 mL sin L : Applying the initial displacement, y ( x; 0) = 1 X r =1 r 2 mL sin L B r : Multiplying both sides by mY p ( x ) and integrating over 0 to L; we have Z L 0 y ( x; 0) m r 2 mL sin L dx = 1 X r =1 B r Z L 0 m r 2 mL sin L ! r 2 mL sin L ! dx: The only non-zero term on the right-hand side occurs when r = p; B p = Z L 0 y ( x; 0) m r 2 mL sin L dx: (1) A p = L Z L 0 _ y ( x; 0) m r 2 mL sin L dx: (2) (a) y ( x; 0) = 0 , _ y ( x; 0) = g ( x ) Zero initial displacement gives B r = 0 for all r: A r : Substituting _ y ( x; 0) with g ( x ) in Equation (2), we have A p = L m r 2 mL Z L 0 g ( x )) sin L dx:
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4 CHAPTER 10 CONTINUOUS MODELS (b) y ( x; 0) = sin x , _ y ( x; 0) = 0 Zero initial velocity results in A p = 0 : the initial displacement does not satisfy the boundary conditions. We can proceed with the math- ematics with the current initial displacement, but it will not make physical sense. A similar initial y ( x; 0) = sin Let us use this initial B p = Z L 0 y ( x; 0) m r 2 mL sin L dx B 1 = r mL 2 . All the other coe¢ cients except for B 1 are zero. This makes sense because the initial displacement present. Applying the second initial condition results in A p = 0 : (c) y ( x; 0) = 0 , _ y ( x; 0) = sin x This problem is the same as (b) with g ( x ) = sin x: Again, the initial condition does not satisfy the boundary condition. If we let _ y ( x; 0) = sin &x=L; A 1 = q mL 2 L and all the other coe¢ cients are zero. (d)
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This note was uploaded on 02/26/2012 for the course 650 443 taught by Professor Benaroya during the Fall '11 term at Rutgers.

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ch+10+sol+CLASS (1) - 3 2. Solve for the string response,...

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