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Unformatted text preview: Homework 10.1 A fixedfixed string having a tension of P and mass/length ρ is excited by a force/length of f ( x,t ) = f g ( t ), as shown below with f being constant in both x and t . If g ( t ) = sin Ω t , the particular solution for the EOM for this system at x = L/ 3 can be written as u ( L/ 3 ,t ) = U (Ω) sin Ω t a) Determine an expression for U (Ω) using modal uncoupling. This result will be in terms of an infinite summation. b) Identify the modes that do NOT contribute to U (Ω). c) Make a hand sketch of U (Ω) vs. Ω for Ω up through the first four resonance frequencies that appear in the response at x = L/ 3. (Note that not all modes will contribute to this response.) Clearly indicate all frequencies of resonance and antiresonance in this frequency range. f g t ( ) u x , t ( ) SOLUTION The natural frequencies and massnormalized modal functions for the fixedfixed string are given by: ω j = jπ s P ρL 2 ˆ U ( j ) ( x ) = r 2 ρL sinjπ x L The following modal coordinate transformation using these modes: u ( x,t ) = ∞ X j =1 ˆ U ( j ) ( x ) p j ( t ) produces the following uncoupled modal EOMs: ¨ p j + ω 2 j p j = Z L ˆ U ( j ) ( x ) f ( x,t ) dx = ˆ f j sin Ω t where ˆ f j = f Z L L/ 2 ˆ U ( j ) ( x ) dx = f r 2 ρL Z L L/ 2 sinjπ x L dx = f r 2 ρL L jπ cosjπ x L x = L x = L/ 2 = f r 2 ρL L jπ cosj π 2 cosjπ The particular solutions of the modal equations are: p j ( t ) = ˆ f j ω 2 j Ω 2 sin Ω t = P j sin Ω t Therefore, u ( L 3 ,t ) = ∞ X j =1 ˆ U ( j ) ( L 3 ) P j sin Ω t = U (Ω) sin Ω t where U (Ω) = f r 2 ρL 2 L π ∞ X j =1 ( cosj π 2 cosjπ ) sinj π 3 j ( ω...
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This document was uploaded on 12/23/2011.
 Fall '09

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