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hw_5_solu

# hw_5_solu - ME163 Mechanical Vibrations(Winter 2009 HW-5...

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ME163 Mechanical Vibrations (Winter 2009) HW-5 Due 2.12.09 in class -25% if late within 24 hours -50% if later than that 1. Forced Vibration, The Complete Solution. Consider the direct forced vibration system: ¨ x + 0 . 5 ˙ x + 25 x = 1 . 55sin(8 t ) (1) (a) Derive the analytical expression for the total solution (both transient and steady state) For the steady state portion we have φ ( ω ) = arctan (2 ζω/ω n ) (1 - ( ω/ω n ) 2 ) g ( ω ) = ( F/k ) 1 r ((1 - ( ω/ω n ) 2 ) 2 + (2 ζω/ω n ) 2 ) x s ( t ) = g ( ω )sin( ωt - φ ( ω )) and for the transient solution we have φ t = arctan( x (0) ω n r (1 - ζ 2 ) / ( ˙ x (0) + ζω n x (0))) X t = r ( x (0) 2 + (( ˙ x (0) + ζω n x (0)) / ( ω n r (1 - ζ 2 ))) 2 ) x t ( t ) = X t exp( - ζω n t )sin( r (1 - ζ 2 ) ω n t + φ t ); and of course the total solution is the sum of the two solutions or x ( t ) = x s ( t ) + x t ( t ) (b) Solve the equations numerically in Matlab using ODE45 for a time span [0 , 30] and initial conditions x (0) = 1, ˙ x (0) = 1. Plot three graphs (1) evaluated the transient response derived above versus time and create a plot (2) evaluate the steady state response versus time as derived above. (3) plot the numerical solution from ODE45 next to the total solution (steady state + transient). Include all code The code is below clear all close all c=.5; k = 25; om = 8; m = 1; F = 1.55; tt = linspace(0,30,1000); x0 = 1; xdot0 = 1; [t,X] = ode45(@equation,tt,[x0 xdot0],[],m,c,k,om,F); omega = om; omega_n = sqrt(k/m); ME163 Mechanical Vibrations 1 HW5

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zeta = c/(2*omega_n*m); THETA = 0; PhaTF = atan2((2*zeta*omega/omega_n),(1 - (omega/omega_n)^2)); MagTF = (F/k)./sqrt( (1 - (omega/omega_n)^2)^2 + (2*zeta*omega/omega_n)^2 ); xsteady = MagTF*sin(omega.*t - PhaTF ) ;
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hw_5_solu - ME163 Mechanical Vibrations(Winter 2009 HW-5...

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