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, Free vibration

, Free vibration. Eigenvalue equation. EIGENVALUE EQUATION. Example 1.1 ( Contiue ):. Eigenvalue E quation. Eigenvalue E quation. m=0.184 kg, L =0.24 m, k=201.1 N/m, c=2.34 Ns/m. MatLAB Code:. a=[0.0001828,0.05,4.46,182.96,5241.18];p=roots(a);vpa(p,4). clc;clear

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, Free vibration

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  1. , Free vibration Eigenvalue equation EIGENVALUE EQUATION Example 1.1 (Contiue): Eigenvalue Equation

  2. Eigenvalue Equation m=0.184 kg, L =0.24 m, k=201.1 N/m, c=2.34 Ns/m MatLAB Code: a=[0.0001828,0.05,4.46,182.96,5241.18];p=roots(a);vpa(p,4) clc;clear m0=0.184;l0=0.24;k0=201.1;c0=2.34; m=[m0*l0^2/8,0;0,3*m0/4]; c=[27*c0*l0^2/16,-9*c0*l0/4;-9*c0*l0/4,6*c0]; k=[9*k0*l0^2/8,-3*k0*l0/2;-3*k0*l0/2,4*k0]; syms s;p=solve(det(m*s^2+c*s+k));vpa(p,4) Eigenvalues: -18.5+42.1i, -18.5-42.1i, -97.4, -139

  3. p=-σ+iω rad/s iω ω0 φ -σ Eigenvalues : -18.5+42.1i, -18.5-42.1i, -97.4, -139 The form of free vibration response : Initial conditions determine the values of A1, φ1, A2 and A3 Free responses approach zero as t  . Steady state response becomes zero A system is stable if its whole eigenvalues have negative real parts. f0=1/T0 -97.4 için Δt=0.0033 , t∞=0.0645 p=-σ -139 için Δt=0.0023 , t∞=0.0452 For the system Δt=0.0023 , t∞=0.34

  4. x(t) x(t) 1 0.5 5 0.2 3 ξ=0.1 t t

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