Question 4.5.2: Consider a 2-dof system whose matrix equation of motion is g......

a. Since the system is undamped and therefore the responses have no delay or phase difference with respect to the applied force, one can write

${\binom{x_{1}}{x_{2}}}={\binom{X_{1}}{X_{2}}}{\cos{\omega{t}}}=$${\binom{X}{\Theta}}\cos\omega t.$

In this equation, X and Θ are the amplitudes of instantaneous displacements $x_{i}.$

Substituting the above solution and its derivatives into Equation (i) and upon simplification, one has

$\begin{bmatrix}(k_{11}-m_{11}\omega^{2}) & k_{12} \\ k_{21} & (k_{22}-m_{22}\omega^{2}) \end{bmatrix}{\binom{X}{\Theta}}={\binom{0}{0}}$                  (ii)

or writing it in a more concise form

$\left[Z(\omega)\right]\left(_{X_{2}}^{X_{1}}\right)={\binom{0}{0}}$

where the so-called impedance matrix has been defined by Equation (4.18) as

$\begin{bmatrix}(k_{11} – m_{1}\omega^{2}) & k_{12} \\ k_{21} & (k_{22} – m_{2} \omega^{2}) \end{bmatrix}{\binom{X_{1}}{X_{2}}}={\binom{F_{1}}{0}}$            (4.18)

$[Z(\omega)]= \begin{bmatrix}(k_{11} – m_{11} \omega^{2}) & k_{12} \\ k_{21} & (k_{22} – m_{22} \omega^{2}) \end{bmatrix}$

The frequency equation is

$\left|{\begin{matrix}{(k_{11}-m_{11}\omega^{2})}&{k_{12}}\\ {k_{21}}&{{}(k_{22}-m_{22}\omega^{2})}\end{matrix}}\right|=0.$

Substituting the system parameters given above and writing λ = ω²,

$\left|{\begin{array}{ll}40,000 – 4000\lambda&10,000\\ 10,000&55,400−2560λ\end{array}}\right|=0.$

Operating on this equation, one has

$(40,000-4000\lambda)(55,400-2560\lambda)-(10,000)^{2}=0.$

This quadratic equation in λ gives $\lambda_{1}=9.2173,\,\lambda_{2}=59.8457,$ so that the two natural frequencies are

$\omega_{1}={\sqrt{9.2173}}=3.036\,\mathrm{rad/s},$       (iia)

$\omega_{2}={\sqrt{59.8457}}=4.736{\frac{\mathrm{rad}}{s}}$             (iib)

In order to obtain the mode shapes, one first substitutes $\omega_{1}$ into Equation (ii), resulting in

$\left({\frac{X}{\Theta}}\right)_{(1)}=-3.18\,\mathrm{m/rad}$             (iva)

where the subscript (1) on the lhs of the equation denotes the amplitude ratio associated with the first natural frequency.

Similarly, substituting ${{\omega}}_{2}$ into Equation (ii), one has

$\left({\frac{X}{\Theta}}\right)_{(2)}=0.202\,\mathrm{m/rad}.$         (ivb)

Equations (iva) and (ivb) give the two mode shapes of the 2-dof system.