Question 6.9: Determine the forced response of the undamped two degree of ...
Determine the forced response of the undamped two degree of freedom system shown in Fig. 6.9, assuming that the force F_{2}(t) = 0.
In this case, we have
F = \begin{bmatrix} F_{1} \\ 0 \end{bmatrix}, M = \begin{bmatrix} m_{1} & 0 \\ 0 & m_{2} \end{bmatrix}, K = \begin{bmatrix} k_{1}+k_{2} & -k_{2} \\-k_{2} & k_{2} \end{bmatrix}
that is,
m_{11} = m_{1}, m_{22} = m_{2}, m_{12} = m_{21} = 0\\[0.5 cm]k_{11} = k_{1} + k_{2}, k_{22} = k_{2}, k_{12} = k_{21} = −k_{2}
The constant a of Eq. 6.108 is given by
Therefore, the amplitudes \overline{X_1} \text{and} \overline{X_2} of Eq. 6.110 are, in this case, given by
\overline{X_{1}} = \frac{1}{a} \frac{F_{1}(k_{22} − ω^{2}_{f} m_{22})}{(ω^{2}_{f} − ω^{2}_{1})(ω^{2}_{f}− ω^{2}_{2})} = \frac{F_{1}(k_{2} − ω^{2}_{f} m_{2})}{m_{1}m_{2}(ω^{2}_{f} − ω^{2}_{1})(ω^{2}_{f}− ω^{2}_{2})}\\[0.5 cm] \overline{X_{2}} = -\frac{1}{a} \frac{F_{1}(k_{21} − ω^{2}_{f} m_{21})}{(ω^{2}_{f} − ω^{2}_{1})(ω^{2}_{f}− ω^{2}_{2})} = \frac{F_{1}k_{2}}{m_{1}m_{2}(ω^{2}_{f} − ω^{2}_{1})(ω^{2}_{f}− ω^{2}_{2})}where the natural frequencies ω_1 \text{and} ω_2 are obtained using the constants of Eq. 6.108 as
The forced responses, x_1(t) \text{and} x_2(t), of the two masses m_1 \text{and} m_2 can then be written as
x_1(t) = \overline{X_{1}} \sin ω_ft, \qquad x_2(t) =\overline{X_{2}} \sin ω_ft
Plots of the steady state amplitudes \overline{X_{1}} \text{and} \overline{X_{2}} versus the frequency \omega_f are shown in Fig. 6.10. In this figure, the amplitude \overline{X_{1}} approaches zero when ωf approaches \sqrt{k_2/m_2}.
