Question 6.10: Determine the steady state response of the two degree of fre...

Assuming θ_{2} > θ_{1} and \dot{θ}_{2} > \dot{θ}_{1}, and using the free body diagram shown in the figure, one obtains the following two differential equations
m_{1}l^{2} \ddot{θ} = kl^{2}_{1}(θ2 − θ1) − m_{1}glθ_{1} + Fl sin ωft
m_{2}l^{2}\ddot{θ}_{2} = −kl^{2}_{1}(θ_{2} − θ_{1}) − m_{2}glθ_{2} + T sin ωft
which can be written in matrix form as

in which
m_{11} = m_{1}l^{2},\qquad  m_{22} = m_{2}l^{2},\qquad m_{12} = m_{21} = 0\\k_{11} = kl^{2}_{1}+ m_{1}gl, \qquad k_{22} = kl^{2}_{1} + m_{2}gl, \qquad k_{12} = k_{21} = −kl^{2}_{1}

F_{1} = Fl, F_{2} = T

The constants of Eq. 6.99 are then defined as
a = m_{1}m_{2}l^{4}
b = −[m_{1}l^{2}(kl^{2}_{1}+ m_{2}gl) + m_{2}l^{2}(kl^{2}_{1}+ m_{1}gl)]

= −2m_{1}m_{2}gl^{3} − (m_{1} + m_{2})kl^{2}l^{2}_{1}
c = (kl^{2}_{1} + m_{1}gl)(kl^{2}_{1}+ m_{2}gl) − k^{2}l^{4}_{1}

= m_{1}m_{2}(gl)^{2} + kl^{2}_{1}gl(m_{1} + m_{2})
In terms of these constants, the natural frequencies ω_1 \text{ and } ω_2 are given by
ω_1 = \frac{−b +\sqrt{b^2 − 4ac}}{2a}\qquad, ω_2 =\frac{−b −\sqrt{b^2 − 4ac}}{2a}
In order to obtain the steady state response, assume a solution in the form

Substituting this assumed solution into the differential equations, the amplitudes Θ_{1} \text{ and } Θ_{2} can be determined using Eq. 6.110 as
Θ_1 = \frac{1}{m_1m_2l^4} \frac{F_1l[kl^2_1+ m_2gl − ω^2_f m_2l^2] + T kl^2_1}{(ω^2_f− ω^2_1)(ω^2_f− ω^2_2)}\\[0.5 cm] Θ_2 = \frac{1}{m_1m_2l^4} \frac{T[kl^2_1+ m_1gl − ω^2_f m_1l^2] +Fl kl^2_1}{(ω^2_f− ω^2_1)(ω^2_f− ω^2_2)}