Question 2.2.3: Compute the constants of integration A and ϕ of equation (2.......
Compute the constants of integration A and \phi of equation (2.37) and compare these values to the values of A and \phi for the unforced case given in Window 2.2 for the parameters \omega_n=10\ rad / s , \omega=5\ rad / s , \zeta=0.01, F_0=1000\ N , m=100\ kg, and the initial conditions x_0=0.05\ m and v_0=0.
x(t)=A e^{-\zeta \omega_n t} \sin \left(\omega_d t+\phi\right)+X \cos (\omega t-\theta) (2.37)
Solution First, compute the values of X and \theta from equation (2.36) for the particular solution (forced response). These are
x_p(t)=\overbrace{\frac{f_0}{\sqrt{\left(\omega_n^2-\omega^2\right)^2+\left(2 \zeta \omega_n \omega\right)^2}}}^X \cos (\omega t-\overbrace{\tan ^{-1} \frac{2 \zeta \omega_n \omega}{\omega_n^2-\omega^2}}^\theta) (2.36)
X=\frac{f_0}{\sqrt{\left(\omega_n^2-\omega^2\right)^2+\left(2 \zeta \omega_n \omega\right)^2}}=0.133\ m \quad \text { and } \quad \theta=\tan ^{-1}\left(\frac{2 \zeta \omega_n \omega}{\omega_n^2-\omega^2}\right)=0.013\ rad
so that the phase for the particular solution is nearly zero \left(0.76^{\circ}\right). Thus the solution given by equation (2.37) is of the form
x(t)=A e^{-\zeta \omega_n t} \sin \left(\omega_d t+\phi\right)+X \cos (\omega t-\theta) (2.37)
x(t)=A e^{-0.1 t} \sin (9.999 t+\phi)+0.133 \cos (5 t-0.013)
Differentiating this solution yields the velocity expression
v(t)=-0.1 A e^{-0.1 t} \sin (9.999 t+\phi)+9.999 A e^{-0.1 t} \cos (9.999 t+\phi)-0.665 \sin (5 t-0.013)
Setting t=0 and x(0)=0.05 in the expression for x(t) and solving for A yields
A=\frac{0.005-0.133 \cos (-0.013)}{\sin (\phi)}=\frac{-0.083}{\sin (\phi)}
Substitution of this value of A, setting t=0 and v(0)=0 in the expression for the velocity yields
0=-0.1(-0.083)+9.999(-0.083) \cot (\phi)+0.665 \sin (0.013)
Solving for the value of \phi yields \phi=1.55\ rad \left(88.8^{\circ}\right) and thus A=-0.083\ m, the values of the amplitude and phase of the transient part of the solution (including the effects of the initial conditions and the applied force).
Next, consider the coefficients A and \phi evaluated for the homogenous case F_0=0, but keeping the same initial conditions. Using these values (see Window 2.2), the incorrect magnitude and phase become
A=\sqrt{(0.05)^2+\left[\frac{(0.01)(10)(0.05)}{10 \sqrt{1-(0.01)^2}}\right]^2}=0.05\ m
and
\phi=\tan ^{-1}\left(\frac{9.999}{0.05(10)}\right)=1.521 rad \left(87.137^{\circ}\right)
Comparing these two sets of values for the magnitude A and the two sets of values for the phase \phi, we see that they are very different. Thus the values of the constants of integration are greatly affected by the forcing term. In particular, the amplitude of the transient is greatly increased and its phase is reduced by the influence of the driving force.