Question 2.3.2: Consider the system of Figure 2.12. Let J denote the inertia......

Modeling the system as a torsional vibration problem and summing moments about the shaft, the equation of motion becomes

\ddot{J \theta}+k \theta=a F_0 \sin \omega t

Taking the Laplace transform (see Appendix B) of the equation of motion yields

J s^2 X(s)+k X(s)=a F_0 \frac{\omega}{s^2+\omega^2}

where X(s) is the Laplace transform of \theta(t). Algebraically solving for X(s) yields

X(s)=a \omega F_0 \frac{1}{\left(s^2+\omega^2\right)\left(J s^2+k\right)}

Next use a table (see Appendix B) to compute the inverse Laplace transform to get

\begin{aligned} \theta(t) & =L^{-1}(X(s))=a \omega F_0 L^{-1}\left(\frac{1}{\left(s^2+\omega^2\right)\left(J s^2+k\right)}\right) \\ & =\frac{a \omega F_0}{J} L^{-1}\left(\frac{1}{\left(s^2+\omega^2\right)\left(s^2+\omega_n^2\right)}\right)=\frac{a \omega F_0}{J} \frac{1}{\omega^2-\omega_n^2}\left(\frac{1}{\omega} \sin \omega t-\frac{1}{\omega_n} \sin \omega_n t\right) \end{aligned}

Here L^{-1} denotes the inverse Laplace transform and the natural frequency is

\omega_n=\sqrt{\frac{k}{J}}

Note that the solution computed here using the Laplace transform agrees with the solution obtained using the method of undetermined coefficients expressed in equation (2.25) for the case of zero initial conditions.
The transfer function for the system is simply

x(t)=x_0 \cos \omega_n t+\left(\frac{v_0}{\omega_n}-\frac{\omega}{\omega_n} \frac{f_0}{\omega_n^2-\omega^2}\right) \sin \omega_n t+\frac{f_0}{\omega_n^2-\omega^2} \sin \omega t     (2.25)

\frac{X(s)}{F(s)}=H(s)=\frac{1}{J s^2+k}

This result is in agreement with equation (2.59) for the case c=0.

\frac{X(s)}{F(s)}=\frac{1}{\left(m s^2+c s+k\right)}=H(s)     (2.59)