Question 9.6: Find the maximum response of an undamped single-degree-of-fr...

The natural period of the system is T=2 \pi / \omega=1 \mathrm{~s}. The maximum response may be attained either within the duration of the forcing function or in a free-vibration era after the force has ceased to act. If the maximum does occur within the free-vibration era, it will be attained within a half-cycle of free motion. We should therefore compute the response of the system for a length of time equal to at least 1.6+0.5 T=2.1 \mathrm{~s}.

We select a time duration of 2.4 \mathrm{~s} for the calculation of response. This also becomes the time period T_{0} for a frequency-domain analysis using discrete Fourier transform. The forcing function is extended by adding a sufficient number of zeros to make up the period. It is then sampled at 0.1 \mathrm{~s} intervals, as shown in Figure E9.6a. The number of samples N is equal to 24 .

The periodic impulse response function \bar{h}(t), obtained from Equation 9.80,

\bar{h}(t)=\frac{1}{2 m \omega}\left(\sin \omega t+\frac{\cos \omega t \sin \omega T_0}{1-\cos \omega T_0}\right) \qquad (9.80)

is shown in Figure E9.6b. The response of the system to a periodic version of the forcing function with a period of T_{0} is obtained by convolving the sampled forcing function and \bar{h}(t) through the frequency domain using Equation 9.82.

\bar{u}(k \Delta t)=\frac{1}{2 \pi} \sum\limits_{n=0}^{N-1} G(n \Delta \Omega) \bar{H}(n \Delta \Omega) e^{2 \pi i k n / N} \Delta \Omega \qquad (9.82)

The resulting periodic response \bar{u}(t) is shown in Figure E9.6c and Table E9.6. The periodic response is expected to be different from the transient response because in the former the displacement and velocity at the beginning of a period are not zero. The displacement at t=0 in the periodic response is found to be 0.0793 in., while the velocity at the same time calculated from Equation 9.85

\dot{\bar{u}}(0)=-\frac{4 \pi}{\left(T_0\right)^2} \sum\limits_{n=0}^{N / 2} n \operatorname{Im}\{U(n \Delta \Omega)\} \quad(9.85)

works out to 0.4105 \mathrm{in} . / \mathrm{s}.

The corrective responses are obtained from Equations 9.89 \mathrm{a} and 9.89 \mathrm{b}

\begin{aligned}&\eta_1(t)=\Delta u(0) r(t) \quad(9.89 \mathrm{a}) \\&\eta_2(t)=\Delta \dot{u}(0) s(t) \quad \text { (9.89b) }\end{aligned}

with \Delta u(0)= -0.0793 and \Delta \dot{u}(0)=-0.4105 in./s. The estimate of transient response obtained by the superposition of periodic response \bar{u}(t) and the corrective responses is shown in Figure E9.6d and Table E9.6. For the purpose of comparison, the exact response obtained from Equations b, c, and \mathrm{d} of Example 9.4 is also shown.

Table E9.6 Response calculation through superposition of corrective responses based on initial conditions.