IVP’s With Step Functions — Question 3

Question 3 – Solution

Strategy. Removing the force subtracts a delayed step response. The two surviving state coordinates determine whether motion persists.

Step 1: Transform and invert. For $s>0$, $$Y=\frac{F(1-e^{-Ts})}{s(s^2+\omega^2)},\qquad \boxed{y=\frac F{\omega^2}\left[1-\cos\omega t -H_T(t)(1-\cos\omega(t-T))\right].}$$ Before $T$ this has zero initial data and satisfies the forced equation. The subtracted response and its first derivative vanish at activation, so $y,y'$ match at $T$; after $T$ the equation is homogeneous.

Step 2: Recover the shutoff state and amplitude. The matching state is $$y(T)=\frac F{\omega^2}(1-\cos\omega T),\qquad y'(T)=\frac F\omega\sin\omega T.$$ For $t\ge T$, subtraction and the cosine difference identity give $$y(t)=\frac{2F}{\omega^2}\sin(\omega T/2) \sin(\omega(t-T/2)).$$ Its amplitude is therefore $\boxed{R=2F|\sin(\omega T/2)|/\omega^2}$.

Step 3: Classify complete return to rest. A homogeneous oscillator remains zero exactly when both state coordinates are zero. Here that happens precisely for $$\boxed{T=\frac{2\pi n}{\omega},\qquad n=1,2,\ldots.}$$ In this particular pulse family, $y(T)=0$ forces $\cos\omega T=1$ and hence also $y'(T)=0$. For a general forcing history, zero displacement alone is insufficient: a nonzero velocity launches a new oscillation.

Step 4: Compute energy and convergence. After shutoff, $$\boxed{E=\tfrac 12\omega^2R^2 =\frac{2F^2}{\omega^2}\sin^2(\omega T/2).}$$ If $R>0$, the oscillatory tail has an ordinary Laplace transform exactly for $s>0$: at zero its primitive does not converge, and negative $s$ gives growing integrals over fixed sign intervals. If $R=0$, the output has compact support, so its transform exists for every real $s$. Apparent poles in the displayed formula then cancel. The graph compares $F=\omega=1$ at two durations.