Mechanical Vibrations — Question 9

Question 9 – Solution

Strategy. Position and velocity matching ensure a continuous state, but continuous actuator force also requires compatible acceleration.

Step 1: Fit the cubic. Write $x=a_0+a_1t+a_2t^2+a_3t^3$. The initial conditions give $a_0=0.10$, $a_1=0$; the final conditions give $a_2+a_3=-0.10$, $2a_2+3a_3=0$. Hence $$\boxed{x_c=0.10(1-3t^2+2t^3).}$$ The nonsingular two-equation system makes this cubic unique.

Step 2: Check the actuator joins. Differentiating gives $x_c'=0.60(t^2-t)$ and $x_c''=1.20t-0.60$. Thus $$\boxed{F_c=x_c''+2x_c'+5x_c=t^3-0.30t^2-0.10\quad\mathrm N.}$$ The holding force before zero is $5(0.10)=0.50\,\mathrm N$. Yet $F_c(0)=-0.10\,\mathrm N$ and $F_c(1)=0.60\,\mathrm N$, so both joins require force jumps. Acceleration jumps from zero to $-0.60$ initially and from $0.60$ to zero finally. The joined trajectory is $C^1$ and piecewise $C^2$, not globally $C^2$.

Step 3: Add acceleration matching. For a polynomial of degree at most five, the initial conditions fix $a_0=0.10$, $a_1=a_2=0$. The endpoint conditions become $$a_3+a_4+a_5=-0.10,\quad 3a_3+4a_4+5a_5=0,\quad 6a_3+12a_4+20a_5=0.$$ Solving gives $a_3=-1$, $a_4=1.5$, $a_5=-0.6$, uniquely. Therefore $$\boxed{x_q=0.10(1-10t^3+15t^4-6t^5).}$$

Step 4: Verify force continuity and rest. Substitution into the model yields $$\boxed{F_q=0.50-6t+12t^2-5t^3+1.5t^4-3t^5\quad\mathrm N.}$$ It has $F_q(0)=0.50$ and $F_q(1)=0$, matching the constant forces outside the motion interval. All six trajectory conditions hold, so the joined motion is $C^2$. For $t>1$, zero forcing and zero value and slope at one have the unique solution $x=0$. The improvement is continuous force and acceleration; no claim of a force-minimizing trajectory is needed.