Mechanical Vibrations — Question 10

Question 10 – Solution

Strategy. The work done by the actuator splits into stored energy and damping loss. Periodicity removes the net storage change over a cycle.

Step 1: Derive the power balance. For $E=x'^2/2+9x^2/2$, multiplying $x''+2x'+9x=F$ by $x'$ gives $$\boxed{E'=Fx'-2x'^2.}$$ The steady energy is periodic, so integrating over one period gives $\langle Fx'\rangle=2\langle x'^2\rangle$. This is an average statement for general $\omega$.

Step 2: Compute the mean input. Write $x=A\cos\omega t+B\sin\omega t$. Coefficient matching yields $$\Delta=(9-\omega^2)^2+4\omega^2,\quad A=\frac{6(9-\omega^2)}{\Delta},\quad B=\frac{12\omega}{\Delta}.$$ Since $x'=-\omega A\sin\omega t+\omega B\cos\omega t$, averaging the products gives $$\boxed{\overline P=\langle 6\cos\omega t\,x'\rangle =3\omega B=\frac{36\omega^2}{\Delta}\quad\mathrm W.}$$ Also $2\langle x'^2\rangle=\omega^2(A^2+B^2)=36\omega^2/\Delta$, independently checking the energy balance.

Step 3: Maximize average power. For $\omega>0$, $$\overline P=\frac{36}{(9/\omega-\omega)^2+4}\le 9\,\mathrm W.$$ Equality holds uniquely when $\omega^2=9$. Thus $\boxed{\omega=3\,\mathrm{s^{-1}},\quad\overline P_{\max}=9\,\mathrm W}$.

Step 4: Account for the work at the maximum. At this frequency $A=0$, $B=1$, so $x=\sin 3t\,\mathrm m$ and $x'=3\cos 3t\,\mathrm{m/s}$. Therefore $$E=\tfrac 92(\cos^23t+\sin^23t)=\boxed{4.5\,\mathrm J}.$$ The instantaneous input and damping powers are both $18\cos^23t\,\mathrm W$: here $E'=0$ at every instant, not just on average. Over $T=2\pi/3\,\mathrm s$, the supplied work is $9T=\boxed{6\pi\,\mathrm J}$, exactly the energy dissipated. Constant stored energy does not mean zero energy throughput.