Mechanical Vibrations — Question 8

Question 8 – Solution

Strategy. The damper responds to velocity relative to the support. Distinguish motion of the mass from deformation of the suspension.

Step 1: Derive the two models. Newton’s law gives $x''=-9(x-b)-2(x'-b')$. Therefore $$x''+2x'+9x=2b'+9b, \qquad \boxed{z''+2z'+9z=-b''=B\omega^2\cos\omega t.}$$ The inertial term in the relative equation comes from $x''=z''+b''$; it is not an extra spring force.

Step 2: Compute both amplitude ratios. Let $\Delta=(9-\omega^2)^2+4\omega^2$. Matching sine and cosine in the relative equation gives $$z=A\cos\omega t+D\sin\omega t,\quad A=\frac{B\omega^2(9-\omega^2)}{\Delta},\quad D=\frac{2B\omega^3}{\Delta}.$$ Thus $Z/B=\omega^2/\sqrt\Delta$. Since $x=z+B\cos\omega t$, its cosine coefficient is $B+A$, not its amplitude $B+Z$. Simplifying $(B+A)^2+D^2$ gives $$\boxed{\frac XB=\frac{\sqrt{81+4\omega^2}}{\sqrt\Delta},\qquad \frac ZB=\frac{\omega^2}{\sqrt\Delta}.}$$ The homogeneous roots $-1\pm 2\sqrt 2i$ ensure the initial-data transients decay.

Step 3: Find the isolation threshold. Since $\Delta>0$, $X/B<1$ exactly when $$\Delta-(81+4\omega^2)=\omega^2(\omega^2-18)>0.$$ For $\omega>0$, this means $\boxed{\omega>3\sqrt 2\,\mathrm{s^{-1}}}$. Equality holds at $3\sqrt 2$; below that positive frequency the absolute motion is amplified.

Step 4: Interpret the limiting motions. As $\omega\to 0^+$, $X/B\to 1$ and $Z/B\to 0$: the mass follows the slowly moving support. As $\omega\to\infty$, $X/B\to 0$ but $Z/B\to 1$. In fact $A/B\to-1$ and $D/B\to 0$, so $z\approx-b$ while $x\approx 0$. An almost stationary mass can still have relative deformation comparable to the base motion.