Mechanical Vibrations — Question 4

Question 4 – Solution

Strategy. Critical damping fixes the repeated root, but the sign of the linear factor still depends on the initial velocity.

Step 1: Fit the repeated-root solution. The characteristic polynomial is $(r+3)^2$. Thus $$\boxed{x(t)=e^{-3t}[a+(v+3a)t],\qquad x'(t)=e^{-3t}[v-3(v+3a)t].}$$ Here the numerical rate $3$ has units $\mathrm{s^{-1}}$. The two data follow immediately, and substitution gives $x''+6x'+9x=0$.

Step 2: Classify crossings. Since $e^{-3t}>0$, a positive-time zero occurs exactly when $v+3a<0$. In that case it is unique and $$\boxed{t_c=\frac{a}{-v-3a}.}$$ The linear factor changes sign there, so this is a crossing, not a touch. If $v\ge-3a$, the factor remains positive for all finite $t\ge 0$, including the equality case $x=ae^{-3t}$.

Step 3: Classify monotone positive return. For $v\ge-3a$, the displacement is positive. Its derivative is nonpositive for all $t\ge 0$ exactly when $v\le 0$, because $v-3(v+3a)t$ then begins nonpositive and decreases. Thus the required range is $$\boxed{-3a\le v\le 0.}$$ For $v>0$, the mass first moves farther from equilibrium. For $v<-3a$, it crosses equilibrium and cannot be a positive monotone return.

Step 4: Exhibit overshoot. For the stated data, $x=0.10e^{-3t}(1-3t)$ and $x'=e^{-3t}(-0.60+0.90t)$. Therefore $$\boxed{t_c=1/3\,\mathrm s,\quad t_{\min}=2/3\,\mathrm s, \quad x_{\min}=-0.10e^{-2}\,\mathrm m.}$$ The derivative changes from negative to positive at the minimum. Critical damping eliminates sustained oscillations, but sufficiently negative initial velocity still causes one overshoot. The comparison uses $\tau=3t$ and normalized displacement $x/a$.