Mechanical Vibrations — Question 5

Question 5 – Solution

Strategy. Compare complete release responses, rather than assuming a larger damping force always gives a faster return.

Step 1: Solve both release problems. For $c=4$, the repeated root is $-2$, giving critical damping. For $c=5$, roots $-1,-4$ give overdamping. Fitting the data yields $$\boxed{x_4=a(1+2t)e^{-2t},\qquad x_5=\frac a3(4e^{-t}-e^{-4t}).}$$ Their characteristic forms verify the equations; both have value $a$ and slope zero at the release.

Step 2: Check positivity and decrease. The first expression is positive, and the second is positive because $4e^{-t}>e^{-4t}$. Their derivatives are $$x_4'=-4at e^{-2t}<0,\qquad x_5'=\tfrac{4a}{3}(e^{-4t}-e^{-t})<0\quad(t>0).$$ Both tend to zero.

Step 3: Prove the comparison for every time. Applying $D^2+5D+4$ to $w=x_5-x_4$ gives $$w''+5w'+4w=-x_4',\qquad w(0)=w'(0)=0.$$ Variation of parameters with $e^{-t},e^{-4t}$ yields $$w(t)=\int_0^t\frac{e^{-(t-s)}-e^{-4(t-s)}}3[-x_4'(s)]\,ds.$$ For $0<s<t$, both factors are strictly positive. Hence $\boxed{x_5(t)>x_4(t)}$ for every $t>0$. This integral proof avoids relying on a finite plot or only on large-time rates.

Step 4: Interpret settling and heavy damping. Each response reaches any $\varepsilon\in(0,a)$ exactly once, by continuity and strict decrease. At the time $x_4=\varepsilon$, one still has $x_5>\varepsilon$, so the more heavily damped response reaches the level later. For general $c>4$ in this SI model, $$r_{\mathrm{slow}}=\frac{-c+\sqrt{c^2-4mk}}{2m} =\frac{-2k}{c+\sqrt{c^2-4mk}}\longrightarrow 0^-.$$ The release excites this mode with coefficient $-ar_{\mathrm{fast}}/(r_{\mathrm{slow}}-r_{\mathrm{fast}})>0$. Excessive damping therefore introduces a slow return, with $r_{\mathrm{slow}}\sim-k/c$; damping strength alone is not a measure of settling speed.