Mechanical Vibrations — Question 3

Question 3 – Solution

Strategy. Track energy in both position and velocity. Displacement can grow while total mechanical energy falls.

Step 1: Solve the free motion. The equation is $x''+2x'+5x=0$, with roots $-1\pm 2i$, so the motion is underdamped. The data give $$\boxed{x=\tfrac 12e^{-t}\sin 2t,\qquad x'=e^{-t}[\cos 2t-\tfrac 12\sin 2t].}$$ These expressions have $x(0)=0$, $x'(0)=1$ and residual zero.

Step 2: Derive the dissipation law. Multiplying the equation by $x'$ gives $$E'=x'(mx''+kx)=-cx'^2=\boxed{-2x'^2\le 0}.$$ Nevertheless, $x'(0)=1$, so $x$ and $|x|$ initially increase from zero. Kinetic energy is partly converted into spring energy while the damper removes energy. Monotonic total energy is not monotonic displacement.

Step 3: Find the first turning point. The first root of $x'=0$ satisfies $\tan 2t=2$, in $0<2t<\pi/2$. Set $a=\tfrac 12\arctan 2$. Then $$\boxed{t_*=a,\qquad x(t_*)=e^{-a}/\sqrt 5\,\mathrm m.}$$ The velocity changes from positive to negative, so this is a positive maximum. Initially $E(0)=1/2\,\mathrm J$. At the turning point, $$E(a)=\tfrac 52x(a)^2=\tfrac 12e^{-2a}\,\mathrm J, \quad \boxed{E(0)-E(a)=\tfrac 12(1-e^{-2a})\,\mathrm J.}$$

Step 4: Interpret zero instantaneous loss. At a turning point, $x'=0$, so damper power is instantaneously zero. But $x\ne 0$ there and $x''=-5x\ne 0$, so motion resumes immediately. The displayed response and velocity both tend to zero; hence $E\to 0$ and total dissipated energy is $1/2\,\mathrm J$. In the figure, $X=\sqrt{k/m}\,x$ and $V=x'$ both have velocity units, and $E=m(X^2+V^2)/2$.