Vertical motion of a unit mass on a certain coil spring.

Question

Question: The differential equation for the vertical motion of a unit mass on a certain coil spring in a certain medium is $$ \frac{d^2x}{dt^2}+2b\frac{dx}{dt}+b^2x=0 $$ where $b>0$. The initial displacement of the mass is $A$ feet and its initial velocity is $B$ feet per second.

i) Show that the motion is critically damped and that the displacement is given by $$x(t)=(A+Bt+Abt)e^{-bt}$$ ii) If $A$ and $B$ are such that $-\frac{A}{B+bA}$ and $\frac{B}{b(B+bA)}$ are both negative, show that the mass approaches its equilibrium position monotonically as $t$ goes to infinity without either passing through this equilibrium position or attaining an extreme displacement from it for $t>0$.

iii) If $A$ and $B$ are such that $-\frac{A}{B+bA}$ is negative and $\frac{B}{b(B+bA)}$ is positive, show that the mass doesn't pass through its equilibrium position for $t>0$ that its displacement from this position attains a single extremum at $t=\frac{B}{b(B+bA)}$ and that thereafter the mass tends to its equilibrium position monotonically as $t$ goes to infinite.

iv) If $A$ and $B$ are such that $-\frac{A}{B+bA}$is positive, show that the mass passes through its equilibrium position at $t=-\frac{A}{B+bA}$ attains an extreme displacement at $t=\frac{B}{b(B+bA)}$ and that thereafter the mass tends to its equilibrium position monotonically as $t$ goes to infinite.

Attempt: For part (i) here is what I did: The characteristic equation is $$ \lambda^2+2b\lambda+b^2=0 $$ Solving this we have $$ \lambda_{1,2}=-b $$ Hence the motion is critically damped and the displacement is given by $$ x(t)=(c_1+c_2t)e^{-bt} $$ Using the initial condition $x(0)=A$ and $x'(0)=B$ we have $$x(t)=(A+Bt+Abt)e^{-bt}$$ But I don't know what to do for part (ii),(iii) and (iv). Any help?

Answer 1

We can get the equilibrium position enforcing $x(t)=0$, so:

$$0=[A+t(B+bA)]e^{-bt}\Longrightarrow t=-{A\over B+bA}<0$$

hence the mass can't pass through the equilibrium position because $t<0$ in the case ii), but it approaches this equilibrium position when $t\to+\infty$:

$$\lim_{t\to+\infty}[A+t(B+bA)]e^{-bt}=0;$$

making the derivative:

$$x'(t)=(B+bA)e^{-bt}-b[A+t(B+bA)]e^{-bt}=[B-bt(B+bA)]e^{-bt}$$

and requiring $x'(t)=0$ we get $t^*={B\over b(B+bA)}$. For $t>t^*$ the derivative is always positive (if $t^*$ is a minimum) or negative (if $t^*$ is a maximum), so the mass approaches its equilibrium position monotonically. For the case ii) $t^*={B\over b(B+bA)}<0$, thus the mass can't attain an extreme displacement from its equilibrium position for $t>0$. Cases iii) and iv) should be easy now.