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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?

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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.

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  • $\begingroup$ The reason for not attaining the extreme point is that $t^*=\frac{B}{b(B+bA)}<0$. (or as I understand, the sign of the derivative doesn't change for $t>0$. Am I right?) Then when I move to iii) the only difference is it is given that $t^*=\frac{B}{b(B+bA)}>0$ how this give us a single sign change in the derivative? I'm too confused with parameter and variable changes. Any help please? $\endgroup$ – marya Dec 23 '16 at 7:14
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    $\begingroup$ Yes it is right, the reason for not attaining the extreme point is that $t^*<0$. The sign of the derivative is another thing and it say if the function is increasing or decreasing. For iii) $t^*>0$ and we know that we have a max or min, actually it doesn't matter if max or min, the important is that after a max the function increases and after a min the function decreases so anyway the function is monotone. Why is it so? This is because the derivative is linear in $t$ (apart from the factor $e^{-bt}$ always positive), so there is only a solution to $x'(t)=0$. $\endgroup$ – MattG88 Dec 23 '16 at 12:41
  • $\begingroup$ Remember that a negative time doesn't make sense physically so this is the reason for not attaining the extreme point in ii) where $t^*<0$, but the function after $t^*$, in particular for $t>0$, must decrease or increase, anyway it is monotone and it tends to zero. $\endgroup$ – MattG88 Dec 23 '16 at 12:53
  • $\begingroup$ I was thinking only in what condition the sign of the derivative change. I didn't thought about the physical meaning of $t^*$. Thank you! $\endgroup$ – marya Dec 23 '16 at 13:00
  • $\begingroup$ Mathematically there aren't problems and the derivative changes sign in a neighborhood of $t^*$ and in this particular case it is monotone before and after $t^*$, but you are solving a physical problem so you must pay attention to the physical meaning of your quantities: $t\ge 0$!! $\endgroup$ – MattG88 Dec 23 '16 at 13:09

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