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Spring Dashpot Exercises Worked


Spring Mass Dashpot

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1. A 64 lb object stretches a spring 4 ft in equilibrium. It is attached to a dashpot with damping constant {$c=8$} lb-sec/ft. The object is initially displaced 18 inches above equilibrium and given a downward velocity of 4 ft/sec. Find its displacement and time--varying amplitude for {$t>0$}.

First, we get the spring constant.

{$$ 64{\rm lb.} = k \varDelta l = k 4{\rm ft.};\quad k=16{\rm lb./ft.} $$}

This gives the homogeneous equation for the system:

{$$ 64y^{\prime\prime} + 8y^\prime + 16y = 0 $$}

where y is understood to be a function of t in seconds.

Dividing by 64 to get to standard form:

{$$ \begin{align} y^{\prime\prime} + Ay^\prime + By &= 0 \cr y^{\prime\prime} + \frac{1}{8}y^\prime + \frac{1}{16}y &= 0 \tag{eq. a}\end{align}$$}

What we mean by a solution to this equation is to find values which make the left side and the right side equal, which in this case means to find values for which the left side is 0.

Now to get the characteristic equation, {$ e^{rt} $} is substituted for y and then divided out, steps that are usually performed by inspection.

{$$ \begin{align} y^{\prime\prime} + \frac{1}{8}y^\prime + \frac{1}{4}y &= 0 \cr (e^{rt})^{\prime\prime} + \frac{1}{8}(e^{rt})^\prime + \frac{1}{4}e^{rt} &= 0 \cr r^2e^{rt} + \frac{1}{8}re^{rt} + \frac{1}{4}e^{rt} &= 0 \cr \left( r^2 + \frac{1}{8}r + \frac{1}{4}\right)e^{rt} &= 0 \tag{eq.b}\cr r^2 + \frac{1}{8}r + \frac{1}{4} &= 0 \cr \end{align} $$}

{$ {\rm Eq. b}$} is telling us that {$ y = e^rt $} is a solution for values of r which make {$ r^2 + \frac{1}{8}r + \frac{1}{4} $} equal to 0, since {$ e^rt $} can never be 0.

This does not seem to yield to factoring, so it is necessary to resort to the quadratic formula:

{$$ \begin{align} r =& \frac{{ - \frac{1}{8} \pm \sqrt {\frac{1}{8}^2 - 4(1)\frac{1}{4}} }}{{2(1)}} \cr =& \frac{{ - \frac{1}{8} \pm \sqrt {\frac{1}{64} - \frac{64}{64}} }}{2} \cr =& - \frac{1}{16} \frac{{ \pm \sqrt {\frac{1}{64} - \frac{64}{64}} }}{2} \cr =& - \frac{1}{16} \pm i\frac{3}{16} \sqrt {7}\cr \end{align} $$}

From Eq. b we know {$ e^{rt} $} is a solution to the equation. So for the complex solution z:

{$$ \Large z(t) = e^{(\frac{1}{16} \pm i\frac{3}{16} \sqrt {7})t} $$}

Taking only the solution with the positive sign on i:

{$$ \Large \begin{align} z(t) &= e^{(\frac{1}{16} + i\frac{3}{16} \sqrt {7})t} \cr &= e^{(\frac{1}{16})t} \left(e^{(i\frac{3}{16}\sqrt{7})t}\right) \cr &= e^{(\frac{1}{16})t} \left( \cos(\frac{3}{16}\sqrt{7}t) +i\sin(\frac{3}{16}\sqrt{7}t)\right) \cr \end{align}$$}

But it is easily shown that each of the real and imaginary parts are independent solutions to the equation, and if the imaginary part by itself is a solution, it can be divided through by i to be a real solution. So by superposition:

{$$ \Large \begin{align} y(t) &= c_1e^{(\frac{1}{16})t}cos(\frac{3}{16}\sqrt{7}t) + c_2e^{(\frac{1}{16})t}sin(\frac{3}{16}\sqrt{7}t) \cr \end{align}$$}


Sources: Illustration By LP (Self-published work by LP)Attribution-Share Alike 3.0 Unported license.

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This is a student's notebook. I am not responsible if you copy it for homework, and it turns out to be wrong.

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