A spring with a spring constant k of 20 pounds per foot is loaded with a 10-pound weight and allowed to reach equilibrium. It is then displaced 1 foot downward and released. If the weight experiences a retarding force in pounds equal to four times the velocity at every point, find the equation of motion.


1 Answer 1


Newton's second ($F = ma$) gives $my'' = -4y' - 20y \implies my'' + 4y' + 20y = 0$. The mass $m$ and weight $W$ are related by $W = mg$ where $g$ is acceleration due to gravity. We know that

$$W=10 \text { pound} \implies m=\frac Wg=\frac{10}{32} \,\text {slug}$$ being $g=32 \:\text {feets}/s^2$. Remember that the spring constant $k$ of $20$ pounds per foot and $W=kx$ (Hook's law).


$$\frac{10}{32}y'' +4y' + 20y=0, \quad y(0)=1, y'(0)=0 $$

It is a ODE of the 2nd order where $y(t)=e^{-\frac{32t}{5}}\left(c_1\cos \left(\frac{24t}{5}\right)+c_2\sin \left(\frac{24t}{5}\right)\right) \tag 1.$ In fact

$$\frac{d^2y^2}{dt^2} +4\left(\frac{32}{10}\right)\frac{d y }{dt } + \left(\frac{32}{10}\right)\cdot 20y=0$$

With some steps

$$\frac{d^2y^2}{dt^2} +\left(\frac{64}{5}\right)\frac{d y }{dt } + 64 y=0$$

The ausiliary equation of the differential equation is:

$$\lambda^2+\frac{64}5\lambda +64=0 \iff 5\lambda^2+64\lambda+320=0$$ Solving for $\lambda$ we will have

$$\lambda=\frac{-64\pm 48 i}{10}=-\frac{32}5\pm \frac{24}5 i$$ i.e. the $(1)$.

  • $\begingroup$ And so the equation of motion would be: y = [4sin(24t / 5) / 3 + cos(24t / 5)]/e^(32t / 5) $\endgroup$
    – Rajapeg
    Nov 18, 2022 at 20:42
  • $\begingroup$ @Rajapeg Yes. It is correct. $\endgroup$
    – Sebastiano
    Nov 18, 2022 at 20:44

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