# Find the equation of motion using Hooke's Law. [closed]

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.

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).

Hence

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

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 $$(1)$$.

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