A spring with a mass of $2 \mathrm {~kg}$ has damping constant 8 and spring constant 80 . Graph the position function of the mass at time $t$ if it starts at the equilibrium position with a velocity of $2 \mathrm {~m} / \mathrm { s }$.

Question

Quizplus · Accepted Answer

The position function of the mass is given by:
$$x(t) = e^{-\frac{b}{2m}t} (A \cos(\omega t) + B \sin(\omega t))$$
where
$$\omega = \sqrt{\frac{k}{m}-\left(\frac{b}{2m}\right)^2}$$
$$b=8, m=2, k=80, A=0, B=\frac{2}{\omega}$$
Initial conditions are $x(0)=0$ and $x'(0)=2$. Using these and simplifying, we get:
$$x(t) = \frac{2}{\sqrt{39}} e^{-4t/39} \sin\left(\sqrt{\frac{1999}{390}}t\right)$$
which matches graph B.

A Spring with a Mass Of 2 kg2 \mathrm {~kg}2 kg Has Damping Constant 8 and Spring Constant 80

A Spring with a Mass Of $2 \mathrm {~kg}$ Has Damping Constant 8 and Spring Constant 80