Answer
a) k=3
b)
equation of motion:
$cos(\frac{\sqrt 3}{2}t)+\frac{\sqrt 3}{3}sin(\frac{\sqrt 3}{2}t)$ or $\sqrt \frac{4}{3}sin(\frac{\sqrt 3}{2}t+\sqrt 3)$
Circular Frequency:
$\sqrt \frac{3}{4}$
Amplitude:
$\sqrt \frac{4}{3}$
Phase Angle:
$\sqrt 3$
Period:
$\frac{4\sqrt 3\pi}{3} seconds$
Work Step by Step
a) Hooke's Law states $F=ks$. Force equals spring constant times elongation.
Plugging the numbers into the equation gives us $(3)=k(1), k=3$
b) With mass = 4, damping = 0, and spring constant = 3, we get the equation $4\frac{d^{2}x}{dt^{2}}+3 = 0 $ and the corresponding Auxiliary Equation $4m^{2}+3=0$
This gives us the general solution $c_{1}cos(\frac{\sqrt 3}{2}t)+c_{2}sin(\frac{\sqrt 3}{2}t)$.
The problem gives initial conditions $x(0)=1, x^{'}(0) = 0.5$, which show $c_{1}=1, c_{2}=\frac{\sqrt 3}{3}$
equation of motion:
$cos(\frac{\sqrt 3}{2}t)+\frac{\sqrt 3}{3}sin(\frac{\sqrt 3}{2}t)$ or $\sqrt \frac{4}{3}sin(\frac{\sqrt 3}{2}t+\sqrt 3)$
Circular Frequency:
$\sqrt \frac{k}{m} = \sqrt \frac{3}{4}$
Amplitude:
$\sqrt {(1)^{2}+\frac{\sqrt 3}{3}^{2}})$ = $\sqrt \frac{4}{3}$
Phase Angle:
$tan^{-1}(\frac{c_{1}}{c_{2}}) = \sqrt 3$
Period:
$\frac{2\pi}{\frac{\sqrt 3}{2}} = \frac{4\sqrt 3\pi}{3} seconds$
