Going through some problem sheets from previous semesters and can't find a full solution for this question so was wondering what the answers might be.
A particle of mass $m$ moves on the $x$ axis under the action of the force $F_1(x) = −kx$ , where $k$ is a positive constant.
(a) (i) Employ Newton’s 2nd law to write down a differential equation of motion.
(ii) Find the general solution to the equation of motion.
(iii) Determine the potential $V(x)$ of the force $F_1$ assuming that $V$(0) = 0.
(iv) Employ conservation of energy to find a relation between position $x$ and velocity $\dot x$ of the particle for a prescribed value $E_0$ of the total energy.
(b) Assume that in addition to the force $F_1(x)$ there is also a strong damping force F2($\dot x$) = −$ν$$\dot x$ , $v$ = const, $ν$ > 2sqrt(km)
(i) Employ Newton’s 2nd law to write down a differential equation of motion.
(ii) Find the general solution to the equation of motion.
I understand how to use Newton's Laws but with differential equations I keep getting mixed up,thanks in advance!
Newton's second law for the system gives $$ F_1 = m a = m \frac{\mathrm{d}^2x}{\mathrm{d}t^2} = m\ddot{x} = -k x$$ that is $$\ddot{x} +\omega^2 x=0$$ where $\omega = \sqrt{\frac{k}{m}}$. Solving this differential equation, we find that the motion is described by the function $$ x(t) = A\cos\left( \omega t+\phi\right) $$ The potential energy is $V = \frac{1}{2}kx^2$.
The kinetic energy is $K = \frac{1}{2} mv^2(t) = \frac{1}{2}m\dot x^2$ and from the conservation of energy we have $$E_0=\frac{1}{2}kx^2+\frac{1}{2}m\dot x^2$$
With the force $F_2$ the balance of forces (Newton's second law) gives $F=F_1+F_2$ that is $$m\ddot x=-kx-\nu\dot x$$ which can be rewritten into the form $$ \ddot x + 2\zeta\omega\dot x + \omega^{2} x = 0 $$ where $ \omega = \sqrt{\frac{k}{m}}$ and $\zeta = \frac{\nu}{2 \sqrt{mk}}$ is called the damping ratio.
we can solve the equation by assuming a solution $x$ such that $x(t) = e^{\gamma t}$ where the parameter $\gamma$ is, in general, a complex number. Substituting this assumed solution back into the differential equation gives $$\gamma^2 + 2 \zeta \omega \gamma + \omega_0^2 = 0$$ which is the characteristic equation. Solving the characteristic equation will give two roots $$\gamma_{\pm}=-\zeta \omega \pm \omega \sqrt{\zeta ^2 - 1}.$$
The solution to the differential equation is thus $$ x(t) = Ae^{\gamma_+ t} + Be^{\gamma_- t}$$
If ${\nu}>{2 \sqrt{mk}}$ we have $\zeta>1$ and the system is over-damped and there are two different real roots. The solution to the motion equation is $$ x(t) = Ae^{\gamma_+ t} + Be^{\gamma_- t}=e^{-\zeta \omega t}\left(Ae^{\omega \sqrt{\zeta ^2 - 1}\, t} + Be^{-\omega \sqrt{\zeta ^2 - 1}\, t}\right)$$