A damped spring mass system is modelled below: $$m\frac{d^2y}{dt^2}=F_s+F_d\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space t>0$$ $$y(0)=0\space\space\space\space\space\space\space\space\space\frac{dy}{dt}=v_0$$ where $y(t)$ is displacement, $m$ is mass and $t$ is time. $F_d=-ky$ and $F_s=-c\frac{dy}{dt}$ are the restoring and damping forces, respectively.
- Nondimensionalize the equation.
- Find $\epsilon$, a measure of the strength of the damping. When $c$ is small $\epsilon$ is small. The system has weak damping. Include $\epsilon$ in the equation.
Here is my attempt.
$$\left[m\frac{d^2y}{dt^2}\right] = \left[-ky\right]\implies[k]=MT^{-2}$$ $$\left[m\frac{d^2y}{dt^2}\right] = \left[-c\frac{dy}{dt}\right]\implies[c]=LT^{-1}$$
Let $y=y_cx$ and $t=t_cv$. Then $$\frac{d}{dt}=\frac{d}{dv}\frac{dv}{dt}=\frac{d}{dv}\frac{1}{t_c}=\frac{1}{t_c}\frac{d}{dv}$$ $$\frac{d^2}{dt^2}=\frac{d}{dt}\left(\frac{d}{dt}\right)=\frac{d}{dt}\left(\frac{1}{t_c}\frac{d}{dv}\right)=\frac{1}{t_c}\frac{d}{dv}\left(\frac{d}{dv}\right)\frac{dv}{dt}=\frac{1}{t_c^2}\frac{d^2}{dv^2}$$
So substituting back into the original equation: $$m\frac{1}{t_c^2}\frac{d^2(y_cx)}{dv^2}=-k(y_cx)-c\frac{1}{t_c}\frac{d(y_cx)}{dv}\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space (t_cv)>0$$
$$m\frac{y_c}{t_c^2}\frac{d^2x}{dv^2}=-k(y_cx)-c\frac{y_c}{t_c}\frac{dx}{dv}\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space v>0$$
$$\frac{d^2x}{dv^2}=-\frac{kt_c^2x}{m}-c\frac{t_c}{m}\frac{dx}{dv}\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space v>0$$
And
$$x(0)=0\space\space\space\space\space\frac{dx}{dv}=\frac{t_c}{y_c}v_0$$
Now this is the part I am not sure of:
Since $\frac{dx}{dv}$ is dimensionless: $$\frac{t_c}{y_c}v_0=1$$ $$t_cv_0=y_c$$
And since terms in an equation have the same dimension: $$-\frac{kt_c^2}{m}=-c\frac{t_c}{m}$$ $$t_c=\frac{c}{k}$$
So finally: $$y_c=\frac{cv_0}{k}$$
$$\frac{d^2x}{dv^2}=-\frac{c^2}{mk}x-\frac{c^2}{mk}\frac{dx}{dv}\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space v>0$$ $$x(0)=0\space\space\space\space\space\frac{dx}{dv}=1$$
Now I'm not sure what the question is alluding to when it says "when $c$ is small $\epsilon$ is small"; not sure if that means the relationship is linear or quadratic. $$\epsilon=\frac{c}{\sqrt{mk}} \space\space\space OR \space\space\space \epsilon=\frac{c^2}{{mk}}$$
My questions:
- Is my final non dimensional form for part 1 correct?
- Should $\epsilon$ be taken as linear or quadratic?