Question

Ok this is a long one so please bear with me.
I have this equation for y0 (it's the equilibrium distance for some sistem of springs and a mass).
\[ 0 = m g +2k(1-\frac{l}{\sqrt{y_0^2+l^2}})y_0 \]
So m is mass, k is spring constant and l is lenght of one spring. All well and good.
I solved the equation for y_0 with Mathematica and got 4 solutions {y01,y02,y03,y04}.
I am interested in the limiting value of y0 for m,k,g and l. I used Mathematica again to get:
\[ m \rightarrow 0 \quad \left\{0,0,0,0\right\} \]

Answer 1

(lol i ran out of space)
\[ m \rightarrow \infty \quad \left\{ComplexInfinity, ComplexInfinity, -\infty, -\infty \right\} \]
\[ g \rightarrow 0 \quad \left\{0,0,0,0\right\} \]
\[ g \rightarrow \infty \quad \left\{ComplexInfinity, ComplexInfinity, -\infty, -\infty \right\} \]
\[ k \rightarrow 0 \quad \left\{ComplexInfinity, ComplexInfinity, -\infty, -\infty \right\} \]
\[ k \rightarrow \infty \quad \left\{0,0,0,0\right\} \]
\[ l \rightarrow 0 \quad \left\{ DirectedInf\left[\frac{1}{2}(1-i \sqrt{3})\right],DirectedInf\left[\frac{1}{2}1+i \sqrt{3}\right],-\infty ,ComplexInfinity \right\} \]
\[ l \rightarrow \infty \quad \left\{0,0,-\frac{g m}{2 k},-\frac{g m}{2 k}\right\} \]

All the limits except the second last one make sense to me. Bigger mass or stronger g makes the ball drop lower. And stronger spring (k)
holds the ball higher up (closer to 0). (By looking at the limits I deduced that solution number 3 is the right one).

Finally my question is why when I nondimensionalize the equation with:
\[ \gamma = \frac{y}{y_c} \]
\[ y_c = \frac{mg}{2k} \]
\[ \tau = \frac{t}{t_c} \]
\[ t_c = \sqrt{\frac{m}{2k}} \]

I get the equation,

\[ (1-\frac{1}{\sqrt{(\frac{mg}{2kl})^2 \gamma_0^2 + 1}})\gamma_0 + 1 = 0 \]

and define a new parameter,
\[ \alpha = (\frac{mg}{2kl})^2 \]

and look at the limiting values of alpha I get different results for gamma0???
I would think that a large alpha would mean stronger gravitational force therefore a lower
equilibrium distance gamma0. And with a smaller alpha I would assume to get gamma0 = 0. But
\[ \alpha \rightarrow 0 \quad \left\{ DirectedInf\left[\frac{1}{2}(1-i \sqrt{3})\right],DirectedInf\left[\frac{1}{2}1+i \sqrt{3}\right],-\infty ,-\frac{1}{2} \right\} \]
\[ \alpha \rightarrow \infty \quad \left\{ 0,0,-1,-1 \right\} \]

Meaning that a big alpha means y0 goes to -(mg)/(2k) and a small alpha means y0 goes to -inf. This is exactly the opposite of
what I got before nondimensionalization...

Someone please help me.

Answer 2

@hartnn and @UnkleRhaukus

Answer 3

@Callisto @UnkleRhaukus @ganeshie8 @AccessDenied @ghazi

Answer 4

@shfreeman i appreciate your hard work :) ...

Answer 5

Thank you. Can you help out a fellow hard worker?
I would appreciate if someone could try the things I did and report on their results. So we can at least see that there is a mistake somewhere..

Answer 6

i have to go through this...because i can see there is a mistake whilst you were doin nondimensionalization ...but i am not sure...so i'll take time thank you