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I'm trying to find the function $ y = y(m) $ in terms of $k, \alpha \text{ and } y_0$ given the force function for a real spring $F = -kx + \alpha x^2$. I also know that $x = y_0 + y$ and $F = ma$ where $a$ is the second derivative of $x$. The problem is, whe I substitute

$m \frac{d^2(y_o-y)}{dy^2} = -k(y_0 - y) + \alpha(y_0 - y)^2$

I'm not sure whether the term $\frac{d^2(y_o-y)}{dy^2}$ is correct and if I can really take the derivative there.

Edit: I think I came up with a solution.

since $x = y_0 - y$ then $\dot{x} = -1 $ and $\ddot{x} = 0$. With that in mind We can move to:

$$ 0 = -k(y_0 - y) + \alpha(y_0 - y)^2 $$ which will take us to $$ y(y\alpha - 2y_0\alpha +k) = -y_0(y_0\alpha -k ) \text{ or } y^2\alpha - y(2y_0 + k) = -y_0(y_0\alpha - k) $$ But, once again I'm stuck here... I don't know how to get away with $y$ as a function here.

hcp
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  • "where $a$ is the second derivative of $x$" $;-;$ Derivative with respect to what? Certainly not $y$. – dxiv Jan 06 '22 at 08:33
  • I imagine that since I'm looking for a function y(m) the derivative is in respect to m. Am I correct? – hcp Jan 06 '22 at 08:35
  • actually. It's an accelaration, so it has to be in respect to time, t. – hcp Jan 06 '22 at 08:35
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    Right, it is time. – dxiv Jan 06 '22 at 08:36
  • @dxiv I think I've made some progress with your hint... but I'm stuck again – hcp Jan 06 '22 at 11:42
  • "since $x=y_0−y$ then $\dot{x}=−1$" $;-;$ No, then $,\dot{x} = - \dot{y},$. First thing, though, I think you should edit the question and explain what those variables are, what the equations describe, and what exactly is the problem you are trying to solve. For example, if you are looking for the value $y(m)$ at equilibrum vs. the equation of motion $y(t)$ given some initial conditions. – dxiv Jan 07 '22 at 07:58

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Instead of $\frac{d^2(y_0 - y)}{dy^2}$ does not make sense as it stands and instead use$m\ddot{x}$.

I believe you should use $\frac{d^2}{dt^2}$ instead.

You can actually do a first integral $$ m\ddot{x} = m\dot{x}\frac{d}{dx}\dot{x} = -kx + \alpha x^2 $$ or $$ m\frac{d}{dx}\frac{\dot{x}^2}{2}= -kx + \alpha x^2 $$ integrating w.r.t $x$ $$ \frac{m\dot{x}^2}{2} = -\frac{k}{2}x^2 + \frac{\alpha}{3}x^3 + C $$ Then you can use $x = y_0 - y$ etc.

Chinny84
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