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I'm finding the equations of motion of a mass attached to four springs in a box. See picture:

enter image description here

In the prompt, we're instructed to use "the small-oscillations approximation, and neglect terms of order $\frac{x^2}{a^2}$ , $\frac{y^2}{a^2}$ , and $\frac{xy}{a^2}$". This all makes perfect sense to me.

Using both force diagrams and the Lagrangian approach, I find the equations of motion. I have the solution, but I do not see how it is possible to reach that solution.

For example, let's find the x-component of the force from the spring at the "top" of the box. The length of the spring for an arbitrary x, y is $\sqrt{x^2 + (a-y)^2}$, and so our total force vector is $F_1$ = $K_2\left(a-\sqrt{x^2 + (a-y)^2}\right)$. And taking the x-component we have:

$$ F_{1x} = K_2\left(a-\sqrt{x^2 + (a-y)^2}\right) \frac{x}{\sqrt{x^2 + (a-y)^2}} $$

And I am told from the solutions that $F_{1x} \approx 0$. I cannot see how this is possible. I've tried using the approximation $(1+x^2)^{-1/2} \approx (1-\frac{1}{2}x^2)$, but it seems no matter what I do I fail to reach 0.

Does anyone see how small angle approximation can lead to getting $F_{1x} = 0$ here?

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  • $\begingroup$ Could you more precisely restate that as $F_{1x}\in o(f(a,\,x\,y))$ for some function $f$? Note$$a-\sqrt{a^2+k}=a(1-\sqrt{1+k/a^2})\sim-k/(2a)\in o(1)$$if $k\in o(a)$. $\endgroup$
    – J.G.
    Commented Aug 24, 2021 at 16:12
  • $\begingroup$ expand $F_{1x}$ into a Taylor series wrt $x$ and the first term is $-(y-2a)x/(y-a)^2$ $\endgroup$
    – hyportnex
    Commented Aug 24, 2021 at 16:19
  • $\begingroup$ @J.G. I'm sorry, I'm not sure what you're asking me there. If $k = (a-y)^2$, continuing on your equation, I don't get 0, just some combination of a's, y's, and x's that you cannot use the small approximation to eliminate. $\endgroup$
    – Mattkwish
    Commented Aug 24, 2021 at 16:36
  • $\begingroup$ @hyportnex Carrying out the Taylor expansion, I do not get that, and neither does wolframalpha - did you make a mistake? Even if you do get that, expansion, once again I don't see how that leads to 0. $\endgroup$
    – Mattkwish
    Commented Aug 24, 2021 at 16:38
  • $\begingroup$ Take $y = 0$. For $x> 0$ you have the left spring stretched which returns to it and the right spring, compressed, which pushes. The two forces are in the same direction and you have to find $-2kx$. So there is an error in your calculation. $\endgroup$
    – Vincent Fraticelli
    Commented Aug 24, 2021 at 16:48

1 Answer 1

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the force $F_{1x}$ is:

$$F_{1x}={\frac {K_{{1}} \left( \sqrt {{x}^{2}+{y}^{2}-2\,ya+{a}^{2}}-a \right) x}{\sqrt {{x}^{2}+{y}^{2}-2\,ya+{a}^{2}}}} $$

take the Taylor series for the denominator

$$\sqrt {{x}^{2}+{y}^{2}-2\,ya+{a}^{2}}\overset{\text{Taylor}}{\mapsto}=a$$ and for the nominator $$K_{{1}} \left( \sqrt {{x}^{2}+{y}^{2}-2\,ya+{a}^{2}}-a \right) x\overset{\text{Taylor}}{\mapsto}=-K_{{1}}yx=0$$

thus $F_{1x}=\frac{0}{a}=0$

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  • $\begingroup$ Thank you. Two questions - in my Taylor series expansion for the denominator, I get $a-y$, did you expand only in terms of x? When I included y I got the extra -1 term from the partials wrt y. Second question is how legitimate is an approximation that is a ratio of two Taylor series? Seems borderline to me, but you got the answer so I'm assuming this is what was intended by the question. $\endgroup$
    – Mattkwish
    Commented Aug 24, 2021 at 18:06
  • $\begingroup$ the Taylor series expansion for the denominator second order is $~\sqrt {a}\sqrt {a-2\,y}$ , you obtain also zero. your second question I am not sure about this ? $\endgroup$
    – Eli
    Commented Aug 24, 2021 at 18:29
  • $\begingroup$ Taylor expansion second order is $f(x,y)=f \left( 0,0 \right) +D_{{1}} \left( f \right) \left( 0,0 \right) x+D _{{2}} \left( f \right) \left( 0,0 \right) y~$ where $D_1=\frac{\partial f(x,y)}{\partial x}~$ and $~D_2=\frac{\partial f(x,y)}{\partial y}$ $\endgroup$
    – Eli
    Commented Aug 24, 2021 at 18:35

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