# How does the small angle approximation lead to 0 here?

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

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?

• 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 $k\in o(a)$.
– J.G.
Commented Aug 24, 2021 at 16:12
• expand $F_{1x}$ into a Taylor series wrt $x$ and the first term is $-(y-2a)x/(y-a)^2$ Commented Aug 24, 2021 at 16:19
• @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. Commented Aug 24, 2021 at 16:36
• @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. Commented Aug 24, 2021 at 16:38
• 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. Commented Aug 24, 2021 at 16:48

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

• 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. Commented Aug 24, 2021 at 18:06
• 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 ?
– Eli
Commented Aug 24, 2021 at 18:29
• 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 $~D_2=\frac{\partial f(x,y)}{\partial y}$
– Eli
Commented Aug 24, 2021 at 18:35