Force on Body Attached to Spring at Displacement x - A.P. French

[Slimy0233]

Source: A.P. French's Vibrations and Waves

I do not recognize the first equation, can someone explain how it came to be? The reasoning behind it.

How can force on a body attached to a spring at small displacement x be represented as

? I know recognize F = - kx (restoring force)

I realize that the mass is at equilibrium and not rest, thus there were/are multiple forces acting on the spring, thus, I guess my question simplifies, what is the nature of the forces

if -kx is restoring force, what are the rest of the forces, can someone please state an example for better understanding?

edit: Good God, creating a post is no joke :')
edit 2: The math which was visible at first is not visible now.

 

[Ibix]

The first equation is just a Taylor series. Whatever form the restoring force takes (not just Hooke's law), you can Taylor expand it into a series, a polynomial of some high (possibly infinite) order. But if we agree to restrict ourselves to the region where ##x## is small then ##x^2##, ##x^3##, etcetera, must be really small and we can neglect all terms except the ##x## one. Then the force reduces to the linear restoring force you are familiar with.

(That's a paraphrase of the paragraph between the two marked equations, by the way.)

 

[Slimy0233]

The first equation is just a Taylor series. Whatever form the restoring force takes (not just Hooke's law), you can Taylor expand it into a series, a polynomial of some high (possibly infinite) order.

I am sorry, can you please explain this more. Especially the "whatever form the restoring force takes" part.

 

[Ibix]

Well, a restoring force is just how strongly a system resists deformation or displacement. It doesn't have to be directly proportional to the displacement. An obvious example is a pendulum, where the restoring force is proportional to the sine of the displacement angle, ##\theta##. But you can expand that sine as a Taylor series, and as long as you keep the angle small then you can neglect the ##\theta^3## and higher terms (the even power terms are zero in this case). This is the formal justification for writing "##\sin\theta\approx\theta## for small ##\theta##". Once you have done that, you have justified modelling a small-amplitude pendulum as a simple harmonic oscillator.

(And, although French doesn't mention it above, you can find out how big ##\theta## has to be for the ##\theta^3## term to matter, and hence how small a "small" amplitude actually must be.)

 

[Ibix]

So, in the case of the pendulum, the first equation you have marked would be $$\begin{eqnarray*}
F(\theta)&=&-mg\sin\theta\\
&=&-mg\left(\theta-\frac{\theta^3}{3!}+\frac{\theta^5}{5!}-\ldots\right)
\end{eqnarray*}$$As long as ##\theta## is small this is approximately ##F(\theta)\approx-mg\theta##. This is the right hand side of the second equation you have marked, which would therefore be$$ml\frac{d^2\theta}{dt^2}=-mg\theta$$

 

[Slimy0233]

@Ibix These are one of the best answers I have ever received. Thank you very much!!

Beautifully explained!

 

[Ibix]

You're very welcome. I don't think the explanation is quite as unparalleled as you say, but I'm glad it helped you.