# What are the missing steps in the derivation of this equation?

I am trying to understand the model of a piano hammer described here. The relevant excerpt is:

As I understand it, they are saying in (12a) that if the hammer spring's uncompressed end's y position is greater than the string's y position, then the force of the hammer/spring will be:

$$F = Kh(yh(t) - ys(t))^{Ph}$$

(12b) is simply saying force is also equal to negative mass of the mass-spring combo times its acceleration. This is Newton's law I believe, where the spring's force (12a) must equal the mass's inertial force (12b).

The derivation of (13a) is straightforward since there's no change.

(13b) I'm having trouble understanding though. The best I can do is use the differentiation theorem for Laplace transforms described here:

$$f(t) = m*a(t)$$

$$f(t) = m*s*v(t)$$

$$f(t) = m*s^{2}*y(t)$$

$$y(t) = f(t)/(m*s^{2})$$

Is this what they did? If so, what is the type of substitution they did for $$s$$ here? I have been told about the Euler and Tustin method in another thread. What is this?

Also what is meant by the extra $$z^{-1}$$ at the end of (13b)? I was told multiplying by $$z^{-1}$$ means you just delay the whole equation by 1 sample. Is that what they're referring to to solve the problem they describe at the end?

Lastly, how would you rewrite/adapt these equations to accommodate an initial velocity for the mass component? ie. An excitation impulse velocity at time 0, ie. v0?

I have this example for a simple mass with v0 as the initial velocity:

But I'm not exactly sure how that really works. I mean, you end up with:

$$a(s) = s*V(s) - v0$$

How would you go one step further to describe $$a(s)$$ in terms of $$y(s)$$ while still maintaining the $$v0$$ parameter in there?

Thanks for any guidance.