# Interpreting derivatives

So we have a function such that the distance moved by a particle (say $$s$$) is proportional to $$sin(Ct)$$ where $$C$$ is a constant. Now i needed to show that the rate of change of velocity is directly proportional to the distance of the particle. Which was easy.

But the question stated 'distance of the particle measured along its path from a fixed position', now i don't know about this (maybe it's related to the negative sign of the second derivative?) But i would appreciate clarifying my doubts about what i really need to show.

Also, what are some examples of a real world particle that follows its path as said in the function?

• This question is not worded well enough to know what you want. Is this from a homework problem? It's possible that the wording is to indicate that the particle's coordinate is measure from its equilibrium position. Please clarify by stating the complete original question. – ggcg Nov 14 '18 at 3:08
• The full question is as follows: show that if a particle moves so that the space descibed is given by s is directly proportional to sin(Ct) where C is a constant, the rate of increase of velocity is proportional to the distance of the particle measured along its path from a fixed position. – user182947 Nov 14 '18 at 3:35
• Thanks. It looks like someone answered it, quite well. They got your intent. – ggcg Nov 14 '18 at 4:21

The potential energy for a spring is $$U = \frac{1}{2}k(x-a)^2$$, where $$a$$ is the equilibrium separation, $$x$$ is the current separation, and $$k$$ is the spring constant. In the potential below, we can calculate an equilibrium distance of $$R = 2^\frac{1}{6}\sigma$$, and an effective spring constant which is really gross, but it is k in this Desmos plot: https://www.desmos.com/calculator/69emlzge0i You can see that it can be approximated by a parabola near the equilibrium point quite accurately for small oscillations. The moral of the story is that any sufficiently small oscillation has roughly that form.