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?

  • $\begingroup$ 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. $\endgroup$ – ggcg Nov 14 '18 at 3:08
  • $\begingroup$ 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. $\endgroup$ – user182947 Nov 14 '18 at 3:35
  • $\begingroup$ Thanks. It looks like someone answered it, quite well. They got your intent. $\endgroup$ – ggcg Nov 14 '18 at 4:21

When the problem is asking for the distance measured along its path from a fixed position, I imagine it's really asking for displacement relative to a fixed position, as the distance it travels is more complicated to calculate, and not really that relevant.

As for examples, my oh my that's a great question. Harmonic oscillators have that path, and harmonic oscillators are used everywhere in physics. If you have a potential energy function that is complicated, usually you're really interested in just the behavior around stable equilibrium points. You can show that for small oscillations, any small oscillation will look just like a harmonic oscillator. This is because at stable equilibrium points, the potentials are like parabolas (the shape of a potential for a harmonic oscillator).

So, if you know the bonding potential energy for a diatomic molecule, you can calculate the frequency that they wiggle towards and away from each other as if they were two masses on the end of a spring!

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.

Lennard Jones Potential


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.