This question is somewhat related to my previous question: What is the time period of an oscillator with varying spring constant?

In that question, time period of mass-spring system with variable stiffness was asked. Answers were really helpful, and with help of answers I managed to derive time period of certain oscillator system $$m \ddot{x}+b \dot{x}+k(x)x=0 \, .$$ This stiffness $k(x)$ had dependence on displacement which was assumed to be second order polynomial. I also put a small attenuation term $b$ to spring equation, but that's just to check how attenuation term effect the system. In reality, this term is quite small and can be dismissed. Equation I got is

$$k(x) = \alpha + \beta x +\gamma x^2$$

from which we find

$$T = 4 \int_{x_0}^{x_\text{max}} \frac{dx}{\sqrt{\left(v_0^2 - \dfrac{2}{m} \left(\dfrac{\alpha x^2}{2} + \dfrac{\beta x^3}{3} + \dfrac{\gamma x^4}{4}\right) - \dfrac{bx^2}{m}\right)}}$$

From this equation I got quite nice results when I calculated it numerically with Wolfram Alpha. However something came to my mind: If we do not know the stiffness of spring a-priori, but we know time period of the spring we get into trouble. Approaches mentioned in last post seem not to work in this situation. If we have variable stiffness mass-spring system $m \ddot{x} + k(x) x = 0$ and we know always it's period of oscillation, initial values $x(0)$ and $\dot{x}(0)$ and even $x(t)$, what is it's stiffness? So what is function $K=f(T)$? I am quite sure that $T=2 \pi \sqrt{\dfrac{m}{K}}$ is no good in this case.

  • $\begingroup$ What do you mean by "we know its period of oscillation with any $x$"? Most oscillations will go through many different values of $x$. Do you mean "as a function of amplitude"? $\endgroup$ – Michael Seifert Aug 27 '15 at 13:58
  • $\begingroup$ I suspect that the answer will not be unique. $\endgroup$ – march Aug 27 '15 at 16:47
  • $\begingroup$ Michael Seifert: Good question. "We know its period of oscillation with any x" is now changed to “we know always it's period of oscillation, initial values x(0) and x'(0) and even x(t)”.. Originally I was trying to say something like “we know always x(t)”, but wrote some weird stuff instead. $\endgroup$ – dr_mushroom Aug 28 '15 at 7:03

Your Answer

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

Browse other questions tagged or ask your own question.