If, for a (not necessarily simple harmonic) oscillator I have that $$\frac{dx}{dt} = G(x)$$ then I can express the period of motion as $$\int_{0}^{T/4} dt = \int_{0}^{X_{max}} \frac{dx}{G(x)}.$$ What if I want to write the LHS in terms of an integral over frequency instead of time? Can I write $dt = - df / f^{2}$? What would the limits of the integral be in this case?


It is not possible to write $dt =-df/f^2$. It is not possible because time $t$ take on any value from $-\infty$ to $\infty$, whereas the frequency $f$ of a harmonic oscillator is a constant which is determined by parameters of the harmonic oscillator

$$ f = \frac{1}{2\pi} \sqrt{\frac{k}{m}}$$

But as it was not clearly stated in the post this might be not the answer which is requested.

The other interpretation of the symbol $f$ would lead to the following conclusion: What is probably meant is that you want to re-parametrize the integral, instead of computing $\int F(x) dx$ you want to compute$\int F(g(s))d (g(s))$ with some appropriate function $g$. If for $g$ $f$ is used one would actually get:

$$\int \frac{dx}{f(x)} = \int \frac{df(x)}{f(f(x))}$$

The new integrand actually looks similar to the proposed one $-\frac{df}{f(x)^2}=-\frac{df}{f(x) f(x)}$ but it is completely different: In the proposal the function $f$ is used squared in the denominator, whereas the correct result is $(f\circ f)(x)$ in the denominator which corresponds to a twofold successive execution of the function on x which is much more complicated. So such a operation will not help to compute the integral.

  • $\begingroup$ Apologies, poor choice of function name. I have edited the question. My idea was to change the function so that I solve not for the evolution of x = x(t), but for the evolution of f = f(x), i.e., how frequency changes with position. For this purpose, you can assume that the oscillator is perturbed and is no longer simple harmonic. $\endgroup$ – Gordon Sep 15 '19 at 21:25

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.