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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?

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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.

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  • $\begingroup$ 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

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