# Oscillator integral for frequency

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.

• 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. – Gordon Sep 15 '19 at 21:25