If you have a low-noise and well-sampled signal, a quick way to estimate it is to find $\sqrt{-f''(t)/f(t)}$. For a signal $f(t)=A sin(\omega t+\phi)$ the second derivative is $-A \omega^2 sin(\omega t+\phi)$ which is $-\omega^2$ times the original. This is useful if you want a quick response and only have part of a cycle, so no zero-crossings. But obviously it depends on being able to get an accurate second derivative numerically.
Nullius in Verba
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