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.

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.