Negative frequency of a Harmonic Oscillator

Question

We have just started learning small oscillations at grad level and our professor pulled out the technique of Fourier Analysis to illustrate its power by applying it to the case of a simple harmonic oscillator.

The differential equation is: $\ddot x+w_o^2x=0$

At this point he said, using the definition of fourier transform i.e. $$x(t)=\int_{-\infty}^{+\infty}\tilde x(w)\exp(iwt)\ dw$$

we get: $$(-w^2+w_0^2)\tilde x(w)=0$$ or $\tilde{x}(w)=A\delta(w-w_o)+B\delta(w+w_o)$

My problems:

1. What does $\tilde x(w)$ mean physically?
2. I have looked at other answers from the site and I can't understand the meaning of a negative $w$. It would be great if this doubt is cleared as well.

Any help is appreciated.

Answer 1

When Fourier transform is used in the form with complex exponential, the Fourier coefficients satisfy $$ [\tilde{x}(\omega)]^*=\tilde{x}(-\omega), $$ which follows from the fact that $x(t)$ is a real function, i.e. $x(t)^*=x(t)$. Thus, the negative frequencies here do not really play independent role. One could write the Fourier transform as an expansion in sines and cosines, considering only positive frequencies, but it would be cumbersome, see here.

In a general setting $\tilde{x}(\omega)$ is the (complex) amplitude of the oscillations, i.e., it contains the information about their real amplitude and their phase. However, in the case of the oscillator this interpretation is only confusing, since there is only one frequency and the delta functions appear.