My lecturer said that when a oscillating body loses energy in the form of radiation the natural frequency of oscillation is equal to the frequency of the emitted radiation. Can someone explain why?


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    $\begingroup$ Hint: try an Ansatz $x=x_0\exp(i\omega t)$ in $m\ddot{x}+m\gamma\dot{x}+kx=F_0\exp(i\omega t)$. $\endgroup$ – J.G. Aug 14 at 5:41

I don't see what this has to do with the title.

Anyway, if you oscillate one end of a long rope (the equivalent of the electromagnetic field in this case) with your hand (the equivalent of the oscillating body in this case) at frequency $\omega$ then the propagating wave in the rope will oscillate at the same frequency. This is an intuitive analogy to oscillating charged particles and electromagnetic waves.


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