Recall r2=x2+y2,
and θ=tan−1yx.
If
ω=dθdt,
we have
ω=(dxdt,dydt)⋅(−yx2+y2,xx2+y2)=v⋅k,
where k:=(−yx2+y2,xx2+y2)
so
ω=vkcosϕ,
which gives
v=ωk,
for ϕ=2nπ.
As ‖k‖=1√x2+y2=1r,
this means to each trajectory x(t)=(x(t),y(t)), we can associate a harmonic wave with frequency ω=‖v(t)‖‖x(t)‖.
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