gw=geiwt=−goei(−xre+wt)
ix is a circular wave front parallel to the surface of Earth.
∂gw∂t=−iwgeiwt
which makes ∂gw∂t also parallel to the surface of Earth. From which we formulate a gravitation potential term GPE, the rate of change of gravitational potential,
dGPEdt=m∂gw∂th=−iwmgheiwt
where h is the reference height along g, and m the mass of a body.
Then we consider the same body in motion on the surface of Earth with speed v,
KE=12mv2
dKEdt=12mdv2dt=mv.a
Since both KE and GPE, is in the same direction, because of the term i, we can have resonance when,
dKEdt=dGPEdt
12mdv2dt=−iwmgheiwt
Since, g=d2hdt2,
12dv2dt=−iwhd2hdt2eiwt
Multiplying i to both sides,
i.12dv2dt=whd2hdt2eiwt ---(1)
i.vdvdt=i.dxdtd2xdt2=whd2hdt2eiwt
Because of the factor i that rotates dv2dt to the direction of h, dv2dt acts like a driving force along h. Changing the rate of change of v2 can drive the system (a car travelling perpendicular changing gravitational wave) to resonance.
We know that x is not oscillating; not a sinusoidal, Therefore, h(t) must have a sinusoidal time component to cancels the eiwt factor in the differential equation (1). Consider,
h=ho(t)ebt then,
h′=h′o(t)ebt+bho(t)ebt
h″=h″o(t)ebt+2bh′o(t)ebt+b2ho(t)ebt={h″o(t)+2bh′o(t)+b2ho(t)}ebt
then,
whd2hdt2eiwt=who(t){h″o(t)+2bh′o(t)+b2ho(t)}e2bteiwt
when,
e2bt=e−iwt
the sinusoidal time component in the differential equation cancels, ie,
h=ho(t)e−i(w/2)t
We can simplify further,
i.12dv2dt=whd2hdt2eiwt
i.12dv2dt=who(t){h″o(t)+2(−iw2)h′o(t)+(−iw2)2ho(t)}
i.12dv2dt=who(t){h″o(t)−iwh′o(t)−w24ho(t)} ---(*)
i.12dv2dt is along real h and −iwh′o(t) is along x.
Equating the rotated parts,
From (*),
12dv2dt=who(t)h″o(t)−w34h2o(t) ---(2)
and
w2ho(t)h′o(t)=0
12w2ddt{ho(t)}2=0=d(−A2)dt
where −A2 is a contant
h2o(t)=−2A2w2
ho(t)=i√2Aw
From (2), since, h′o(t)=0, h″o(t)=0
12dv2dt=vdvdt=v.a=0−w34−2A2w2
a=+wA22v this is the condition for resonance
And,
h=i√2Awe−i(w/2)t
Aw has the unit of m when A has the unit of ms-1, a velocity, which is why it is multiplied by the factor i.
h=√2Awe−i{(w/2)t−π/2}
When t=0, at the on set of resonance, considering magnitudes only,
h(t=0)=√2Aw
12mv2=mgh(0)=mg√2Aw
A=w2g√2v2, h=v22ge−i{(w/2)t−π/2}
The condition for a car flip becomes,
a=wA22v=w{w2g√2v2}22v=w316g2v3
So, when the acceleration a is,
a=(2*pi*7.489)^3/(16*9.087^2)*v3=78.86v3 in ms-2
we will experience oscillation perpendicular to the direction of travel at a frequency of 7.489/2 = 3.745 Hz. Since, this is an exchange of KE and GPE, the greater the velocity the greater the amplitude of this resonance will be. This often happens at low speed, when the acceleration is pressed on hard.
More importantly and more frequently, this can also happen when,
a=−wA22v
ie. in sudden deceleration, the car can also flip when,
a=-78.86v^3
Note; With the initial condition of h=re when t=0 at the onset of resonance, the result is a huge number that do not make sense. This is because h is the change in height not the absolute height.