This post is related to the one on *circular orbits*. I’ll continue to follow the exposition in *Elements of Newtonian Mechanics* by J.M. Knudsen and P.G. Hjorth (Spriner, 1995), this time starting with page 33. As before, the point is to derive a temporo-spatial theory that is symmetric with the spatio-temporal one. Although the parallel theory is shown for 1D space + 1D time, it may be expanded to 1D space + 3D time.

A harmonic oscillator is like the spring below with a bob (weight):

The force is an example of Hooke’s law. Let’s first look at the horizontal case:

F_{x} = –*kx*,

where *k* is a constant determined from the properties of the spring. From Newton’s second law:

*m d²x*/*dt²* = –*kx*,

where *m* is the mass of the bob, *x* is the displacement, and *t* is the time. The solution to this differential equation can be written in the form:

*x*(*t*) = *A _{s}* cos(

*ωt + θ*),

_{s}where *ω* = √(*k*/*m*), *A* is the *spatial *amplitude, and *θ* is the spatial phase angle. The period T must satisfy

cos(*ωt + θ*) = cos(*ω*(*t* + T)* + θ*).

Since the cosine is periodic with the period 2π, we have

*ω*T = 2π,

or

T = 2π/*ω* = 2π √(*m*/*k*).

Thus the period is independent of the amplitude. Note that

* d²x*/*dt²* = –*ω²x*.

Now consider a bob (weight) on a vertical spring under gravity. The total force acting on the bob is:

F = –*ky* + *mg*,

where k is the spring constant. The equilibrium position is *y _{0}* such that:

*y _{0}* =

*mg*/

*k*.

The equation of motion is

*m d²y*/*dt²* = –*ky* + *mg*,

which can be written in the form

*d²y*/*dt²* *+ (k*/*m)(y – y _{0}*) = 0.

The solution to this differential equation is

*y – y _{0}* =

*A*cos(

_{s}*ωt + θ*).

This shows that the bob will oscillate around the equilibrium position with the period, T, which is the distime:

T = 2π √(*m*/*k*).

Now let’s look at simple temporo-spatial harmonic motion. Hooke’s law in the horizontal case is this:

Γ_{t} = –*k′t*,

where *k′* is a constant determined from the properties of the spring. From Newton’s second law:

*ℓ d²t*/*dx²* = –*k′t*,

where ℓ is the *vass* of the bob, *t* is the time, and *x* is the displacement. The solution to this differential equation can be written in the form:

*t*(*x*) = *A _{t}* cos(

*ψx + φ*),

_{t}where *ψ* = √(*k*/*m*), *A _{t}* is the

*temporal*amplitude, and

*φ*is the

_{t}*temporal*phase angle. The displacement extent S must satisfy

cos(*ψx + φ*) = cos(*ψ*(*x* + S)* + φ*).

Since the cosine is periodic with the period 2π, we have

*ψ*S = 2π,

or

S = 2π/*ψ* = 2π √(ℓ/*k′*).

Thus the extent S is independent of the temporal amplitude. Note that

* d²t*/*dx²* = –*ψ²t*.

Now consider a bob (weight) on a vertical spring under levity. The total *release *acting on the *vass* is:

*Y* = –*k′t* + *ℓh*,

where k′ is a spring constant and *h* is the levitational relentment. The equilibrium time position is *t _{0}* such that:

*t _{0}* =

*ℓh*/

*k′*.

The equation of motion is

*ℓ d²t*/*dy²* = –*k′t* + *ℓh*,

which can be written in the form

*d²t*/*dy²* *+ (k′*/*ℓ)(t – t _{0}*) = 0.

The solution to this differential equation is

*t – t _{0}* =

*A*cos(

_{t}*ψy + φ*).

_{t}This shows that the bob will oscillate around the equilibrium time position with the displacement extent, S:

S = 2π √(ℓ/*k′*).