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Simple harmonic motion

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 the theory for time-space that is symmetric with space-time. 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):

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

Fx = –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) = As cos(ωt + θs),

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

ωT = 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 y0 such that:

y0mg/k.

The equation of motion is

m d²y/dt² = –ky + mg,

which can be written in the form

d²y/dt² + (k/m)(y – y0) = 0.

The solution to this differential equation is

y – y0As cos(ωt + θ).

This shows that the bob will oscillate around the equilibrium position with the period (distimement extent), T, which is the extent of the distimement:

T = 2π √(m/k).


Now let’s look at simple harmonic motion from the perspective of time-space. Hooke’s law in the horizontal case is this:

Γt = –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) = At cos(ψx + φt),

where ψ = √(k/m), At is the temporal amplitude, and φt is the 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

ψS = 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 gravity. The total surge acting on the vass is:

Γ = –k′t + ℓh,

where k′ is a spring constant and h is the gravitational prestination. The equilibrium time position is t0 such that:

t0ℓh/k′.

The equation of motion is

ℓ d²t/dy² = –k′t + ℓh,

which can be written in the form

d²t/dy² + (k′/ℓ)(t – t0) = 0.

The solution to this differential equation is

t – t0At cos(ψy + φt).

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

S = 2π √(ℓ/k′).

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