SMH from force acting at a distance

[Tigers!]

I occasionally tutor school students at our church in algebra, trig, logarithms, exponents etc. Very rarely do I drift into calculus.

However, a girl and I got to talking about the classis problem of boring a tunnel from one side of the earth to the other through the centre. We discussed it's SMH (Simple Harmonic Motion) behaviour under the influence of just gravity. She asked would SMH would occur if a chord was bored through the earth from, say, Australia to England and a mass dropped in said chord? I had never considered that before and have spent a few weeks thinking about it. I have reached the end of my mathematical ability.

With reference to the attached picture
The path of the particle is considered to be on the X-axis and gravity is pointed towards the centre of the earth and is thus -Y. The length of the path is 2L with range of -L to +L. The centre of the path can be considered to be the origin at (0,0). The motion of the particle is SMH. The total angle covered by the path is 2ɵ.

Some equations are

Fx = g cosɵ = max = m.d2x/dt2

Fy = g sinɵ = may = m. d2y/dt2

I could convert into polar form.

Main question:

Could g acting on a particle at angle ɵ give rise to SMH?

Next question:

If so, what are the next steps to solve? Has this been considered somewhere by someone that I could look at?

 

[bilby]

I occasionally tutor school students at our church in algebra, trig, logarithms, exponents etc. Very rarely do I drift into calculus.

However, a girl and I got to talking about the classis problem of boring a tunnel from one side of the earth to the other through the centre. We discussed it's SMH (Simple Harmonic Motion) behaviour under the influence of just gravity. She asked would SMH would occur if a chord was bored through the earth from, say, Australia to England and a mass dropped in said chord? I had never considered that before and have spent a few weeks thinking about it. I have reached the end of my mathematical ability.

With reference to the attached picture View attachment 39431
The path of the particle is considered to be on the X-axis and gravity is pointed towards the centre of the earth and is thus -Y. The length of the path is 2L with range of -L to +L. The centre of the path can be considered to be the origin at (0,0). The motion of the particle is SMH. The total angle covered by the path is 2ɵ.

Some equations are

Fx = g cosɵ = max = m.d2x/dt2

Fy = g sinɵ = may = m. d2y/dt2

I could convert into polar form.

Main question:

Could g acting on a particle at angle ɵ give rise to SMH?

Next question:

If so, what are the next steps to solve? Has this been considered somewhere by someone that I could look at?

The answer is yes, assuming a frictionless spherical horse in a vacuum.

Not only would you see SHM, but the period would be identical for motion along a chord as for a diametrical borehole.

It’s about 90 minutes for a cycle, for a hypothetical Earth mass and density sphere - even if your “borehole” is just a short chord joining two points very close together on the surface of your homogeneous and perfectly spherical planet. The period is also identical to the period of a satellite in low orbit.

The period of a pendulum doesn’t vary with the length of the swing arc, for the same reason. If the arc is short, the pendulum just moves slower, and the period stays constant - unless your planet has an atmosphere, or your pendulum components exhibit friction, of course.

 

[Swammerdami]

What is the best way to eliminate most dissipative energy (frictions) and how close can you get to zero?

Nitpick: If the tunnel goes far below the surface, you'll want to replace force g, with gr/r' where r' denotes the distance to the Earth's center.
(Newton found a geometric demonstration of this fact, which is regarded as one of his several pièces de résistance.)

 

[bilby]

What is the best way to eliminate all dissipative energy (frictions) and how close can you get to zero?

Magnetic levitation, in as hard a vacuum as you can achieve.

And it depends how much money you’re throwing at the problem.

 

[Tigers!]

I occasionally tutor school students at our church in algebra, trig, logarithms, exponents etc. Very rarely do I drift into calculus.

However, a girl and I got to talking about the classis problem of boring a tunnel from one side of the earth to the other through the centre. We discussed it's SMH (Simple Harmonic Motion) behaviour under the influence of just gravity. She asked would SMH would occur if a chord was bored through the earth from, say, Australia to England and a mass dropped in said chord? I had never considered that before and have spent a few weeks thinking about it. I have reached the end of my mathematical ability.

With reference to the attached picture View attachment 39431
The path of the particle is considered to be on the X-axis and gravity is pointed towards the centre of the earth and is thus -Y. The length of the path is 2L with range of -L to +L. The centre of the path can be considered to be the origin at (0,0). The motion of the particle is SMH. The total angle covered by the path is 2ɵ.

Some equations are

Fx = g cosɵ = max = m.d2x/dt2

Fy = g sinɵ = may = m. d2y/dt2

I could convert into polar form.

Main question:

Could g acting on a particle at angle ɵ give rise to SMH?

Next question:

If so, what are the next steps to solve? Has this been considered somewhere by someone that I could look at?

The answer is yes, assuming a frictionless spherical horse in a vacuum.

Not only would you see SHM, but the period would be identical for motion along a chord as for a diametrical borehole.

It’s about 90 minutes for a cycle, for a hypothetical Earth mass and density sphere - even if your “borehole” is just a short chord joining two points very close together on the surface of your homogeneous and perfectly spherical planet. The period is also identical to the period of a satellite in low orbit.

The period of a pendulum doesn’t vary with the length of the swing arc, for the same reason. If the arc is short, the pendulum just moves slower, and the period stays constant - unless your planet has an atmosphere, or your pendulum components exhibit friction, of course.

Thank you

 

[Swammerdami]

The period of a pendulum doesn’t vary with the length of the swing arc, for the same reason. If the arc is short, the pendulum just moves slower, and the period stays constant - unless your planet has an atmosphere, or your pendulum components exhibit friction, of course.

Nitpick: The period of a friction-free pendulum DOES vary, though the variance is tiny for all but large arcs. The constant-period path ( Tautochrone curve) is the cycloid, not the circle. As seen in the animated gif at the Wiki page, the cycloid is close to a circle until it does its flip. It was Galileo who thought the tautochrone was the circle and invented the pendulum clock, though he didn't build one. But it was Christiaan Huygens who actually built successful pendulum clocks based on circular arcs; AND proved that the cycloid, not the circle, solved the tautochrone problem. (He even attempted to build pendulum clocks which followed a cycloidal arc but was unsuccessful.)

The previous paragraph applies to friction-free pendulums. In fact — because both friction and friction-free frequency increase with increasing arc length — the friction in a pendulum clock may tend to cancel the slight deviation of the circle from the tautochrone.