It’s been a long time since physics class, but… At the moment you jump down your hole, don’t you already have significant angular momentum due to the earth’s rotation? And since the sections of the earth that are closer to the center are rotating more slowly, I think you’re going to crash into the side of your tunnel long before you get to the middle.
Typically, a pole-to-pole hole is assumed to avoid that complication.
Chronos, thanks for asking this question. I just worked it out, the answer is yes, and I think it is quite nice.
The key for me was to realize that, if the motion is harmonic along a chord, it must have exactly the same spring constant, since harmonic oscillators are isochronous (as someone with your name knows well).
A picture helps. Draw the chord, the diameter parallel to it, and the angle theta between the entrance to the hole and the diameter. Think about the force and distance when you jump in the hole. The component of force into the chordal hole is Fcos(Theta), where F is the normal force of gravity at the surface of the Earth. The distance to the middle of the chord is Rcos(Theta), where R is the radius of the Earth. Thus the ratio of force to distance at the surface of the Earth is the same for the chordal hole and the diametrical one.
Now, consider the case in which you have traveled some distance into the hole. Remove the spherical shell further from the center than you and consider the remaining sphere. The same analysis applies, thus verifying that, since the force law is linear in distance for the diametrical hole, it must obey the same law for the chordal hole, and with the same force constant and thus the same period of oscillation.
Which, of course, makes it quite obvious. If I’d actually sat down and set up the problem, I would have seen to cancel the cosines… But I didn’t actually set it up, so I didn’t see it.
There’s a lesson to be learned here, I’m sure.
To clarify for people who might not have realized it, all the results showing equal periods for a circular orbit and for tunnels through the Earth are assuming an Earth with uniform density. The result isn’t correct for the actual Earth, since its density varies with depth.
Get someone else to do your work? 
Right, the better approximation for the Earth is actually a uniform-strength gravitational field (little-g), all the way down until you get to the boundary of the core (at which point uniform density does become pretty good). But the uniform-density case is easier to calculate, and isn’t an absolutely terrible approximation.
Given that air molecules have a finite size, I think your oscillations truly will cease. A pendulum provides an analogy.
Jasper Fforde’s series of Thurday Next books (that’s the character’s name) is about an alternate 1980s universe where fixed wing crafts don’t exist (they fly by blimp), Wales is a People’s Republic separated from the rest of Great Britain by barbwire, Literary special detective agents can jump in and out of books.
One of the ways to travel in this series is by earth tube. The author points out it takes (I think 23 minutes) to go from any point to any other point in the Earth.
Doesn’t the air pressure calculations also have to take into consideration the decrease in gravity as we move towards the center of the earth?
My second attempt (in which I arried at 5M psi) does indeed take into account the decrease in gravity.
Any real pendulum arrives at a dead stop eventually because of friction in its pivot, long before the effects of air molecules (on an individual basis) will be felt.
If we’re going to include such small effects, then it’s safe to say the oscillations won’t ever come to a definite end; at some point their magnitude will be smaller than the brownian motion of the vehicle as it is buffeted about by the individual air molecules, but it’ll be a gradual transition, like slowly decreasing the volume on your car stereo until you finally realize you can’t hear it over the wind noise.
I wouldn’t advise a two meter hole diameter at 125MPH, it better be miles across if you want to free fall tens of thousands of miles back and forth through it with no danger of losing an arm or leg. Speaking of hole diameter, the perfect zero gravity inside the sphere would be distorted by the missing mass of the polar hole. Assuming you’re motionless at the center, would the resulting insignificant imbalance (or significant if the hole were large enough or the Earth dense enough) try to align your body equatorially and tug your hands and feet away from your center or would it just make you drift to whichever side of the hole was fractionally closer?
The easiest approach to this is to realize that the gravity at the center of the tube is the same as the gravity of the entire Earth minus the gravity of the portion that was removed to make the tube. The gravity of the removed cylinder will pull you towards its axis, although not nearly as strongly as the surface gravity of the Earth. In the center of the tube, therefore, you’ll be pulled slightly towards the near wall of the tube.
If you were right on the axis with your head towards the Eastern hemisphere and you feet towards the Western, you’d be being pulled apart, but I doubt you could feel it.
An overestimate of the effect would be to assume that most of your body weight was concentrated in your head and in your feet. Assuming 50 pounds on each end and a height of 6 feet, the tidal force at any position in the tube would be the the equivalent of your head and feet feeling the same pull (in opposite directions) that a three milligram mass would feel at the surface of the Earth. No, you would not be able to feel it.
I see Cecil neglected to take account of the tidal forces affecting bodily fluids, which would decelerate the participant at some point before forever. What an amateur.
Never doubt the wisdom of Cecil. His discussion was about the uniform density model, in which the force is a linear function of distance; thus the tidal force does not change as your body oscillates back and forth. If the tidal force is constant, there is no dissipation of energy.
You’re forgetting the bunny rabbit. And the tidal forces lead to the dissipation of energy as heat. How are you suggesting this heat gets turned back into kinetic energy?
ETA: Not to mention that Cecil doesn’t state he’s dealing with uniform density, it’s just been assumed in this thread for ease of calculation.
Thanks for the confirmations, I was just making sure there wasn’t a glitch in my model of reality. I assumed the force was insignificant, my main worry was all the rethinking I was in for if it didn’t exist at all. I’ve had educated people argue there’s a perfect zero gravity at the center even after I’ve increased the hole to 5,000 miles and remade the Earth from neutronium.
rickbdotcom gave a fairly decent estimate of how long it would take to get from the North Pole to the South Pole, but the lack of accuracy clouds a lot of the history associated with this problem. The letter at the beginning of Cecil’s article alludes to why you need a slightly more accurate answer:
"You may recall that Alice wondered about how deep the tube was while she was falling down the rabbit hole into Wonderland. "
Lewis Carroll took this idea one step further to calculate how long it would take to go from one end of a straight tube to the other no matter where the tube were located (obviously, this would only work if the Earth didn’t rotate, but …). If the tube is straight, then a traveler would be falling downhill at some angle for half the trip and their momentum would carry them back up the hill on the other half the trip.
It turns out that, no matter where the ends of the tube are located, the duration of the trip is the same. The time to go from the North Pole to the South Pole by gravity train? 42 minutes. The time to go from New York to London by gravity train? 42 minutes. The time to go from London to Sydney by gravity train? 42 minutes. Etc.
Lewis Carroll was fascinated by this idea. It’s the reason that the number 42 constantly pops up in his Alice books - including one of my favorite puns: Why was the mad hatter mad? He planned tea for two, but Alice showed up for tea, too.
I imagine other readers might think of at least one other author that seemed to like the number 42.
I assumed you were talking about the tidal force from the Earth’s gravity causing the blood to slosh back and forth with each oscillation. My point is that no sloshing would occur and no heat would be generated since the tidal force in the uniform density model is constant. If you were talking about the tidal force of the moon, yes that does oscillate every twelve hours but that is much much smaller than the already negligible tidal force from the Earth. The only reason the Moon has observable tidal effect on the Earth is because the Earth is so much bigger than a person.
You’re absolutely right! My bad.