PHYS 0475 Lecture Notes - Lecture 2: Materiel, University Of Manchester
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Let x(t) be the coordinate along the tunnel with the origin placed in the middle, r the distance from the train to the center of the earth. Then x = r cos , (1) where is the angle between the direction to the middle of the tunnel, and vertical direction. The gravity force acting on the train is equal to. M (r) = (4 /3) r3, where is the density of the earth. Substituting this expression for m (r), we obtain. We only need the component of this force, acting along the tunnel; the rest is compensated by the rail reaction. F cos , where means the same as in (1). Thus the equation of motion (newton"s second law) and (1) give mx = f cos = . Mx, where we have chosen the minus sign from the evident physical considera- tions. Thus our train behaves like a linear oscillator x = k2x, where k2 =
