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SHM: Gravity-Powered Train (Brace Yourself)

  1. Apr 10, 2016 #1
    1. The problem statement, all variables and given/known data

    Two cities are connected by a straight underground tunnel, as shown in the diagram. A train starting from rest travels between the two cities powered only by the gravitational force of the Earth, [itex]F = - \frac{mgr}{R}[/itex].
    Find the time [itex]t_1[/itex] taken to travel between the two cities (i.e. half the period). The distance between the two cities is d and the radius of the Earth is R. Now, suppose the train is given an initial an initial velocity [itex]v_0[/itex]. What is [itex]v_0[/itex] if the time taken to reach the other end of the tunnel is now [itex]t_2 = \frac{t_1}{2}[/itex].

    Hint: Since [itex]t_1 = \frac{T_1}{2} [/itex] that means [itex]t_1 = \frac{1}{2} \frac{2 \pi}{w} = \frac{\pi}{w}[/itex] and therefore [itex]t_2 = \frac{\pi}{2w}[/itex].

    Capture.PNG

    I have the answer to this question I just don't know how to do it.

    2. Relevant equations

    Simple Harmonic Motion, Separable DE, Second-order DE, Newton's Second Law, Chain Rule

    3. The attempt at a solution

    I can get an equation for [itex]v_0[/itex] in terms of x and v but I don't know how to get that in terms of t or if I'm even approaching this the right way. 20160410_175252.jpg
     
  2. jcsd
  3. Apr 10, 2016 #2

    berkeman

    User Avatar

    Staff: Mentor

    Your picture of your work is very hard to read. Can you type your work into the forum? Have you learned how to use LaTeX here yet? There is a tutorial on it in the Help/How-To section (hover over INFO at the top right of the page). :smile:
     
  4. Apr 10, 2016 #3
    the 2nd part of the question is still SHM, which is known to have a sinusoidal solution A*cos(ωt+Φ), where A and Φ are to be found by fitting the initial conditions given above. With a complete trajectory it should be trivial to work out the required v0. Chain rule is not required to solve the equation.
     
  5. Apr 10, 2016 #4
    Hiya, sorry about that. It wasn't going down the right path so I tried it again below.
    Thanks. Overthinking can be the worst. However, I'm still doing something wrong I believe. This is my attempt:
    [tex]\frac{\mathrm{d^2} x}{\mathrm{d} t^2} = \frac{-g}{r} x[/tex]Therefore the solution is of type: [itex]x = Asin(wt + \phi)[/itex] and also since [itex]x(0) = Asin(\phi) = \frac{-d}{2}[/itex], and [itex]x(t_2) = x \left( \frac{\pi}{2w} \right )= Asin\left( \frac{\pi}{2} + \phi\right ) = \frac{d}{2}[/itex] this gives a value of [itex]\phi = - \frac{\pi}{4}[/itex]. Plugging this back into the original equation solves for the amplitude: [itex]Asin(\frac{-\pi}{4}) = \frac{-d}{2}[/itex]. Therefore [itex]A = \frac{d}{\sqrt{2}}[/itex]. Since [itex]v_0 = \frac{\mathrm{d} x(0)}{\mathrm{d} t} = Awcos\left(\phi \right )= Awcos\left( - \frac{\pi}{4}\right )= \frac{d}{\sqrt{2}} \frac{1}{\sqrt{2}}w[/itex]. Thus: [tex]v_0 = \frac{d}{2} w[/tex].
     
  6. Apr 10, 2016 #5
    This is the result I obtained as well
     
  7. Apr 10, 2016 #6
    Thanks for your help guys. I have a much better understanding of the problem now (it's kinda neat actually) but the answer given in the solution page is [itex]v_0 = Rw[/itex], which I'm not sure how to obtain.
     
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