# SHM: Gravity-Powered Train (Brace Yourself)

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1. Apr 10, 2016

### lowea001

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, $F = - \frac{mgr}{R}$.
Find the time $t_1$ 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 $v_0$. What is $v_0$ if the time taken to reach the other end of the tunnel is now $t_2 = \frac{t_1}{2}$.

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

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 $v_0$ 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.

2. Apr 10, 2016

### 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).

3. Apr 10, 2016

### throneoo

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.

4. Apr 10, 2016

### lowea001

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:
$$\frac{\mathrm{d^2} x}{\mathrm{d} t^2} = \frac{-g}{r} x$$Therefore the solution is of type: $x = Asin(wt + \phi)$ and also since $x(0) = Asin(\phi) = \frac{-d}{2}$, and $x(t_2) = x \left( \frac{\pi}{2w} \right )= Asin\left( \frac{\pi}{2} + \phi\right ) = \frac{d}{2}$ this gives a value of $\phi = - \frac{\pi}{4}$. Plugging this back into the original equation solves for the amplitude: $Asin(\frac{-\pi}{4}) = \frac{-d}{2}$. Therefore $A = \frac{d}{\sqrt{2}}$. Since $v_0 = \frac{\mathrm{d} x(0)}{\mathrm{d} t} = Awcos\left(\phi \right )= Awcos\left( - \frac{\pi}{4}\right )= \frac{d}{\sqrt{2}} \frac{1}{\sqrt{2}}w$. Thus: $$v_0 = \frac{d}{2} w$$.

5. Apr 10, 2016

### throneoo

This is the result I obtained as well

6. Apr 10, 2016

### lowea001

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 $v_0 = Rw$, which I'm not sure how to obtain.