SHM: Gravity-Powered Train (Brace Yourself)

[lowea001]

Homework Statement

[/B]
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.

Homework Equations

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

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.

 

[berkeman]

Homework Statement

[/B]
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 $t_2 = \frac{\pi}{2w}$.

View attachment 98901

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

Homework Equations

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

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. View attachment 98904

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).

 

[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.

 

[lowea001]

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).

Hiya, sorry about that. It wasn't going down the right path so I tried it again below.

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.

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: $$v_0 = \frac{d}{2} w$$.

 

[throneoo]

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: $$v_0 = \frac{d}{2} w$$.

This is the result I obtained as well

 

[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.