SHM: Gravity-Powered Train (Brace Yourself)

In summary: 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.
  • #1
lowea001
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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, [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.

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 [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
 
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  • #2
lowea001 said:

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, [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].

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 [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. 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). :smile:
 
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  • #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.
 
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  • #4
berkeman said:
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:
Hiya, sorry about that. It wasn't going down the right path so I tried it again below.
throneoo said:
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:
[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].
 
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  • #5
lowea001 said:
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].
This is the result I obtained as well
 
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  • #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.
 

FAQ: SHM: Gravity-Powered Train (Brace Yourself)

How does the gravity-powered train work?

The gravity-powered train utilizes the concept of simple harmonic motion (SHM) to move forward. The train is suspended from a track and as it moves, it creates a back and forth motion due to the force of gravity acting on it. This motion is converted into kinetic energy, propelling the train forward.

What is simple harmonic motion?

Simple harmonic motion is a type of periodic motion in which a body moves back and forth in a straight line due to a restoring force. This force is proportional to the displacement from the equilibrium position, resulting in a sinusoidal motion.

What are the advantages of using a gravity-powered train?

One major advantage is that the train does not require any external power source, making it environmentally friendly and cost-effective. It also has a simple design, reducing maintenance costs and increasing efficiency.

How does the train maintain its speed?

The train maintains its speed by utilizing the principle of conservation of energy. As it moves along the track, it converts potential energy into kinetic energy, and vice versa. The train's speed remains constant as long as the track has a consistent slope and no external forces act on it.

Can the gravity-powered train be used for long-distance transportation?

Theoretically, yes, but in practice, it may not be the most efficient mode of transportation for long distances. The train's speed is limited by the slope of the track, and it may not be practical to construct a track with a significant slope over long distances. Additionally, the train's speed may be affected by external factors such as wind resistance and friction, making it less reliable for long-distance transportation.

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