Mass attached to two opposing strings

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Homework Statement


http://faculty.mint.ua.edu/~pleclair/ph125/Misc/PH105_exam2_1Nov_soln.pdf
Similar to question two on this site. Except I'm looking for the velocity as the ball crosses the origin.


Homework Equations


Listed in the question.


The Attempt at a Solution


I got the answer, I'm just stumbling on the reasoning.

The answer says to add the potential energy of the two springs together and that they are the same. So U1+U2=2U1. So solving for velocity is a matter of 1/2 m v^2 = 2U1

However, if you were to break the components of the force into vector form, you would have the y components opposing each other. So the velocity would only be increased by the horizontal component of the force.

Adding the two potential energy sums in such a manner together would be if they were pulling the same direction, would it not?

The energy from the two stretched strings would partially go into the mass but not all of it being that the forces are opposing so that the mass stays at a neutral y position.
 

Answers and Replies

  • #2
SammyS
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Energy is a scalar quantity, so it doesn't have components like a vector does.
 
  • #3
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Right, I guess what I don't understand is the reasoning as to why two forces acting at an angle that creates opposition are going to apply all of their energy into an object. Everything prior to this told me that two forces opposing would just waste the energy into some other effect rather then moving, yet this example seems to be saying that the magnitude of both will be applied to the object.
 
  • #4
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Also, couldn't you sum the potential energy in the x direction separate from the potential energy in the y direction? And similarly for both objects as separate entities. Technically, we are ignoring the Z direction potential energy in this setup, so they are separable, right?

So you would have x for the springs with the same magnitude and both negative. Y would have equal magnitudes and one positive energy one negative energy.

Logically, you are summing the energy of this system as the integral of the force where x traverses in the x and y direction. Yet applying all of the force to the x direction?
 
  • #5
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Okay so I found a link that seems to disagree with the previous link...
http://www.music.mcgill.ca/~gary/618/week5/lumped.html [Broken]

2. Transverse motion says that the restoring force in the y direction is what I've calculated. The integral with respect to y should be the potential energy that would be traversed in the x direction.

I'm confused...
 
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  • #6
SammyS
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The solution in the link you give in your Original Post, addresses most of the issues you bring up. It says that you can go ahead and find the forces for displacement in the x-direction. Then integrate to get the potential energy function, U(x), for this system.

[tex]U(x)=-\int_0^x{\vec{F}(x)\cdot\hat{i}}dx [/tex]

If you do this, you find that the result is the same as that you get if you simply add the potential energy of the two springs based only on the amount they're stretched, not on the components of the stretching.
 
  • #7
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Ughh, well doing it all a few times I get

kx^2-2kl(x^2+l^2)^1/2+2kl^2
and
kx^2-2kl(x^2+l^2)^1/2

Finding the restoring force uses
([tex]F=-kx=-k(\sqrt{l^2+x^2}-l)cos(acos(\frac{x}{\sqrt{l^2+x^2}}[/tex]

This agrees with the equation listed in the second link. I imagine the 2kl is lost from the cosine.

If you work out the math, you get that the first equation takes F=-kx and integrates to U=1/2*-kx^2 and subs in delta x. The second equation takes F=-kx and subs in delta x but takes the cosα and then integrates. This results in different answers.
 
  • #8
SammyS
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Ughh, well doing it all a few times I get

kx^2-2kl(x^2+l^2)^1/2+2kl^2  This is U(x) in solution key you linked to in the OP.
and
kx^2-2kl(x^2+l^2)^1/2  This appears to be in error. U(x) can be written as U(x)=kx^2+2kL[L-(x^2+L^2)^1/2] , as given in the solution key.
               (I changed lower case "l" to upper case "L" for readability.)


Finding the restoring force uses
([tex]F=-kx=-k(\sqrt{l^2+x^2}-l)cos(acos(\frac{x}{\sqrt{l^2+x^2}}[/tex]

This agrees with the equation listed in the second link. I imagine the 2kl is lost from the cosine.

If you work out the math, you get that the first equation takes F=-kx and integrates to U=1/2*-kx^2 and subs in delta x. The second equation takes F=-kx and subs in delta x but takes the cosα and then integrates. This results in different answers.
Some comments are in red above.

Regarding your expression, which I suppose should be:

[tex]F=-kx=-k(\sqrt{L^2+x^2}-L)\cos\left(\arccos\left(\frac{x}{\sqrt{L^2+x^2}}\right)\right)[/tex]

It is well known from trig., that [tex]\cos(\arccos(t))=t\,,[/tex] for all t such that -1 ≤ t ≤ 1.
Using this you will get:

[tex]F=-k(x-L\frac{x}{\sqrt{L^2+x^2}})\,.[/tex]

The force, F, is the x component of the force due to one of the springs.

Multiply by 2 and integrate to get U(x). You will get the same U(x) that is given above and in the solution key.
 
  • #9
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[tex]F=-k(x-L\frac{x}{\sqrt{L^2+x^2}})\,.[/tex]

The force, F, is the x component of the force due to one of the springs.

Multiply by 2 and integrate to get U(x). You will get the same U(x) that is given above and in the solution key.
Here lies the problem. This integral is NOT
kx^2-2kl(x^2+l^2)^1/2+2kl^2  

This integral yields
kx^2-2kl(x^2+l^2)^1/2

int -kx = 1/2 k x^2
kl * int x(l^2+x^2)^(-1/2) = kl(l^2+x^2)^.5

Which proves the missing 2kl^2 that differentiates from the first equation to the second.
 
  • #10
SammyS
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How about a constant of integration?

The potential U(x) is defined up to an added constant.

If you take your U(x) and look at ΔU = U(x) - U(0), you will get the same result that you get if you use the potential in the posted Solution, USol.(x), and find ΔUSol. = USol.(x) - USol.(0). In other words, USol.(0), whereas your U(x) gives: U(0) = -2kL2.

If you find the potential from the following, it agrees explicitly with that in the posted solution.

[tex]U(x)=-\int_0^x-k\left(x-L\frac{x}{\sqrt{L^2+x^2}}\right)dx\,.[/tex]
 
  • #11
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Hrmmm.... sounds possible.

But wouldn't that be implying that the system would be perpetually sitting at -2kl^2 potential energy? Which if x=0 the -2kl^2. And in the case of my question, when x=0 you would have a remainder potential energy equivalent to -2kl^2?

Thanks a ton, by the way.
 
  • #12
SammyS
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The position at which you define potential energy to be zero is arbitrary. However, once you decide on that location for a given system, you must be consistent. All that maters in working out a problem (using conservation of energy) is the difference in potential energy from one location to another.
 

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