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Mass on a Spring

  • Thread starter erok81
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Homework Statement



Assume that the spring is initially compressed by a quantity x0 from its equilibrium value at x=0. The particle is initially at x=-x0. The mass is released and when the particle crosses x=x0 it detaches from the spring.

Homework Equations



x''+γx'+ω2x=0

Another question asked to obtain the general solution for x(t), which I've done, but I don't think it's needed yet.

γ=0, therefore x''+ω2x=0

The Attempt at a Solution



a)What is the total energy of the system.

This is Etot=KE+U. Since the spring is compressed and the mass isn't moving, it should all be potental. I'm not sure how to express this with what I am given though. :confused:

b) When the spring is released, what is the speed of the particle when it crosses x=0. I am not sure how to begin on this one. I would assume it is all kinetic since it is no longer attached to the spring. But again I am not sure how to express this either.
 

Answers and Replies

  • #2
rock.freak667
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a)What is the total energy of the system.

This is Etot=KE+U. Since the spring is compressed and the mass isn't moving, it should all be potental. I'm not sure how to express this with what I am given though. :confused:
Good! So if the mass is not moving, then it has no KE, and you know the 'extension' for this. So Etot = U. So U is ?

b) When the spring is released, what is the speed of the particle when it crosses x=0. I am not sure how to begin on this one. I would assume it is all kinetic since it is no longer attached to the spring. But again I am not sure how to express this either.
No, it is not attached at x=x0, at x=0, it is still attached.

Etot is the same throughout at any point within the motion, so yes, it is entirely kinetic, with U = 0.

What is the formula for kinetic energy?
 
  • #3
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Good! So if the mass is not moving, then it has no KE, and you know the 'extension' for this. So Etot = U. So U is ?



No, it is not attached at x=x0, at x=0, it is still attached.

Etot is the same throughout at any point within the motion, so yes, it is entirely kinetic, with U = 0.

What is the formula for kinetic energy?
For potential it's U=1/2kx2 and kinetic would be KE=1/2mv2.
 
  • #4
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The part I am confused on is how those two energies relate to the original equation. There is no way it's just KE and U...is it?
 
  • #5
Unless I'm very much mistaken, that is indeed what this problem is asking for... since the only given is a variable. It's probably meant to be just a quick conceptual problem.
 
  • #6
rock.freak667
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The part I am confused on is how those two energies relate to the original equation. There is no way it's just KE and U...is it?
Conservation of mechanical energy (which applies for this question), states that the

Etot = T+ U

T = KE and U = PE.
 
  • #7
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Maybe the reason I am confused is because it references the first question that I had to solve. Then the first part of this one says γ=0, therefore x''+ω2x=0 which, what I am not seeing, doesn't help me finding Etot = KE+ U. But as usual I might just be over complicating it.

To further my confusion there is one more part to it that says plot x(t) indicating when the laws of motion change. This seems to indicate I plot my original solution for x(t) - which doesn't have Etot = KE+ U.
 
  • #8
rock.freak667
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Maybe the reason I am confused is because it references the first question that I had to solve. Then the first part of this one says γ=0, therefore x''+ω2x=0 which, what I am not seeing, doesn't help me finding Etot = KE+ U. But as usual I might just be over complicating it.

To further my confusion there is one more part to it that says plot x(t) indicating when the laws of motion change. This seems to indicate I plot my original solution for x(t) - which doesn't have Etot = KE+ U.
Without using the differential equation you can do the first part by just considering energy.

Conservation of energy states that in a closed system, the energy remains constant. For mechanical energy this means that

Etot= KE + PE

Initially when x=-x0, there is no velocity so KE = 0, giving you Etot and earlier you stated that U = 1/2kx2.

So you can get Etot in terms of x0.

If you want to plot x(t), you need to solve the DE given

x''(t) + ω2x=0

Does this remind you of any sort of motion?
 
  • #9
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Without using the differential equation you can do the first part by just considering energy.

Conservation of energy states that in a closed system, the energy remains constant. For mechanical energy this means that

Etot= KE + PE

Initially when x=-x0, there is no velocity so KE = 0, giving you Etot and earlier you stated that U = 1/2kx2.

So you can get Etot in terms of x0.

If you want to plot x(t), you need to solve the DE given

x''(t) + ω2x=0

Does this remind you of any sort of motion?
That makes sense. I always get stuck on problems like this where the answer is obvious, but I have to make it harder than it is so I don't trust the obvious answer.

As for the motion...I'm not sure. If I solve the DE I have a cosine and sine term. That's about all I have.

I can plot the cos/sin wave. Set the turning point at x0 but then would the graph just flatline afterward? Since the potential goes down as the mass moves away from the initial point and once it is release from the spring it would have a constant KE assuming a frictionless surface.

I'm not sure if that's right though.
 
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  • #10
rock.freak667
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As for the motion...I'm not sure. If I solve the DE I have a cosine and sine term. That's about all I have.
So your solution to that DE is a combination of sine and cosine, good.

I can plot the cos/sin wave. Set the turning point at x0 but then would the graph just flatline afterward? Since the potential goes down as the mass moves away from the initial point and once it is release from the spring it would have a constant KE assuming a frictionless surface.

I'm not sure if that's right though.
Remember, the potential energy is being converted to kinetic energy and vice versa throughout the motion.

The potential energy decreases yes, but remember, that the KE is max at x=0. As x goes towards x=x0, the PE increases while the KE decreases.
 
  • #11
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When the KE is max PE is zero, right?
And since the object leave the spring, once it reaches that point wouldn't PE stay zero and KE become a constant?

I am having trouble translating that to my x(t) graph.

And yep, my DE solution is x(t)=c1cos(ωx)+c2sin(ωx).
 
  • #12
rock.freak667
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Well your DE and its solution would only be valid for when the mass is attached. After that, the DE does not describe its motion, so if the surface is frictionless, then yes the KE would be constant.

If you rewrite your x(t) as a single trig function, you can easily draw its graph.
 

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