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A particle of mass m is contrained to lie on along a frictionless,horizontal plane

  1. Sep 30, 2012 #1
    1. The problem statement, all variables and given/known data
    a particle of mass m is contrained to lie on along a frictionless,horizontal plane subject to a force given by the expression F(x)=-kx. It is projected from x=0 to the right along the positive x direction with initial kinetic energy T(o)=1/2kA^2, k and A are positive constants. Find

    a) The potential energy function V(x) for this force.
    b)The kinetic energy
    c) The total energy of the particle as a function of its position.
    d) Find the turning points of the motion


    2. Relevant equations

    Fx**=F(x)
    -dV(x)/dx=F(x)
    integral of F(x) from x to x0=T-T(o)
    T(o)+V(Xo)=E=T+V(x)

    velocity=dx/dt=+- sqrt(2/m(E-V(x)))

    integreal of dx/+- sqrt(2/m(E-V(x))) from X to Xo=t-t0


    3. The attempt at a solution

    ok part A,B,C i think i got, but i would want someone to double check for me just in case i made mistakes....As for D, my book doesn't really explain the turning points to well so im not exactly sure about how to find it...

    A)-dV(x)/dx=F(x)=-kx
    dV(x)/dx=kx
    dV(x)=kx dx

    i just integrate both sides here from x to x0 to get
    V(x)-V(x0)=1/2kx^2-1/2k(x0)^2

    im guessing x0 is = 0 so

    V(x)=1/2kx^2

    B) integral of F(x) dx= T-T(o)
    integral from x to x(o) of -kx dx=-1/2kx^2=T-T(o)
    1/2kx^2=T(o)-T
    1/2kx^2=1/2kA^2-T
    T=1/2kA^2-1/2kx^2

    now im a little bit confused here,is this supposed to mean that x=a making T=0? how am i supposed to know that?

    C) Assuming i did the other 2 parts properly for the total energy i get:

    1/2kA^2=E

    D) This part I have no idea how to do....i dont understand exactly what this turning point is really, i see in the book it shows a graph of V(x) vs x and theres a region in the middle called the allowed region,the two points of the curve where they are both exactly equal to E are said to be the turning points,but how am i supposde to find them?

    if i use velocity=+- sqrt(2/m(E-V(x))) E cant be smaller than V(x) and what about if they are both euqal? the velocity would be 0...so is it like the point in which E is slightly bigger than V(x) how do i figure this out??
     
  2. jcsd
  3. Sep 30, 2012 #2

    Simon Bridge

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    Re: a particle of mass m is contrained to lie on along a frictionless,horizontal plan

    You should show your reasoning when you present your work. eg. F(x)=-dV/dx = -kx and at x=0 the total energy is entirely kinetic so V(0)=0 ... why not do B by conservation of energy?
    Don't guess.

    ... the turning points are the x positions where the particle comes to rest and turns around. What is the kinetic energy at a turning point? What does this mean about the potential energy?
     
  4. Sep 30, 2012 #3
    Re: a particle of mass m is contrained to lie on along a frictionless,horizontal plan

    ok the problem states that the mass is projected from x=0 so x0 has to be 0
    A)
    but i forgot that x and x0 were supposed to represent lol....i just remembered that they represent the beginning and end points of movements so by that we can say that

    T(o)+V(xo)=E=1/2 Ka^2 since the potential is 0 initially....

    the change in potential energy V(x)-V(x0)= 1/2 Kx^2 by integrating the force with respect to the x-direction at X=final position and Xo=initial position=0

    V(x)=1/2Kx^2

    B) T(o)+V(Xo)=T+V(x)

    and its furthest point on the x axis, there is no longer any kinetic energy remaining so T=0

    1/2Ka^2+0-V(x)=T=0
    1/2Ka^2-1/2Kx^2=0=T

    1/2Ka^2=1/2Kx^2

    so regardless,at the end T is still=0

    C) E=1/2 Ka^2=1/2Kx^2 going from one endpoint to the other make the kinetic and potential become equal when X=0 and when X=X(final displacement)

    D)turning points are when the Kinetic energy is =0?
    so that must mean turning points are when E=+-V(x)?

    D)
     
  5. Oct 1, 2012 #4

    Simon Bridge

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    Re: a particle of mass m is contrained to lie on along a frictionless,horizontal plan

    The relation F=-kx only makes sense if the uncompressed position is at x=0.
    Since F(0)=0, V(0)=0, and all the energy is kinetic E=T0.

    It is useful to reserve the subscript 0 to refer to some kind of initial condition, use the unsubscripted x for the general x axis coordinate. Initial and final x can be xi and xf for example ... or just a and b to avoid using too many subscripts.

    Anyway - when you say:
    ... you must mean that the other way around: x is the end and x0 is the start. Otherwise what follows makes no sense.

    Math is communication - it is important to be careful with your definitions and stick to them.
    You have also been changing variable names... try to stick to one name for one thing.

    I think you are getting a bit tangled there:
    $$\int dV = \int kxdx \Rightarrow V + c_1 = \frac{1}{2}kx^2 + c_2 \Rightarrow V = \frac{1}{2}kx^2 + c$$
    ... you get a constant of integration on both sides, but, they are both constants so just combine them into one that you can find from the initial conditions. $$F(0)=0 \Rightarrow V(0)=0 \Rightarrow c = 0$$
    ... and what is E equal to?
     
  6. Oct 1, 2012 #5
    Re: a particle of mass m is contrained to lie on along a frictionless,horizontal plan

    I think you are getting a bit tangled there:

    lol thats what i meant to say,im just not good at typing everything here.... ok xi=initial position=0 xf=final position

    and what about the integration of Dv from xf-xi?

    i thought the contant terms would cancel eachother out like this

    integral Dv= [V(final)+C]-[V(initial)+C]= V(final)-V(initial)

    V(initial)=V(0)=0

    same thing with the integral of F(x)

    integral Kx=[1/2Kx^2(initial)+C]-[1/2Kx^2(final)+C]

    integral -Kx= T-T(o) integral Kx=T(o)-T

    D) turning points of motion would be E=1/2Kx^2=1/2Ka^2

    now is everything done properly?
     
  7. Oct 1, 2012 #6
    Re: a particle of mass m is contrained to lie on along a frictionless,horizontal plan

    V(X final)=1/2Kx^2
    V(X initial)=V(0)=0

    T(final)=0
    T(initial)=1/2Ka^2
     
  8. Oct 1, 2012 #7

    Simon Bridge

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    Re: a particle of mass m is contrained to lie on along a frictionless,horizontal plan

    The thing about solving DEs is that you don't need to do a definite integral - that's what leaving the "+ c" out the end is for: you can find the "c" from the boundary conditions.

    You can set up a definite integral if you really want to - it's just more work than you need to do.

    None of the other parts need an integration - you can just cite "by conservation of energy" since that is The Law. So ##E=T(x)+V(x)=\text{constant} = T(0)## since V(0)=0 ... and everything else is basic algebra from there :)

    eg. T(x)=E-V(x) = T(0)-V(x) = (K/2)(a2-x2) <---<<< what it looks like without the LaTeX markup ... but it looks better with:
    $$ T(x)=E-V(x)=T(0)-V(x) = \frac{K}{2}\left ( a^2-x^2 \right )$$ Using your last notes to the effect that:
    ##T(0)=\frac{1}{2}Ka^2##
    ##V(x)=\frac{1}{2}Kx^2##
    (uses K as the compression coefficient or "spring constant")

    As we've already discussed, the turning points are at "x-positions such that the kinetic energy is zero" which, in math, is written: ##x:T(x)=0## ... so put T(x)=0 and solve for x to get two values. Normal algebra see?

    Note:
    You mean that

    V(X final)=(1/2)K(X final)^2 ??

    V varies with x as (1/2)Kx^2
    Learning LaTeX math markup is really really worth the effort ... especially if you intend to continue in any kind of science. It lets you write the nice formatted equations I did before and stuff like the equation above come out like this: ##V(x)=\frac{1}{2}kx^2##

    To see how I did that, just click the "quote" button in the bottom-left of this post.

    There is also an advanced editor that has point-and-click tools to help layout - I think it's a bit of a pain tbh. If you have to rely on just plain text, you have to be very careful.

    I think your physics is doing fine - you are just having a bit of fun translating the physical ideas into the language of math... so most of what I've been telling you is like getting a language lesson ;)

    So you would normally do ##\Delta x = x_f-x_i## and $$A=\int_{x_i}^{x_f}y(x)dx$$
     
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