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## Homework Statement

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

## Homework 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

## 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??