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

## 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??

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Simon Bridge
Homework Helper

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....
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?
im guessing x0 is = 0 so
Don't guess.

As for D, my book doesn't really explain the turning points to well so im not exactly sure about how to find it...
... 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?

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)

Simon Bridge
Homework Helper

ok the problem states that the mass is projected from x=0 so x0 has to be 0
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:
... x and x0 ... represent the beginning and end points of movements
... 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 just integrate both sides here from x to x0 to get
V(x)-V(x0)=1/2kx^2-1/2k(x0)^2
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. $$F(0)=0 \Rightarrow V(0)=0 \Rightarrow c = 0$$
D)turning points are when the Kinetic energy is =0?
so that must mean turning points are when E=+-V(x)?
... and what is E equal to?

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?

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

T(final)=0
T(initial)=1/2Ka^2

Simon Bridge
Homework Helper

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

T(final)=0
T(initial)=1/2Ka^2
You mean that

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

V varies with x as (1/2)Kx^2
im just not good at typing everything here
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 $$A=\int_{x_i}^{x_f}y(x)dx$$