Mechanics, PE to position function

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Homework Help Overview

The problem involves a 3 kg object moving along the x-axis with a potential energy function U(x) = 4x². The initial conditions specify the object's position and velocity at x = -0.5 and v = +2, respectively. The goal is to determine the object's position and kinetic energy as functions of time, under the assumption that all forces acting on the object are conservative.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the formulation of the mechanical energy equation and the setup of a second-order differential equation derived from the force equation. There are attempts to solve for position as a function of time, with some participants expressing uncertainty about initial conditions and constants of integration.

Discussion Status

Several participants are actively engaging with the problem, exploring different methods to solve the differential equation. Some have provided guidance on utilizing initial conditions, while others are questioning the assumptions made regarding mechanical energy and the force equations. There is a mix of interpretations regarding the constants involved in the solution.

Contextual Notes

Participants note the challenge of determining constants due to the lack of explicit initial conditions relating time and velocity. There is also a discussion about the correct formulation of the mechanical energy equation, with some confusion regarding its value and the sign of the force derived from the potential energy function.

tjkubo
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Homework Statement


A 3 kg object is moving along the x-axis where U(x) = 4x2. At x = -.5, v = +2. Find the object's position and KE as functions of time. Assume x = 0 at time t = 0. All forces acting on the object are conservative.

Homework Equations


ME = U + K
K = (1/2)mv2
F = dU/dx
F = ma

The Attempt at a Solution


Using initial conditions, ME = 4.
F = dU/dx = 8x
F = ma
8x = (3)d2x/dt2
This is where I got stuck. I was attempting to solve for x(t), find v(t), then use that to find K(t). Assuming everything else is correct, how do you solve a second order differential equation like this? Otherwise, please correct me.
 
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Just solve

[tex]\frac{d^2x}{dt^2}- \frac{8}{3}x=0[/tex]

Do you know how to solve a second order differential equation with constant coefficients?


EDIT: http://www.sosmath.com/diffeq/second/constantcof/constantcof.html"
 
Last edited by a moderator:
I am still having trouble solving for x(t).
I got [tex]x=c_1e^{\sqrt{\frac{8}{3}}t}+c_2e^{-\sqrt{\frac{8}{3}}t}[/tex]
and [tex]c_1+c_2=0[/tex]
but since there is no initial value associating time and velocity, I can't find the constants.
 
tjkubo said:
I am still having trouble solving for x(t).
I got [tex]x=c_1e^{\sqrt{\frac{8}{3}}t}+c_2e^{-\sqrt{\frac{8}{3}}t}[/tex]
and [tex]c_1+c_2=0[/tex]
but since there is no initial value associating time and velocity, I can't find the constants.

How did you get ME=4 by chance?

Also F=-dU/dx not F=+dU/dx
 
Can you use the "At x = -.5, v = +2" condition?
Would it work to begin with
ME = U + K
4 = 4x^2 + 1/2*mv^2 (which includes the x = -.5 condition)
4 = 4x^2 + 1.5(dx/dt)^2
The solution to this differential equation would have only one constant, which you should be able to get using the x=0 at t=0 condition.
 
rock.freak667 said:
How did you get ME=4 by chance?

Also F=-dU/dx not F=+dU/dx

My bad, ME = 7. I forgot to square. (Is is correct to assume that ME is constant?)
Anyway, when I retried solving the differential equation with the initial conditions, I ended up getting 0 = 0 while solving for the constants. ?
 
Delphi51 said:
4 = 4x^2 + 1.5(dx/dt)^2
The solution to this differential equation would have only one constant, which you should be able to get using the x=0 at t=0 condition.

How would you solve this differential equation? The (dx/dt)^2 term throws me off.
 
dx/dt = sqrt(2/3)*sqrt(4 - 4x^2)
sqrt(2/3) dt = dx/sqrt(4 - 4x^2)
Integrate both sides. Doesn't look bad - trig substitution if I'm not mistaken.
 
Yes! I got
[tex]x=\frac{\sqrt{7}}{2}\sin{\sqrt{\frac{4}{3}}\,t}[/tex]
That was much simpler than what I was doing.
Thanks Delphi!
 

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