Mechanics, PE to position function

In summary, we have a 3 kg object moving along the x-axis with a potential energy function of U(x) = 4x^2. At x = -0.5, the object has a velocity of +2. We are asked to find the object's position and kinetic energy as functions of time, assuming x = 0 at time t = 0 and all forces acting on the object are conservative. Using the initial conditions, we get ME = 7 and F = -dU/dx = -8x. This leads to the differential equation dx/dt = sqrt(2/3)*sqrt(4 - 4x^2), which can be solved using trigonometric substitution to get the position
  • #1
tjkubo
42
0

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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  • #2
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"
 
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  • #3
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.
 
  • #4
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
 
  • #5
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.
 
  • #6
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. ?
 
  • #7
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.
 
  • #8
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.
 
  • #9
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!
 

FAQ: Mechanics, PE to position function

What is mechanics?

Mechanics is a branch of physics that deals with the study of motion and the forces that cause it. It is concerned with the physical laws that govern the behavior of objects in motion.

What is PE to position function?

PE to position function is a mathematical relationship between potential energy (PE) and position. It describes how the potential energy of an object changes as its position changes.

How is PE to position function related to mechanics?

PE to position function is an important concept in mechanics because it helps us understand the relationship between energy and motion. By analyzing the potential energy of an object at different positions, we can predict its future motion and behavior.

How is PE to position function used in real-world applications?

PE to position function is used in many real-world applications, such as calculating the trajectory of a projectile, predicting the behavior of a simple pendulum, and understanding the motion of objects in a gravitational field. It is also used in engineering and design to optimize the performance of machines and structures.

What are some common misconceptions about PE to position function?

One common misconception about PE to position function is that potential energy is only dependent on an object's height. In reality, potential energy can also be affected by other factors such as the object's mass and the strength of the gravitational field. Another misconception is that potential energy is the same as kinetic energy, when in fact they are two different forms of energy that can be converted into each other.

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