X(t) of a diatomic molecule with given v_0

[AbigailM]

Homework Statement

Two identical carts (of mass m) are free to move on a frictionless, straight horizontal track. The masses are connected by a spring of constant k and un-stretched length $l_{0}$. Initially the masses are a distance $l_{0}$ apart with the mass on the left having a speed $v_{0}$ to the right and the mass on the right at rest. Find the position of mass on the left as a function of time.

Homework Equations

$q=x_{2}-x_{1}-l_{0}$

$\dot{q}=\dot{x_{2}}-\dot{x_{1}}$

$\ddot{q}=\ddot{x_{2}}-\ddot{x_{1}}$

The Attempt at a Solution

$m\ddot{x_{1}}=-k(x_{2}-x_{1}-l_{0})$

$m\ddot{x_{2}}=k(x_{2}-x_{1}-l_{0})$

remembering $\ddot{q}=\ddot{x_{2}}-\ddot{x_{1}}$

$\ddot{q}=\frac{2k}{m}q=\omega^{2}q$

$q(t)=c_{1}e^{\omega t}+c_{2}e^{-\omega t}$

Just wondering if I'm on the right track? If so I'll do the initial conditions and then solve for $x_{1}(t)$

Thanks for the help!

 

[Infinitum]

Hi AbigailM!

It would help if you state what x₁ and x₂ are... :tongue2:

 

[AbigailM]

oops, x1 is the position of cart 1 and x2 is the position of cart 2.

 

[Infinitum]

What have you chosen as the origin? I believe q is the elongation/compression of the spring? Is cart 1 the left cart, and cart 2 the right?PS : Its good to state all assumptions before solving the problem

 

[paranormal]

I can confirm that you are on the right track. Your approach to solving the problem using Newton's second law and the equations for simple harmonic motion is correct. Once you have the general solution for q(t), you can use the initial conditions to find the specific solution for x1(t). Keep up the good work!