Two masses connected by spring, find x(t) of m1.

[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 the mass on the left as a function of time.

Homework Equations

I'll write the position of m_{1} as x and the position of m_{2} as y.

m_{1}=m_{2}

x(0)=0

\dot{x}(0)=v_{0}

q=y-x-l_{0} and its derivatives.

L=\frac{1}{2}m\dot{x}^{2}+\frac{1}{2}m\dot{y}^{2}-\frac{1}{2}k(y-x-l_{0})^{2}

The Attempt at a Solution

Equations of motion for the Lagrangian
m\ddot{x}=k(y-x-l_{0})

m\ddot{y}=-k(y-x-l_{0})

Using q=y-x-l_{0} and \ddot{q}=\ddot{y}-\ddot{x}
\ddot{q}=\frac{-2k}{m}q=-\omega^{2}q

which has the solution
q(t)=Acos\omega t+Bsin\omega t

We also have equations in the form
\ddot{x}=-\omega^{2}x and \ddot{y}=-\omega^{2}y

Substituting the two previous equations into our equations of motion for the Lagrangian we find that
y=-x

Now we substitute y=-x into q=y-x-l_{0} and get
q=-2x-l_{0}

x(t)=-\frac{1}{2}q(t)-\frac{l_{0}}{2}

x(t)=-\frac{1}{2}Acos\omega t-\frac{1}{2}Bsin\omega t-\frac{l_{0}}{2}

\dot{x}(t)=\frac{\omega}{2}(Asin\omega t - Bcos\omega t)

Apply initial conditions
x(0)=0 so A=-l_{0}

\dot{x}(0)=v_{0} so B=\frac{-2v_{0}}{\omega}

So our solution is
x(t)=\frac{v_{0}}{\omega} sin\omega t + \frac{l_{0}}{2}cos \omega t - \frac{l_{0}}{2}

where \omega=\sqrt{\frac{2k}{m}}

Just to point out I set the initial position of the left mass at the origin as I figured all we need is the position information of the two masses relative to each other. Please comment on this if I did this wrong.

Is this solution close? As always thanks for the help!

 

[TSny]

q=y-x-l_{0} and its derivatives.

L=\frac{1}{2}m\dot{x}^{2}+\frac{1}{2}m\dot{y}^{2}-\frac{1}{2}k(y-x-l_{0})^{2}

The Attempt at a Solution

Equations of motion for the Lagrangian
m\ddot{x}=k(y-x-l_{0})

m\ddot{y}=-k(y-x-l_{0})

Using q=y-x-l_{0} and \ddot{q}=\ddot{y}-\ddot{x}
\ddot{q}=\frac{-2k}{m}q=-\omega^{2}q

which has the solution
q(t)=Acos\omega t+Bsin\omega t

Hello. I think everything's ok so far.

We also have equations in the form
\ddot{x}=-\omega^{2}x and \ddot{y}=-\omega^{2}y

Substituting the two previous equations into our equations of motion for the Lagrangian we find that
y=-x

I don't follow the above. \ddot{x} and \ddot{y} should each depend on both x and y as shown by your equations of motion derived from the Lagrangian.

Also, x = -y would imply \dot{x} = -\dot{y} which would not satisfy the initial condition that \dot{x} = v_o and \dot{y} = 0 at t = 0. (Or have you suddenly switched to the center of mass reference frame?)

Now we substitute y=-x into q=y-x-l_{0} and get
q=-2x-l_{0}

x(t)=-\frac{1}{2}q(t)-\frac{l_{0}}{2}

x(t)=-\frac{1}{2}Acos\omega t-\frac{1}{2}Bsin\omega t-\frac{l_{0}}{2}

\dot{x}(t)=\frac{\omega}{2}(Asin\omega t - Bcos\omega t)

Apply initial conditions
x(0)=0 so A=-l_{0}

\dot{x}(0)=v_{0} so B=\frac{-2v_{0}}{\omega}

So our solution is
x(t)=\frac{v_{0}}{\omega} sin\omega t + \frac{l_{0}}{2}cos \omega t - \frac{l_{0}}{2}

where \omega=\sqrt{\frac{2k}{m}}

I don't believe this is correct. Note that in the frame of reference in which the problem is set up, the system as a whole will slide to the right while the masses oscillate. I think the problem is asking for the position function of the left mass in this frame of reference.

You might try introducing another variable Q = x + y. Derive an expression for \ddot{Q} and solve it. The solutions for q and Q should yield solutions for x and y.

Another approach would be to switch to the center of mass reference frame and solve for the motion of the masses in that frame and then switch back to the original frame.

 

[TSny]

[

The Attempt at a Solution

Equations of motion for the Lagrangian
m\ddot{x}=k(y-x-l_{0})

m\ddot{y}=-k(y-x-l_{0})

Using q=y-x-l_{0} and \ddot{q}=\ddot{y}-\ddot{x}
\ddot{q}=\frac{-2k}{m}q=-\omega^{2}q

which has the solution
q(t)=Acos\omega t+Bsin\omega t

We also have equations in the form
\ddot{x}=-\omega^{2}x and \ddot{y}=-\omega^{2}y

Substituting the two previous equations into our equations of motion for the Lagrangian we find that
y=-x

I'm a little confused as to which frame of reference you are setting up your equations.
If you are working in a "moving" frame of reference in which the CM is at rest, then the equation y = -x would be ok if you are taking the origin at the CM. If you want to take the origin of this moving frame at the initial position of m1, then how would you fix the equation y = -x?

Apply initial conditions
x(0)=0 so A=-l_{0}

\dot{x}(0)=v_{0} so B=\frac{-2v_{0}}{\omega}

Do you have the correct initial velocity for m1 in the frame of reference moving with the CM?