Exploring Oscillatory Motion of Two Connected Masses on a Frictionless Track

[grandpa2390]

Homework Statement

Two masses slide freely in a horizontal frictionless track and are connected by a Spring whose force constant is k. Fid the frequency of oscillatory motion for this system.

Homework Equations

My professor posted the solution but I am having trouble understanding everything that he did.
according to the solution, the first step is defining the Force equation as
F = -k(x2 - x1 - deltaXe) where deltaXe is the equilibrium separation. x1 is the distance to the center of mass of m1 from some point to the left of m1. x2 is the distance from that point to the center of m2. so that x2-x1 equals the distance between the center of m1 and m2.

The Attempt at a Solution

the distance between the atoms at which the force on each atom is zero. Is the force repulsive (atoms are pushed apart) or attractive (atoms are pulled together) if their separation is (b) smaller and (c) larger than the equilibrium separation
is this the distance between the two masses when the spring is neither compressed nor stretched? Wouldn't that just be x2-x1? so the the Force would be 0?

 

[grandpa2390]

and then later on in the solution he takes says that dXe is equal to x2e - x1e (x2e = the distance of m2 from the center of the spring during equilibrium and likewise with x1e) shouldn't dXe = x2e + x1e ?

 

[BvU]

My professor posted the solution but I am having trouble understanding everything that he did

Doesn't look like a relevant equation to me. Don't you have a few things that look like

##m_1\ddot x_1 = - F_s##

##m_2\ddot x_2 = F_s##

## F_s = -k (x_2 - x_1 - \Delta x_{eq})##
​

so that we can do some mathematics on them ?

Because in that case we could easily derive

##m_1\ddot x_1 + m_2\ddot x_2 =0##

and concentrate on

##m_1 \ddot x_1 - m_2 \ddot x_2 + 2k( x_2 - x_1 - \Delta x_{eq})=0## ?

And perhaps the masses happen to be equal ? Saves work and yields a pleasant answer.

Then: In your attempt there is the sudden appearance of atoms that weren't there before. Please read what you wrote as if you were a helper trying to make sense of your post. I am one and I can't.

 

[grandpa2390]

that third equation is the one. I figured out what it was. the origin was defined in the middle of the spring. So that x1 is negative. so it does turn out right.

 

[HalfLostAndSearching]

Yes, you are correct. The equilibrium separation, or the distance at which the force on each atom is zero, is when the spring is neither compressed nor stretched. This would indeed be when x2-x1 is equal to the equilibrium separation, and the force would be 0.

As for the question of whether the force is repulsive or attractive for smaller and larger separations, it depends on the direction of the displacement (deltaXe). If deltaXe is positive (meaning the masses are pulled apart), then the force is repulsive. If deltaXe is negative (meaning the masses are pushed together), then the force is attractive. This is because the force equation is defined as F = -k(x2 - x1 - deltaXe), so a positive deltaXe would result in a positive force, and a negative deltaXe would result in a negative force.