Elementary spring and mass question

[chipotleaway]

Homework Statement

Two masses of equal mass m are attached by a single spring of sprint constant k, what is the resonant frequency of the system?

The Attempt at a Solution

I'm not sure how correct it is to treat the two masses on either end as single mass 2m.

$$2m\ddot{x}=-kx$$
$$\ddot{x}=-\frac{k}{2m}x$$

Therefore, the resonant frequency is $$\omega=\sqrt{\frac{k}{2m}}$$

 

[voko]

No, it is not correct at all.

 

[chipotleaway]

The reasoning is wrong or is the answer wrong as well?
Because I got the same answer but saying that when you stretch the string by x, it's like each side gets displaced is x/2

 

[voko]

What you need to understand here is that the system has two degrees of freedom. They can be, for example, the coordinates of each mass, or the coordinate of the center of mass and the distance between the masses.

 

[gneill]

Hint: By using the center of mass as your origin you can take advantage of symmetry to greatly simplify your analysis.

 

[chipotleaway]

The centre of mass doesn't move, so we can treat each mass as if it were attached by half the length of the spring to a wall at the centre (just realized this part was answered in another post I made)

So then you stretch it on both ends (because applying a force in one direction would only result in acceleration of the centre of mass). If each side is stretched by Δx, then the force on each wold be -kΔx and from the equation of motion, we get just ω=√(k/m)...

Is that right?

I thought again about just pulling on one end of the spring whilst holding onto the other - that would change the position of the centre of mass wouldn't it?

 

[gneill]

The centre of mass doesn't move, so we can treat each mass as if it were attached by half the length of the spring to a wall at the centre (just realized this part was answered in another post I made)

So then you stretch it on both ends (because applying a force in one direction would only result in acceleration of the centre of mass). If each side is stretched by Δx, then the force on each wold be -kΔx and from the equation of motion, we get just ω=√(k/m)...

Is that right?

Not quite. If the whole spring has spring constant k, what's the spring constant for half the spring alone?

I thought again about just pulling on one end of the spring whilst holding onto the other - that would change the position of the centre of mass wouldn't it?

Sure. But from the center of mass frame of reference, it's still halfway between the masses.

 

[chipotleaway]

Hang on - the spring constant changes? I always thought it was always constant for some given spring that obeys Hooke's law.

But now I think about it, if we only look at half the spring, it'll be subject to the same force but will only be extended by half the total displacement. So its spring constant would be double.
Mathematically, if k' is the spring constant of half the length, x is the displacement of the entire spring, then

F=k'(x/2)
2F/x=k'

Substituting in x=F/k gives us 2k=k'.


So now if we apply this one side of the spring in the original problem, then ω=√(2k/m)

 

[voko]

Note you could have derived this result in a slightly different way. Say $x$ is the position of one mass, and $y$ of another. Then the force on the first mass is $- k (x - y - l)$, and by Newton's third law the force on the other mass is $k (x - y - l)$, thus $$

m \ddot x = - k (x - y - l)

\\

m \ddot y = k (x - y - l)

$$ Subtracting the second from the first one, we get $m \ddot z = - 2k z$, where $z = x - y - l$, the deviation of the spring's length from its natural length.