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2 Masses 1 Spring Question Help

by NeedPhysHelp8
Tags: masses, spring
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NeedPhysHelp8
#1
Sep30-10, 11:02 AM
P: 41
1. The problem statement, all variables and given/known data

Two masses m1 and m2 slide freely on a frictionless horizontal plane, and are
connected by a spring of force constant k . Find the natural frequency of oscillation for
this system.


2. Relevant equations

[tex] \ddot{x} + \omega^2x=0 [/tex] where [tex]\omega^2 = \frac{k}{m}[/tex]

[tex] \nu = \frac{\omega}{2\pi} [/tex]

3. The attempt at a solution

So I had no clue where to start with this question so I asked the prof and he told me you don't have to used reduced mass or center of mass for this question. He said just to consider the equations of motion for both masses.
Alright so let's say m1 is displaced by a distance x to the left then
[tex] \ddot{x} + \frac{k}{m1}x = 0 [/tex]
Now I'm not sure if this is right, but m2 also feels a stretch in spring by displacement x but the force is in opposite direction of m1 so:
[tex] \ddot{x} - \frac{k}{m2}x = 0 [/tex]
Now what do I do with both these equations? I'm lost someone please help guide me along here! Thanks
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planck42
#2
Sep30-10, 01:03 PM
P: 82
I believe that doesn't oscillate.
RoyalCat
#3
Sep30-10, 01:21 PM
P: 671
The question asks for the natural frequency of the oscillations, not whether any are excited, planck42.

What is x as you've defined it? If you've defined x as the elongation of the spring, then it is actually [tex]x=\Delta x \equiv x_1-x_2[/tex] in which case you should simply use some algebra to solve for the coordinate [tex]\Delta x[/tex]

Remember, [tex]\ddot {\Delta x} = \ddot x_1 - \ddot x_2[/tex]

Your equations need to be indexed,

[tex] \ddot{x_1} + \frac{k}{m1}x = 0 [/tex]
[tex]\ddot{x_2} - \frac{k}{m2}x = 0 [/tex]

Quinzio
#4
Sep30-10, 03:18 PM
P: 558
2 Masses 1 Spring Question Help

I'm not sure 100% of what I'm saying but searching around the internet I didn't found any relative example, so I will go.

You need to state the Newton 2 law for each mass.
I express something with question marks. If you want you can think what they are, if not, go on.

[tex] a_1 = \frac{-k\ x }{m_1}- ???[/tex]

[tex] a_2 = \frac{-k\ x }{m_2}- ???[/tex]

Then sum the two expressions.
We get
[tex] a_1+a_2 = -\frac{k\ x }{m_1} -\frac{k\ x }{m_2} - ???[/tex]

that gives
[tex] 2(a_1+a_2) = -k\ x \frac{m_1+m_2}{m_1m_2} [/tex]

[tex] a = a_1+a_2 = -k\ x \frac{m_1+m_2}{2m_1m_2} [/tex]

So, the natural oscillator term [tex]\omega^2[/tex] is

[tex]\omega^2 = k \frac{m_1+m_2}{2m_1m_2}[/tex]

[tex]\omega = \sqrt\left(k \frac{m_1+m_2}{2m_1m_2}\right) [/tex]


Assuming the formula is true, use it to deduce the particular case in which one mass is attached to a spring and the other side of the spring is attached to a wall.
The mass of the wall can be thought as ??? ([tex]m_{wall}[/tex].
Resolve the expression
[tex]m = \lim_{m_1 \rightarrow m_{wall}} \frac{m_1+m_2}{2m_1m_2}[/tex]

What happends if one mass if very small, if not zero ?
Quinzio
#5
Sep30-10, 03:25 PM
P: 558
Quote Quote by planck42 View Post
I believe that doesn't oscillate.
Then you have just to try to build one and see what happends.
RoyalCat
#6
Sep30-10, 04:06 PM
P: 671
Quinzio, your derivation makes one critical error, overlooking the algebraic errors, [tex]\ddot x_1 + \ddot x_2=-c (x_1-x_2)[/tex] where [tex]c[/tex] is some positive constant, cannot be solved as harmonic motion, since it is not of the form [tex]\ddot f = -\omega^2 f[/tex]

In order to find the natural frequency of the oscillations, you must bring the two equations to the form [tex]\ddot x_1 - \ddot x_2=-\omega ^2 (x_1-x_2)[/tex]
NeedPhysHelp8
#7
Sep30-10, 04:40 PM
P: 41
Great thanks alot everyone for your help - I'll post my answer later tonight when I have some time to work on it. Much appreciated!
Quinzio
#8
Sep30-10, 05:09 PM
P: 558
Quote Quote by RoyalCat View Post
Quinzio, your derivation makes one critical error, overlooking the algebraic errors, [tex]\ddot x_1 + \ddot x_2=-c (x_1-x_2)[/tex] where [tex]c[/tex] is some positive constant, cannot be solved as harmonic motion, since it is not of the form [tex]\ddot f = -\omega^2 f[/tex]

In order to find the natural frequency of the oscillations, you must bring the two equations to the form [tex]\ddot x_1 - \ddot x_2=-\omega ^2 (x_1-x_2)[/tex]
I'm afraid my equations are right.
(Not because I say it, but because math passages are ok).
I just simulated it and the motion is perfectly sinusoidal.

If I am wrong, no problem, but just show me why.
Solve your equations and you'll find the same what I found.
RoyalCat
#9
Sep30-10, 05:26 PM
P: 671
Quinzio, look over your solution again, it has some major problems.

You have an additional, inexplicable factor of 2, that, and your method is wrong on the conceptual level. You do not have a differential equation in the contraction of the spring ([tex]\Delta x = x_1-x_2[/tex]) but rather a mixed equation in the contraction and the the sum of the accelerations [tex]\ddot x_1 + \ddot x_2[/tex]

Yes, the motion is sinusoidal, but your analysis is incorrect and gives you the wrong oscillation frequency. Your simulation may give you your result, but that would be as a result of writing down Newton's Second Law wrong for the elements in the system (I still don't know what you mean by the question marks).

I'll send you the solution in full in PM form.
planck42
#10
Oct1-10, 06:33 AM
P: 82
Quote Quote by Quinzio View Post
Then you have just to try to build one and see what happends.
Now I get it, I think. That looks kinda like two separate oscillators with their own frequencies that are independent of the presence of the second mass.
RoyalCat
#11
Oct1-10, 07:23 AM
P: 671
Quote Quote by planck42 View Post
Now I get it, I think. That looks kinda like two separate oscillators with their own frequencies that are independent of the presence of the second mass.
No, that is not true. Write out Newton's Second Law for both, and you will see that the equations for the displacement of each are coupled! The force on mass 1, depends on the position of mass 2, and vice versa. :) The meaningful coordinates that can be defined, in this case, are the relative displacement (The distance between the two masses) and the position of the center of mass.
planck42
#12
Oct1-10, 10:43 AM
P: 82
Quote Quote by RoyalCat View Post
No, that is not true. Write out Newton's Second Law for both, and you will see that the equations for the displacement of each are coupled! The force on mass 1, depends on the position of mass 2, and vice versa. :) The meaningful coordinates that can be defined, in this case, are the relative displacement (The distance between the two masses) and the position of the center of mass.
Oh duh, I just wasn't thinking hard enough about this. I thought it was arbitrary how far the masses could go in either direction so long as momentum is conserved, but clearly energy is also conserved. My first post about no oscillations assumes that the spring itself can move.
RoyalCat
#13
Oct1-10, 11:14 AM
P: 671
How springy would such a spring be? :p


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