
#1
Sep3010, 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 



#2
Sep3010, 01:03 PM

P: 83

I believe that doesn't oscillate.




#3
Sep3010, 01:21 PM

P: 672

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_1x_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] 



#4
Sep3010, 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 ? 



#5
Sep3010, 03:25 PM

P: 558





#6
Sep3010, 04:06 PM

P: 672

Quinzio, your derivation makes one critical error, overlooking the algebraic errors, [tex]\ddot x_1 + \ddot x_2=c (x_1x_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_1x_2)[/tex] 



#7
Sep3010, 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!




#8
Sep3010, 05:09 PM

P: 558

(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. 



#9
Sep3010, 05:26 PM

P: 672

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_1x_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. 



#10
Oct110, 06:33 AM

P: 83





#11
Oct110, 07:23 AM

P: 672





#12
Oct110, 10:43 AM

P: 83





#13
Oct110, 11:14 AM

P: 672

How springy would such a spring be? :p



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