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Spring/Mass System with Unequal Masses |
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| Oct17-11, 10:15 PM | #1 |
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Spring/Mass System with Unequal Masses
1. The problem statement, all variables and given/known data
I want to find the equations of motion of two masses [itex]m_1[/itex] and [itex]m_2[/itex] attached to each other by a spring on a smooth surface assuming [itex]m_2[/itex] is given an instantaneous velocity [itex]v_0[/itex] at time zero. Call the unstretched length of the spring [itex]l[/itex]. 2. Relevant equations I want to solve this using purely Newtonian methods. 3. The attempt at a solution The position of [itex]m_1[/itex] in the center of mass frame is given by: [tex] r_{1_{CM}} = r_1 - R_{CM} = \frac {m_2 (r_1 - r_2)}{m_1+m_2} [/tex] Likewise, the position of [itex]m_2[/itex] in the CM frame is: [tex] r_{2_{CM}} = r_2 - R_{CM} = \frac {m_1 (r_2 - r_1)}{m_1+m_2} [/tex] I can write down Newton's equations for each mass using for Hooke's law [itex]r_{2_{CM}} - r_{1_{CM}} - l[/itex] as the displacement of the length of the spring from its equilibrium position. At this point, I get two differential equations that I do not know how to solve. (Not SHM.) Can anybody help me? Thanks. |
| Oct17-11, 11:14 PM | #2 |
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The equation become SHM in the COM frame. In your notation, just make the substitution: [itex]r_{1CM} = - r_{2CM}[/itex]
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| Oct17-11, 11:26 PM | #3 |
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Is that substitution justifiable even though the two masses are unequal? Certainly the distance from the center of mass to [itex]m_1[/itex] need not equal the distance from the center of mass to [itex]m_2[/itex]...
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| Oct17-11, 11:41 PM | #4 |
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Spring/Mass System with Unequal Masses
What are the differential equations you get?
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| Oct17-11, 11:46 PM | #5 |
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The force on mass one will be:
[tex] F_1 = m_1 r''_{1_{CM}} = - k (r_{2_{CM}} - r_{1_{CM}} - l) [/tex] And on mass two: [tex] F_2 = m_2 r''_{2_{CM}} = + k (r_{2_{CM}} - r_{1_{CM}} - l) [/tex] (Please correct me if this is wrong!) |
| Oct18-11, 12:08 AM | #6 |
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Two possible approaches:
1. You should be able to get an equation of the form [tex]\frac{m_1m_2}{m_1+m_2} \ddot{r} = -kr[/tex]with an appropriate definition of r. 2. You could write your equations as single matrix equation and then diagonalize the matrix. That'll decouple the equations for you. I know the first approach is definitely doable because it's a standard result in classical mechanics. The second one might work. I haven't worked it out, so there could be complications I'm not aware of. |
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| center of mass, mass, spring |
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