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Question:

If two identical masses are attached to two walls facing each other and the two masses themselves are connected by a third spring (all springs have same initial length and spring constant k). Now we apply two driving forces to these two masses respectively: one on the left with $F_dcos{2\omega t}$ and the one on the left with $2F_dcos{2\omega t}$. Find a particular solution for $x_1(t)$ and $x_2(t)$.

My solution:

$$m\frac{d^2x_1}{dt^2}=-kx_1+k(x_2-x_1)+F_dcos(2\omega t)$$ $$m\frac{d^2x_2}{dt^2}=-kx_2-k(x_2-x_1)+2F_dcos(2\omega t)$$ $$\rightarrow\frac{d^2}{dt^2}(x_1+x_2)+\omega^2(x_1+x_2)=3\frac{F_d}{m}cos(2\omega t)$$ $$\frac{d^2}{dt^2}(x_2-x_1)+3\omega(x_2-x_1)=\frac{F_d}{m}cos(2\omega t)$$ $$Let ~z_1=x_1+x_2~and~z_2=x_2-x_1~\rightarrow~Guess ~z_1=Ccos(2\omega t)~and ~z_2=Dcos(2\omega t)$$ $$\rightarrow~C=D=-\frac{F_d}{m\omega^2}$$ $$\rightarrow x_1(t)=0,~x_2(t)=Acos(2\omega t)$$

where $A=C=D=-\frac{F_d}{m\omega^2}$.

This solution seems very weird since mass 1 is not moving at all. Could somebody give a hint what is wrong with this anti-intuitive solution?

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closed as off-topic by tpg2114, Brandon Enright, Bernhard, Neuneck, JamalS Nov 12 '14 at 9:08

This question appears to be off-topic. The users who voted to close gave this specific reason:

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  • $\begingroup$ The one one the left and the one on the left? Which one is the right one? $\endgroup$ – LDC3 Nov 12 '14 at 3:42
  • $\begingroup$ hey guys, wake up, you are putting everything on hold without even reading! The OP gave the correct entire solution, he only asked if it was correct because it was not intuitive to him $\endgroup$ – Wolphram jonny Nov 12 '14 at 13:59
  • $\begingroup$ @julianfernandez when someone gives a complete and correct solution, they don't have a question. That's why we put those on hold. $\endgroup$ – David Z Nov 12 '14 at 18:43
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    $\begingroup$ @DavidZ what policy is that? it not certainly what it says above: "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem". And he did have a question. $\endgroup$ – Wolphram jonny Nov 12 '14 at 18:47
  • $\begingroup$ I think the OP's question is more than valid in this case. He wants to know why mass 1 is stationary, and not changing position with time. This is a reasonable point of confusion, seeing as how it is receiving a time-dependent driving force. It may not be immediately obvious to him (nor is it immediately obvious to me after a cursory glance) why mass 1 would necessarily be stationary. Either way, I think his question is non-trivial, and it meets homework policy guidelines. $\endgroup$ – Sean Nov 12 '14 at 19:45
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Your solution is correct. It is not that weird if you think about it. The external force on $x_1$ exactly balances the force from the spring at the center (it has the same frequency but opposite direction, so neither $m_1$ nor the left spring move.

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