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Poll Results
 
 frequency to make m2 stationary?
 Sqrt(K2/<M1+M2>) 0 0%
 Sqrt(K2/M1) 0 0%
 Sqrt(K1/M2) 0 0%
 (Sqrt([<K1*K2>/<K1+K2>]/M2) 0 0%
Total votes: 0. This poll has been closed.


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electricpete

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Reply with quote  #1 

Curran's quiz inspired me to try to think of another.  Here goes: 

Ground === K1 === M1 === K2 === M2  <= Sinusoidal force applied to M2

What is the frequency of sinusoidal force applied to M2 which will result in M2 being stationary? 

electricpete

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Posts: 647
Reply with quote  #2 
Actually I think I'm going to cancel / retire this quiz. It's not that interesting. 

The answer is sqrt(K2/M1).  At that frequency K2 and M1 create a resonant system that would resist movement at the location of M2.  
Curran919

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Reply with quote  #3 
I'd never actual thought about it this way, but if FRF_11 has an antinode (typical PDA), then I guess it makes sense that FRF_22 also has an antinode.

I would not have known where to start approaching the answer conceptually.
electricpete

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Reply with quote  #4 
Attached is my "proof" of my answer to this quiz, using  the electrical analogy for vibration discussed in the  other thread:   https://www.machineryanalysis.org/post/electrical-analogy-for-sinusoidal-steady-vibration-of-discrete-vibrating-systems-10196828?pid=1309085209


 
Attached Files
pdf 2DOF_Quiz2_ByElectricpete.pdf (83.93 KB, 6 views)

electricpete

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Posts: 647
Reply with quote  #5 

Now that I think about it some more, my answer was wrong.  I tried to set it up as an analogy to a dynamic absorber (the dynamic absorber was shown in slide 3 of the electrical analogy to vib thread). 

But the problem here is that there is another element Zk1 connected to the node between Zm1 and Zk2.  The short circuit path through Zk2 and Zm1 does not drag that particular node to zero voltage (zero velocity). So the current (force) flowing through Zk1 is non-zero and we can’t ignore the presence of Zk1. [nono]

In contrast, the dynamic absorber equivalent circuit has extra elements Zk1 and Zm1 that we can ignore, but we can ignore those because they are connected to a node that is dragged to zero voltage by the short circuit Zk2 + Zm2.


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