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gerryliyana

  • 2 years ago

How many number of degrees of freedom of these system??

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  1. gerryliyana
    • 2 years ago
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  2. guyofreckoning2
    • 2 years ago
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    can you provide a higher quality?

  3. gerryliyana
    • 2 years ago
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    what do you mean?

  4. guyofreckoning2
    • 2 years ago
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    I cannot understand the picture... it's too difficult too read

  5. guyofreckoning2
    • 2 years ago
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    *to

  6. guyofreckoning2
    • 2 years ago
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    wait nvm

  7. guyofreckoning2
    • 2 years ago
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    6's? are they 6's in the partial circles?

  8. gerryliyana
    • 2 years ago
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    itsn't 6

  9. guyofreckoning2
    • 2 years ago
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    I am sorry to admit I cannot, I probably just provided false hope and nothing else, I thought it were a pic of something else but I was wrong, not too good at that.. Sorry, I am truly sorry. It's running slow tonight and everything.

  10. guyofreckoning2
    • 2 years ago
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    but at least I can provide a medal

  11. gerryliyana
    • 2 years ago
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    ok no problem ..., :)

  12. guyofreckoning2
    • 2 years ago
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    heh, the place is running slow tonight, kinda mean to drop in and not answer someone's question

  13. kr7210
    • 2 years ago
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    its a coupled oscillator and its degree of freedom is "one"

  14. kr7210
    • 2 years ago
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    2*2-1-1-1=1

  15. Vincent-Lyon.Fr
    • 2 years ago
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    There are two degrees of freedom.

  16. gerryliyana
    • 2 years ago
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    why??

  17. aero_elastic
    • 2 years ago
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    Assuming the two rods that the masses are suspended from are fixed (their length won't change) there is two degrees of freedom in the system. The position of each rod can be written in terms of its angular displacement (1DOF each). The length of the spring too can be written in terms of the two theta terms (assuming one knows the distance between base of the two rods that the masses are suspended from). The system can be described in more terms (x,y coordinates of masses 1&2), but two are all that's needed.

  18. gerryliyana
    • 2 years ago
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    ok, i saw when n independent coordinates are required to specify the position of the masses of a system, the system is of n degrees of freedom. For example if the masses m1 and m2 are contrained to move vertically, at least one coordinate (just call x(t)) is required to define the location of each mass at any time. Thus te system requires altogether two coordinates to specify their positions; it is a two-degree-of-freedom system.., right???

  19. Vincent-Lyon.Fr
    • 2 years ago
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    You're right.

  20. Vincent-Lyon.Fr
    • 2 years ago
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    Is it just a question about degrees of freedom or do you have to derive the equations of motion of this system?

  21. gerryliyana
    • 2 years ago
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    very nice.., thank u @Vincent-Lyon.Fr . it isn't only a question about degrees of freedom, i do have to derive the equations of motion, but in coupled oscillations of a loaded string

  22. gerryliyana
    • 2 years ago
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    until to obtain wave equation

  23. gerryliyana
    • 2 years ago
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    follow me: I finally did read a book "The Physics of Vibrations and Waves) by HJ Pain. It show how the coupled vibrations in the periodic structur of the loaded string become waves in a continuous medium. The equations of motion of the r-th mass to be: \[\frac{ d^{2}y_{r} }{ dt^{2} } = \frac{ T }{ma } (y_{r+1} - 2y_{r} + y_{r-1})\] then separation a = dx and consider the limit dx --> 0 as the masses merge into a continuous heavy string. The: \[\frac{ d^{2}y_{r} }{ dt^{2} } = \frac{ T }{ m } \left( \frac{ y_{r+1}-2y_{r} + y_{r-1} }{ dx} \right) = \frac{ T }{ m }\left( \frac{ (y_{r+1} - y_{r}) }{ dx } -\frac{ (y_{r}-y_{r-1}) }{ dx } \right)\] \[= \frac{ T }{ m } \left[ \left( \frac{ dy }{ dx } \right)_{r+1} - \left( \frac{ dy }{ dx } \right)_{r} \right] \] and \[\left( \frac{ dy }{ dx } \right)_{x+dx} - \left( \frac{ dy }{ dx } \right)_{x} =\frac{ d^{2}y }{ dx^{2} } dx \] I'm a little bit confused by: \[\left( \frac{ dy }{ dx } \right)_{x+dx} - \left( \frac{ dy }{ dx } \right)_{x} =\frac{ d^{2}y }{ dx^{2} } dx \] why \(\left( \frac{ dy }{ dx } \right)_{x+dx} - \left( \frac{ dy }{ dx } \right)_{x} \) is equal to \(\frac{ d^{2}y }{ dx^{2} } dx\) ?????

  24. Aperogalics
    • 2 years ago
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    @gerryliyana it is simple just think practically let dy/dx=t then if i divide it by dx i.e. |dw:1355218737331:dw| then it is derivative :)

  25. gerryliyana
    • 2 years ago
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    what if it's second derivative???

  26. Aperogalics
    • 2 years ago
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    means???????

  27. gerryliyana
    • 2 years ago
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    If dy/dx = t shown by \[\frac{ t_{m+dm}-t_{m} }{ dm } \] how the form formula for second derivative?

  28. Aperogalics
    • 2 years ago
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    it would be |dw:1355219544429:dw| but t=dy/dm so |dw:1355219573174:dw|

  29. Aperogalics
    • 2 years ago
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    @gerryliyana

  30. gerryliyana
    • 2 years ago
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    |dw:1355219655154:dw|

  31. Aperogalics
    • 2 years ago
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    If dy/dx = t shown by tm+dm−tm/dm how the form formula for second derivative? in ur post @gerryliyana can you tell from where dm comes??????????

  32. gerryliyana
    • 2 years ago
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    ok ok i got it.., i'm sorry ..., hehe

  33. Aperogalics
    • 2 years ago
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    no prob. :)

  34. gerryliyana
    • 2 years ago
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    Ok, so \[y' = \lim_{dx \rightarrow 0} \frac{ y_{x+dx} - y_{x} }{ dx } \] and for second derivative \[y'' = \lim_{dx \rightarrow 0} \frac{ y'_{x+dx} - y'_{x} }{ dx }\] \[dx y''= y'_{x+dx} - y'_{x}\] \[dx \frac{ d^{2}y }{ dt^{2} } = \left( \frac{ dy }{ dx } \right)_{x+dx} - \left( \frac{ dy }{ dx } \right) _{x}\] Ok??????

  35. kr7210
    • 2 years ago
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    i'm sorry, a coupled oscillator have 2 degree of freedom coz of have two generalized coordinate to describe the system, sorry again.

  36. gerryliyana
    • 2 years ago
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    ok no problem @kr7210 thank you for coming :)

  37. Aperogalics
    • 2 years ago
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    @gerryliyana it's correct now :)

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