## anonymous 3 years ago How many number of degrees of freedom of these system??

1. anonymous

2. anonymous

can you provide a higher quality?

3. anonymous

what do you mean?

4. anonymous

I cannot understand the picture... it's too difficult too read

5. anonymous

*to

6. anonymous

wait nvm

7. anonymous

6's? are they 6's in the partial circles?

8. anonymous

itsn't 6

9. anonymous

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. anonymous

but at least I can provide a medal

11. anonymous

ok no problem ..., :)

12. anonymous

heh, the place is running slow tonight, kinda mean to drop in and not answer someone's question

13. anonymous

its a coupled oscillator and its degree of freedom is "one"

14. anonymous

2*2-1-1-1=1

15. anonymous

There are two degrees of freedom.

16. anonymous

why??

17. anonymous

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. anonymous

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. anonymous

You're right.

20. anonymous

Is it just a question about degrees of freedom or do you have to derive the equations of motion of this system?

21. anonymous

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. anonymous

until to obtain wave equation

23. anonymous

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. anonymous

@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. anonymous

what if it's second derivative???

26. anonymous

means???????

27. anonymous

If dy/dx = t shown by $\frac{ t_{m+dm}-t_{m} }{ dm }$ how the form formula for second derivative?

28. anonymous

it would be |dw:1355219544429:dw| but t=dy/dm so |dw:1355219573174:dw|

29. anonymous

@gerryliyana

30. anonymous

|dw:1355219655154:dw|

31. anonymous

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. anonymous

ok ok i got it.., i'm sorry ..., hehe

33. anonymous

no prob. :)

34. anonymous

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. anonymous

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

36. anonymous

ok no problem @kr7210 thank you for coming :)

37. anonymous

@gerryliyana it's correct now :)