anonymous
  • anonymous
How many number of degrees of freedom of these system??
Physics
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
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anonymous
  • anonymous
can you provide a higher quality?
anonymous
  • anonymous
what do you mean?

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anonymous
  • anonymous
I cannot understand the picture... it's too difficult too read
anonymous
  • anonymous
*to
anonymous
  • anonymous
wait nvm
anonymous
  • anonymous
6's? are they 6's in the partial circles?
anonymous
  • anonymous
itsn't 6
anonymous
  • 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.
anonymous
  • anonymous
but at least I can provide a medal
anonymous
  • anonymous
ok no problem ..., :)
anonymous
  • anonymous
heh, the place is running slow tonight, kinda mean to drop in and not answer someone's question
anonymous
  • anonymous
its a coupled oscillator and its degree of freedom is "one"
anonymous
  • anonymous
2*2-1-1-1=1
Vincent-Lyon.Fr
  • Vincent-Lyon.Fr
There are two degrees of freedom.
anonymous
  • anonymous
why??
anonymous
  • 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.
anonymous
  • 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???
Vincent-Lyon.Fr
  • Vincent-Lyon.Fr
You're right.
Vincent-Lyon.Fr
  • Vincent-Lyon.Fr
Is it just a question about degrees of freedom or do you have to derive the equations of motion of this system?
anonymous
  • 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
anonymous
  • anonymous
until to obtain wave equation
anonymous
  • 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 \(\frac{ d^{2}y }{ dx^{2} } dx\) ?????
anonymous
  • 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 :)
anonymous
  • anonymous
what if it's second derivative???
anonymous
  • anonymous
means???????
anonymous
  • anonymous
If dy/dx = t shown by \[\frac{ t_{m+dm}-t_{m} }{ dm } \] how the form formula for second derivative?
anonymous
  • anonymous
it would be |dw:1355219544429:dw| but t=dy/dm so |dw:1355219573174:dw|
anonymous
  • anonymous
@gerryliyana
anonymous
  • anonymous
|dw:1355219655154:dw|
anonymous
  • 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??????????
anonymous
  • anonymous
ok ok i got it.., i'm sorry ..., hehe
anonymous
  • anonymous
no prob. :)
anonymous
  • 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??????
anonymous
  • anonymous
i'm sorry, a coupled oscillator have 2 degree of freedom coz of have two generalized coordinate to describe the system, sorry again.
anonymous
  • anonymous
ok no problem @kr7210 thank you for coming :)
anonymous
  • anonymous
@gerryliyana it's correct now :)

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