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

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0can you provide a higher quality?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I cannot understand the picture... it's too difficult too read

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.06's? are they 6's in the partial circles?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I 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
 4 years ago
Best ResponseYou've already chosen the best response.0but at least I can provide a medal

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ok no problem ..., :)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0heh, the place is running slow tonight, kinda mean to drop in and not answer someone's question

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0its a coupled oscillator and its degree of freedom is "one"

VincentLyon.Fr
 4 years ago
Best ResponseYou've already chosen the best response.4There are two degrees of freedom.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Assuming 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
 4 years ago
Best ResponseYou've already chosen the best response.0ok, 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 twodegreeoffreedom system.., right???

VincentLyon.Fr
 4 years ago
Best ResponseYou've already chosen the best response.4Is it just a question about degrees of freedom or do you have to derive the equations of motion of this system?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0very nice.., thank u @VincentLyon.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
 4 years ago
Best ResponseYou've already chosen the best response.0until to obtain wave equation

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0follow 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 rth mass to be: \[\frac{ d^{2}y_{r} }{ dt^{2} } = \frac{ T }{ma } (y_{r+1}  2y_{r} + y_{r1})\] then separation a = dx and consider the limit dx > 0 as the masses merge into a continuous heavy string. The: \[\frac{ d^{2}y_{r} }{ dt^{2} } = \frac{ T }{ m } \left( \frac{ y_{r+1}2y_{r} + y_{r1} }{ dx} \right) = \frac{ T }{ m }\left( \frac{ (y_{r+1}  y_{r}) }{ dx } \frac{ (y_{r}y_{r1}) }{ dx } \right)\] \[= \frac{ T }{ m } \left[ \left( \frac{ dy }{ dx } \right)_{r+1}  \left( \frac{ dy }{ dx } \right)_{r} \right] \] 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
 4 years ago
Best ResponseYou've already chosen the best response.0@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
 4 years ago
Best ResponseYou've already chosen the best response.0what if it's second derivative???

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0If dy/dx = t shown by \[\frac{ t_{m+dm}t_{m} }{ dm } \] how the form formula for second derivative?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0it would be dw:1355219544429:dw but t=dy/dm so dw:1355219573174:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1355219655154:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0If 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
 4 years ago
Best ResponseYou've already chosen the best response.0ok ok i got it.., i'm sorry ..., hehe

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Ok, 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
 4 years ago
Best ResponseYou've already chosen the best response.0i'm sorry, a coupled oscillator have 2 degree of freedom coz of have two generalized coordinate to describe the system, sorry again.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ok no problem @kr7210 thank you for coming :)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@gerryliyana it's correct now :)
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