Question

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

Answer 1

can you provide a higher quality?

Answer 2

what do you mean?

Answer 3

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

Answer 4

*to

Answer 5

wait nvm

Answer 6

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

Answer 7

itsn't 6

Answer 8

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.

Answer 9

but at least I can provide a medal

Answer 10

ok no problem ..., :)

Answer 11

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

Answer 12

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

Answer 13

2*2-1-1-1=1

Answer 14

There are two degrees of freedom.

Answer 15

why??

Answer 16

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.

Answer 17

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???

Answer 18

You're right.

Answer 19

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

Answer 20

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

Answer 21

until to obtain wave equation

Answer 22

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 \(\frac{ d^{2}y }{ dx^{2} } dx\) ?????

Answer 23

@gerryliyana it is simple just think practically let dy/dx=t
then if i divide it by dx i.e.
then it is derivative :)