Calculating Normal Mode: 2 Bars & Metal Spring

[blost]

2 metal, thin, bars (length=l, mass=m) are hunging on same height (distance between bars-d)
Lower end of bars are conected by metal spring (mass=0, k-spring constant, d-length)
My problem is how to calculate a normal mode.

 

[lanedance]

start by writing down the force balance/equations of motion...

 

[blost]

so...

there are 3 diferent forces

1) force which causes tense of springs F=F_0cos(\omega t)

2) force of gravity (I susspect we can pass over it because for small deflection of bars is really little...

3) Restoring force of spring F=-kx

What shall I do next ?

 

[lanedance]

1) is there a sinuosoidal driving force?
2) probably fair assumption, though we''ll keep it in mind
3) ok so this is the main restoring force

now i would pick a co-ordinate system, the angle of each bar makes against vertical seems like a good choice.

Then, for given angles t1, t2 write down the force balance with the equations of motion for each bar...

 

[blost]

1) yes, there is some sinusoidal force but I think we can forget it. I needed it to second part of task...

I have no idea how to do it... I Must write some differential eguation, yes ? I still have big problems with differential equations...

 

[lanedance]

start with an FBD for a single bar, write down the moments about the hinge in terms of the spring force (unknown) then relate that to the angular acceleration about that point... (think about including gravity if needed as well... this will proabbly give you a pendulum mode...)

the other bar should look pretty similar, then using the fact the spring force is the same, couple them together to get a set of differential equations based on the two angles...

 

[blost]

\varepsilon = \frac{M}{I}

I=1/3 ml^2

M=F*l

F=-kx \Rightarrow F=-ksin\alpha l

\varepsilon = \frac{-ksin\alpha l^2}{1/3 ml^2}

\varepsilon =-3 \frac{ksin\alpha}{m}

I try my best but I think it is wrong...

 

[blost]

ok... I tried do It better:

M_1=- \frac{1}{2}d mgsin \theta

M_2=-kxcos\theta d \Leftrightarrow M_2=-ksin\theta d cos \theta d

M_w=- \frac{1}{2}d mgsin \theta-ksin\theta d^2 cos \theta


sin\theta \approx \theta

cos\theta \approx 1

M_w=- \frac{1}{2}d mg \theta-kd^2\theta

I \varepsilon= M_w

\varepsilon= \frac{d^2\theta}{dt^2}

\frac{d^2\theta}{dt^2}+ (\frac{\frac{1}{2}d mg-kd^2}{I})\theta=0

\frac{d^2\theta}{dt^2} + \omega _0^2\theta=0

\omega _0^2= \frac{\frac{1}{2}d mg-kd^2}{I}

\omega _0= \sqrt{\frac{\frac{1}{2}d mg-kd^2}{I}}

is it good ?

 

[lanedance]

so the top ends of the bars are pinned, and they can only roatte about that pin?

its a bit difficult to follow, what exactly is theta here?

but shouldn't you have 2 variables, one to capture the motion of each bar?

 

[blost]

(distance between bars-d)

sory but english is not my first language and... when I put "pin" in my dict there is only "penis" :D

Theta is a delfection angle of bar...

 

[lanedance]

pinned connection means free to rotate about that point, but restrained form any translations

d will be constant at the pinned ends of the bar, but teh bars can rotate inpedendtly, though they wil be affected by the spring. ie you probably need a theta for each bar...

 

[blost]

so... 2 different theta it's a bit too complicated for me... how i must modificated my solution ?