Solving Spring Mass System Homework: Equations of Motion & Linearization

[James_Frogan]

Homework Statement

The system depicted in the attachment consists of a slender rigid link of mass m1 hinged at point O, and supported by a spring of stiffness k2 (200N/m) at point B. The link in turn supports a spring of stiffness k1 (400N/m) at point A connected to a point mass m (2kg).

Determine the equations of motion for the link hinged at O, and the point mass m relative to the static equilibrium of the system. Develop the general equations of motion, and linearise this result for small angle approximations of the link displacement such that sin (theta) ~= (theta) and cos(theta) ~= 1 (valid for angles up to 5 deg).

Homework Equations

The Attempt at a Solution

I haven't gotten very far with this because I don't know how to start solving it. I've come up with the position vector for the mass m, which is:
0.2 cos (theta) i - (0.2 sin (theta) + mg/k) j
and I've tried to do the summation of moments about point O, getting
Mo = 2g cos(theta) - 200*x2 *cos(theta) [x2 is the spring compression/extension, not x^2]
and right about here I get stuck, because I have no idea how to proceed on :(

Any help would be great, I just need to know what are the relevant equations I can use for this.

 

[kuruman]

Hi James_Frogan, welcome to PF!

I sincerely hope that you have seen Lagrangians. Assuming that you have, pick your generalized coordinates and start writing things down. I suggest that you pick x₁ and x₂ for the spring displacements from equilibrium and θ for the angular displacement of the bar. There is a constraint equation relating x₁ and θ. If you have not seen Lagrangians, you need to go ye olde free body diagram way.

 

[James_Frogan]

Hi! Thanks for the welcome. Look forward to helping out some people in my spare time, seems fun :D

Anyway, I haven't done lagrangians before, so I'm stuck with the FBD way. Thing is that I don't know how to get on from there. Do I make a summation of all the Fy Fx Moment forces?What happens after that (because all I get are just some terms, do I equate them to zero and assume that the link is horizontal in the equilibrium position?)

I might have a ton of questions about this one coming cos I entered uni on advanced standing after serving in the army for 2 years, so all my prior knowledge has been eroded haha (plus the lecture notes are pretty bad :( ) is there a website that kind of teaches this stuff in a simple manner?

Thanks

 

[kuruman]

Assume that the top spring is displaced by x₁ and the bottom by x₂. Draw a free body diagram of the bar and put is all the forces acting on it (Don't forget the mass of the bar). Apply Newton's Second Law once for the sum of all the forces and once more for the sum of all the torques. Assume that the link is displaced by a small but non-zero angle θ from the equilibrium position.

I don't know of any website that teaches this stuff. Maybe someone else does.

Good luck.

 

[James_Frogan]

Hi again, I've gotten progress on the question; but I'm stuck at the value of x1 (extension of the lower spring).

It's not a fixed point, but can I use 0.2 sin (theta), since its small deflections? I'm thinking that I can't because the spring extension won't follow the angle theta due to the inertia of the mass.

So what can I use for x1?

 

[kuruman]

If I understand your question correctly, you are worried that the spring is hanging straight down whereas the link is at angle theta with respect to the horizontal. Note however, that if the spring is stretched by amount x₁ in the vertical direction, it will exert a force k₁x₁ in the vertical direction. Treat this force as you would treat the weight of the link except that its point of application different.

 

[James_Frogan]

Hi sorry, i meant that the spring is attached to a mass, not a floor, so i can't use the deflection of the bar as x1? (ie: x1 != 0.2sin(theta)), am i right? in my head i picture what would happen if the bar is moved down, the mass would not move immediately because of inertia, so the lower spring would experience compression. if it is not then what is the value of x1?

 

[kuruman]

Don't worry about how the system starts. Pretend that it has been in motion for a while at which point you take a snapshot. The snapshot shows one spring extended by x₁, the other by x₂ and the link at angle θ with respect to the horizontal. How the system will develop from this point on and how it got to this point is governed by the equations of motion that you are trying to find.

 

[James_Frogan]

hmmm i have derived that x2 is 0.25 sin theta because it is connected to a fixed wall, so is the extension of spring x1 = 0.2 sin theta? or is there a d^2(theta)/dt^2 factor also?

 

[kuruman]

No, x₁ has nothing to do with theta in the sense that it can be anything because it stands for the displacement of the second spring from the equilibrium position. However the position of the hanging mass is 0.2 sinθ + x₁. You will need two equations of motion, one for the link and one for the hanging mass. You will get these (in principle) from the FBD by writing down Newton's Second Law for rotations (link) and linear motion (hanging mass).

 

[CFDFEAGURU]

I solved this problem before in Vierck's book on Vibration Analysis.

See attachment for sketch.

First, represent the spring k1 by an equivalent spring called k3 located directly above k2.

Now allow ke to be equivalent to k3 and k2 in series.

It can be shown that

P3 = ((A+B)/A)*P1

Delta_3 = (A/(A+B))*Delta_1

k3 = P3/Delta_3 = ((A+B)^2/A^2)*P1/Delta_1

Simplyfiying k3

k3 = ((A+B)^2/A^2)*k1

Now putting k3 and k2 in series connected to the mass

ke can be shown to be (I leave this for you to fill in the blanks)

ke = k2*k3/(k2+k3)

Now springs in series add like this

1/ke = 1/k3 + 1/k2

Now substituting in the equations for k3 and k2

ke = k1*k2(A+B)^2/(A+B)^2*k1+A^2*k2

Now for the mass. (Draw a free body diagram for the mass.)

Net force

Fn = mg-ke(delta+x) (delta = rest position to some new position)

Static equilibrium requires

0 = mg-ke*delta

therfore

Fn = -ke*x

Inertial effects

Fn=m*x'' (the double prime indicates the second derivative with respect to time.)

Now

m*x'' +ke*x = 0

x'' + (ke/m)*x = 0

omega = (ke/m)^1/2

Now

x'' + omega^2*x = 0

The general solution to the above ODE is

x = a*Sin(omega*t) + b*Cos(omega*t) and the period is tau = (2*pi)/omega

The frequency is

f = (1/(2*pi))*omega

f = (1/(2*pi))*(ke/m)^1/2

therfore the freqency of the mass is

f = (A+B)/(2*pi)*(k1*k2)/((A+B)^2*k1+A^2*k2))

Sorry the non use of LaTex, but I am horrible with it.

Hope this helps.

Attention mods: I thought that this post was heading in a very confusing direction with the use of Lagrangians to solve such a trivial problem and I didn't want the original OP to lose hope of being helped. That is why I posted the full solution instead of hinting at it through a series of questions and answers.

Thanks
Matt