Undamped Harmonic Motion of a rod

1. Jun 16, 2017

Feodalherren

1. The problem statement, all variables and given/known data

The problem: The mass the m is placed on the rod with the bushing remaining stationary. The end of the rod deflects 2 cm. The bushing is then given a vertical motion y(t) = 0.4 sin (20t) cm. Determine the magnitude of the motion of the mass m (either relative to the bushing or to a fixed frame) for a stable solution.

2. Relevant equations
Undamped harmonic motion. Structural dynamics.

3. The attempt at a solution
I'm a little bit confused regarding how to deal with the fact that no force was given. Since they gave me the position of the bushing as a function of t I differentiated it twice to find the acceleration of the bushing:

a(t)= - 160sin(20t) cm/s^2

so multiplying this by the mass should give me the force that is being applied on the bushing, correct?

However, the rod is clearly not stiff since I am given a deflection, do I assume that the rod follows the motion of the bushing, i.e. that they have the same velocity at some arbitrary time t? If this is the case then I can find another initial condition and the problem becomes trivial. However, if I cannot assume this because of the elasticity of the rod then I am not given enough initial conditions and I get stuck.

edit: I assumed that the rod has negligible mass.

Last edited: Jun 16, 2017
2. Jun 16, 2017

Orodruin

Staff Emeritus
No, this would not be appropriate. You are given a stationary deflection and that tells you something about the properties of the rod that you can use to solve the problem (assuming the rod is not so bent that the problem becomes non-linear).

Initial conditions will not matter at all in this problem. You are looking for the periodic steady state solution only. If it helps, think of a very small damping rather than zero damping and look for the solution after transients die out.

3. Jun 16, 2017

haruspex

What force is the rod applying to the mass when the bushing is not accelerating? Or maybe you meant net force on the mass?

4. Jun 17, 2017

Feodalherren

The way I wanted to set it up was as follows;

ΣFy=my''=mg-ky-F(t)

where k=3EI/L^3 or k=mg/0.02 N/m
and
F(t)=1.60*m*sin(20t) N

and without going through the Laplace transform the solution would be of the form

y(t)= [(P/k) / (1-(Wn/Ω)^2)]sin(Ωt) + Acos(Wn*t) + Bsin(Wn*t)

I was then gonna use the fact that I know that y(0)=2 to find one of the constants but I got stuck in not being able to find B.

I don't think it's supposed to be non-linear. These problems are fairly straight forward I'm just trying to refresh my physics since after graduating I rarely get to use any of the stuff that I learned in college. I realize that I can find k with the deflection but after that I sort of get stuck.

Hmm I've never done any problems with rods and damping. I have no idea how to set up damping for a rod unless something is explicitly given.

5. Jun 17, 2017

Orodruin

Staff Emeritus
You will not be able to set it up like this because the time dependent force will also depend on the position of the mass. I think the easiest way of doing this is to use Lagrangian mechanics to find the equation of motion. Alternatively, you can find an expression for the force, but that expression will necessarily also include the position.

6. Jun 17, 2017

haruspex

Some confusion there?
As given, y is the displacement of the bush. The acceleration of the mass is not y". E.g. y could stop changing, yet the mass still bob up and down.
The mass only feels two forces, gravity and the force from the bar. It does not directly know what the bush is doing.
You might find it easier to think of it as the mass joined to the bush by a vertical spring instead. Create a separate variable for the displacement of the mass (either in the lab frame or relative to the bush).

7. Jun 17, 2017

Feodalherren

Hmm I don't recall much at all about Lagrangian mechanics so I'd rather not venture down that path. Lets try to find an expression for the force as I already started trying to do that.

Ah yeah of course that makes sense. I was actually thinking of it as a vertical spring already! But with waht you said about the the bushing stopping and the mass still bobbing up and down made me realize that I can't use the same variable. but okay, relative to the bushing lets call the motion of the mass x. I still don't see how the summation of the forces changes? The function F(t) should change somehow but I'm not sure how.

8. Jun 17, 2017

Feodalherren

Could it be x(t)=0.02cos(20t)

that would give x(0)=0.02 m

?

9. Jun 17, 2017

Orodruin

Staff Emeritus
No, what you want to do is to find an expression for the force based on what your x is. Then you need to remember that x is the relative motion so it represents a non-inertial frame and so there will be some inertial forces involved as well.

10. Jun 17, 2017

Feodalherren

Well the force when the system is at rest is equal to mg, that is when the beam is deflected 2 cm. Hmm I don't know, this might be above my skill level.

11. Jun 17, 2017

Feodalherren

The force based on what the x is, wouldn't that just be kx? if we take the frame of reference to be on the bushing running along the center of the rod.

Setting a coordinate system along the beam and calling the angle between the mass and the horizontal position of the beam theta,

x=Lsin(θ)

then

F = k*Lsin(θ)

Last edited: Jun 17, 2017
12. Jun 17, 2017

Orodruin

Staff Emeritus
I suggest to take this nice and slow. There are two forces acting on the mass. What are they? What magnitude and direction do they point in? And what is the acceleration of the mass? There is no need to introduce $\theta$ as a new variable, you will just introduce new variables randomly.

13. Jun 17, 2017

Feodalherren

Hmm okay. Do you mean when the mass is at rest? When it is at rest there is the spring force from the beam pointing up and the weight pointing down.

14. Jun 17, 2017

Orodruin

Staff Emeritus
The forces on the mass don't really depend on whether is at rest or not.

15. Jun 17, 2017

Feodalherren

Hmm I think I'm seeing it now. The force on the bushing won't be creating any force on the mass itself, all that it will do is create some movement in the bushing, as the bushing moves it will affect the displacement of the mass and thus the spring force will be varying. So we really only have mg and a time variant kx term. Could it be k(0.02+y(t))?

Last edited: Jun 17, 2017
16. Jun 17, 2017

haruspex

Yes, so why did you write:
?
Was that a mistake?

17. Jun 17, 2017

Feodalherren

No, Im just on a wild goose chase trying to find a way of relating x to y. Im completely clueless on how to do it.

18. Jun 17, 2017

haruspex

You are defining x as the height of the mass relative to the bush (or in the spring model, as the extension of the spring), right?
What does that mean for the compression or tension in the spring? What, then, is the force from that on the mass?

19. Jun 17, 2017

Feodalherren

Yes, that is how I defined x. That means that the compression/tension is kx.

20. Jun 17, 2017

haruspex

Right.
So what is the net force on the mass? (I would define x as an upward displacement. Watch the signs.)