Undamped Harmonic Motion of a rod

[Feodalherren]

Homework Statement

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.

Homework Equations

Undamped harmonic motion. Structural dynamics.

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.

Am I thinking about this in the right way?

edit: I assumed that the rod has negligible mass.

 

[Orodruin]

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?

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).

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.

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.

 

[haruspex]

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

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

 

[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 going to 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.

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).

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.

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.

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.

 

[Orodruin]

my''=mg-ky-F(t)

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.

 

[haruspex]

ΣFy=my''=mg-ky-F(t)

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).

 

[Feodalherren]

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.

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

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).

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 let's 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.

 

[Feodalherren]

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

that would give x(0)=0.02 m

?

 

[Orodruin]

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.

 

[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.

 

[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)

 

[Orodruin]

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.

 

[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.

 

[Orodruin]

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.

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

 

[Feodalherren]

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

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))?

 

[haruspex]

we really only have mg and a time variant kx term

Yes, so why did you write:

Could it be k(0.02+y(t))?

?
Was that a mistake?

 

[Feodalherren]

Yes, so why did you write:

?
Was that a mistake?

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

 

[haruspex]

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

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?

 

[Feodalherren]

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?

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

 

[haruspex]

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

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

 

[Feodalherren]

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

ΣFx=mx''=-kx-mg

 

[haruspex]

ΣFx=mx''=-kx-mg

You went a step too far; I only asked what the force was. ΣF_x=-kx-mg, but x is the relative position. What is the absolute position?

 

[Feodalherren]

That's where I get stuck. I have no clue. x+y(t)?

 

[haruspex]

That's where I get stuck. I have no clue. x+y(t)?

Yes. So in terms of that, what is the acceleration?

 

[Feodalherren]

x''+y''=x''-1.6sin(20t)

 

[haruspex]

x''+y''=x''-1.6sin(20t)

Ok. So what is your ΣF=ma equation now?

 

[Feodalherren]

Hmm

∑F=m(x''-1.6sin(20t))=-kx-mg

 

[haruspex]

Hmm

∑F=m(x''-1.6sin(20t))=-kx-mg

Looks right.

 

[Feodalherren]

So then I end up getting

mx''+kx=-mg+1.6sin(20t)

which is of the form x(t)= [(P/k) / (1-(Wn/Ω)^2)]sin(Ωt) + Acos(Wn*t) + Bsin(Wn*t)

yes?

 

[haruspex]

So then I end up getting

mx''+kx=-mg+1.6sin(20t)

which is of the form x(t)= [(P/k) / (1-(Wn/Ω)^2)]sin(Ωt) + Acos(Wn*t) + Bsin(Wn*t)

yes?

I would expect to see a constant term to deal with the mg.

 

[Feodalherren]

ah yeah, duh!

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

 

[Feodalherren]

Or acutally, couldn't I just find that with

x(0)=-mg/k= [(P/k) / (1-(Wn/Ω)^2)]sin(Ωt) + Acos(Wn*t) + Bsin(Wn*t)

so

A=-mg/k

 

[haruspex]

Or acutally, couldn't I just find that with

x(0)=-mg/k= [(P/k) / (1-(Wn/Ω)^2)]sin(Ωt) + Acos(Wn*t) + Bsin(Wn*t)

so

A=-mg/k

No, that only works at t=0. Over time, the average of that sum of trig functions is zero, not -mg/k.

 

[Feodalherren]

So then by the principle of superpositioning it should be
x(t)= [(P/k) / (1-(Wn/Ω)^2)]sin(Ωt) + Acos(Wn*t) + Bsin(Wn*t) - mg/k

but then I arrive back at my original issue - how do I find A and B. I only have one initial condition?

 

[haruspex]

So then by the principle of superpositioning it should be
x(t)= [(P/k) / (1-(Wn/Ω)^2)]sin(Ωt) + Acos(Wn*t) + Bsin(Wn*t) - mg/k

but then I arrive back at my original issue - how do I find A and B. I only have one initial condition?

No, you have two. At time zero it is at rest in the equilibrium position.