# Homework Help: Driven linear oscillations

1. Oct 11, 2008

### Benzoate

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

A block of mass 2 kg is suspended from a fixed support by a spring of strength 2000 N m^-1. The block is subject to the vertical driving force 36 cos pt N. Given that the spring will yield if its extension exceeds 4 cm, find the range of frequencies that can safely be applied. Take g = 10 m s-2

2. Relevant equations

m*x'' + b*x'+kx= F(t)
x''+2*K*x' + $$\Omega$$2=F0cos(pt)

3. The attempt at a solution

2*x''+ b*x'+ 2000*x=36*cos(pt) N

what is the extension, and how would I find the range of frequencies.

2. Oct 11, 2008

### Hootenanny

Staff Emeritus
Are you sure that you have set up your ODE correctly? There are no velocity dependent forces specified in the question and you don't seem to have any term to account for the weight of the mass.

3. Oct 11, 2008

### Benzoate

I had the last equation written wrong.

m*x'' + b*x'+kx= F(t)
x''+2*K*x'+x*$$\Omega$$2=F0*cos(pt)
I didn't include the weight into the equation because I wasn't sure how to incorporate weight into the oscillation driven equation since there is no weight term.

4. Oct 11, 2008

### Hootenanny

Staff Emeritus
Let's start again from first principles. Newton's second law states that

m*x'' = Fnet

Where Fnet is the sum of all forces acting on the body. So what is the sum of all the forces acting on you block?

5. Oct 11, 2008

### Benzoate

Fnet= Fspring+Fgrav

Fspring= m*x'' + b*x'+kx
Fgrav= m*g

can ignore drag forces for now.

6. Oct 11, 2008

### gabbagabbahey

Are you sure about your F_spring? Is there really a damping force specified in this question? And why is there an mx'' in there?!

7. Oct 11, 2008

### Benzoate

I got the equation
Fspring= m*x'' + b*x'+kx , from the textbook Classical Mechanics written by John R. Taylor.

The problem doesn't specify that there is a damping constant present, but I thought that a damping constant could not be present only when the oscillator is under ideal conditions, i.e. no dissipative force presents and the oscillator oscillates forever.

8. Oct 11, 2008

### Hootenanny

Staff Emeritus
You say that we can ignore drag forces, yet you include a damping term (b*x') in your spring equation. You also pull a second order differential term out of thin air.

One other thing to be careful of is your signs. The weight of the mass will always act down and would there usually be negative is a standard coordinate system.

9. Oct 11, 2008

### Hootenanny

Staff Emeritus
You can't just blindly apply equations from texts without knowing what they mean! Close that textbook and read the question. Now, what are the forces acting on the mass?

10. Oct 11, 2008

### Benzoate

Fnet= Fspring-Fgrav

You are right. drag forces would be present only when a damping constant is present, well at least for a simple harmonic.
x''+2Kx'+(x*(omega)^2)=X''

X=36 cos (pt)
X'= -36 p sin (pt)
X''= -36 p2 cos(pt)

Therefore

x''+2Kx'+(x*(omega)^2)=-36 p2 cos(pt)

11. Oct 11, 2008

### Hootenanny

Staff Emeritus
I still have no idea where you're pulling these equations from. You still have a damping here in there (x'), there is no damping force. As I said above, the best way to start this question is to simply write down the sum of the forces acting on the mass. In other words, write the following expression explicitly.
Fnet = ...

12. Oct 11, 2008

### Benzoate

x'' + 2Kx' + $$\Omega$$2=F(t) is the spring equation for driven harmonic motion. I am not pulling these equations out of thin air. How can there not be a damping present when the only case where a damping constant is not present in the equation for an oscillator is when the linear oscillator is a classical linear oscillator, where external conditions allow oscillator to bounce up and down forever without stopping. And the problem doesn't say that this external driving force will act upon the oscillator forever , so I don't think I can just assume no damping constant is present.

13. Oct 11, 2008

### Benzoate

Fexternal=Fspring -Fgrav ; what other forces could there possibly be?

14. Oct 11, 2008

### gabbagabbahey

There is also a driving force isn't there? ;0)

What are the expressions for each of these three forces,

F_spring=?
F_grav=?
F_driving=?

15. Oct 11, 2008

### Hootenanny

Staff Emeritus
If you assume that there is a damping force, then you can not find a numerical solution (the solution will be a function for the damping parameter). You are not told that the system is damped, nor are you given any information regarding the damping constant, nor can you work out the damping constant with the information provided.

16. Oct 11, 2008

### Benzoate

My external force is my driving force

F_grav=-mg
F_spring=-kx
F_external= 36*cos(pt)
36*cos(pt)= -mg - kx

17. Oct 11, 2008

### gabbagabbahey

Shouldn't F_net=F_grav+F_spring+F_driving? Also since F_spring is presumably the restoring force that the spring imparts onto the mass, shouldn't it be +kx to signify that it is pulling the mass upward (while gravity is pushing it downward)?

18. Oct 11, 2008

### Benzoate

For an damped undriven oscillator

x''+2*K*x' +omega^2*x=0

for a driven harmonic oscillator

x''+2*K*x'+omega^2*x=F_driven

and F_grav i thought was always pointing down so shouldnt F_grav have a negative sign?

19. Oct 11, 2008

### gabbagabbahey

Forget about those two equations for a second and continue on with F_net (the sum of all forces acting on m).

F_net=?

20. Oct 11, 2008

### Benzoate

Okay.

F_net= F_spring + F_driven-F_grav

Why should I be concerned with the net force?

21. Oct 11, 2008

### gabbagabbahey

Because from Newton's second law, the net force is also equal to ma=mx'', so you get a differential equation for x:

mx''=F_net=F_spring + F_driven+F_grav

(this is how the equations of motion for the oscillator in your book are derived!)

Substitute your expressions for F_spring,F_driven,F_grav into this; what do you get?

22. Oct 11, 2008

### Benzoate

F_spring= -kx

F_grav= -mg

F_driven = 36 cos(pt)

mx''= 36 cos(pt)-mg-kx

23. Oct 11, 2008

### gabbagabbahey

Close, F_spring should be positive since it is pulling the mass upwards. So,

mx''= 36 cos(pt)-mg+kx

=> mx''-kx=36cos(pt)-mg

Which is an inhomogeneous ODE. Have you tackled one of these before?

Last edited: Oct 11, 2008
24. Oct 11, 2008

### Benzoate

I think so. I would treat this as an auxillary equation:

mx''-kx= 36 cos(pt)-mg ==>

What should I assume about pt
==> x''-(k/m)*x = 36/m*cos(pt) - g

Should I let x= ceipt?

25. Oct 11, 2008

### Benzoate

should I also take the amplitude into account:

a= F0/(($$\Omega$$2-p2)2-4K2p2)1/2

r^2-1000= 18 cos pt -10

let F= 18e^(ipt)

x'' -1000x = 18 e^(ipt)
x= ce^(ipt)
x'= cip*e^(ipt)
x''= -cp^2*e^(ipt)

e^(ipt) cancel and I am left with :

(cp^2*e^(ipt))-1000(ce^(ipt))= 18e^(ipt) -10

Normally e^(ipt) would cancel , but since I have an extra term where the e^(ipt) is not a coefficient therefore , I can't cancel out e^ipt easily

I don think I will have any qualms with obtaining the complementary solution since:

r^2-1000= 0

therefore

x=Ae^(-31.6277t) + B*t*e^(-31.62777t)

what role will the extension of the spring play in my general equation ?

Last edited: Oct 11, 2008