# Pendulum with two springs and a stick

1. Apr 10, 2012

### dat240

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

A pendulum consists of a stick of length L and mass M that is pivoted about its center, and attached to a pair of springs of stiffness k. As shown in attached pic, the springs are symmetric about the pivot, and the distance between them is l.

a) Find the natural angular frequency ω of this pendulum's small oscillation

2. Relevant equations

I think, T = 2π(m/k)^(1/2) and ω = 2π/T but am not sure.

3. The attempt at a solution

I can see how stored energy is being passed between the springs but am not sure how to setup the problem. Do I need to use COM/torque since mass is distributed throughout the stick? Thanks for your input.

2. Apr 10, 2012

### dat240

Not sure if my attachment worked... going to try to post it again here.

#### Attached Files:

• ###### physics2.JPG
File size:
82 KB
Views:
119
3. Apr 10, 2012

### darkxponent

where is the attempt?

4. Apr 10, 2012

### dat240

I feel like I'm just clutching at equations, but I'll try.

I saw that the physical pendulum equation was T = 2π(I/mgh)^(1/2).

For a thin rod about axis through center perpendicular to length, I = (1/12)ML^2

Since I'm not dealing with gravitational potential, I wanted to substitute the spring energy equation (1/2)kx^2 for mgh.

So, T = 2π(ML^2/6kx^2)^(1/2)

But now that I've written it out, I see that I've added the variable x, which isn't a given. My problem is that I have spring oscillation equations and pendulum equations, but I don't understand how to combine them.

5. Apr 10, 2012

### darkxponent

Seems to me you have lack of practise. Have you studied angular SHM?

6. Apr 10, 2012

### dat240

No. I'm an intro physics student, and we just got to SHM yesterday. I've reviewed my book and notes but am not yet at a level where I understand much about it.

I'm not looking for free answers. I'm looking for help understanding the concepts so I can apply them in this and other problems. I don't know what else to say. If I knew much about angular SHM, I wouldn't be asking for help.

7. Apr 10, 2012

### darkxponent

Most phsics questions are not just formula based. You have to solve it.

Simple harmonc motion is that motion where accelaration is proportional to 'negative of displacement'. When we take angular displacement the anglular acceleration is taken.

Now the best way to solve a SHM question is to displace the body by a distance small 'x' or small 'theeta'(in angular SHM) and then show the forces on the body and hence find its accelation. You will get accelaration as a function of 'x' or 'theeta'(for angular SHM). And if you get accelaration as negative of displacement then only it is SHM.
And if you get so then just use the eqn

-(omega)^2*x=a

In case of angular SHM the eqn becomes

-(omega)^2*(theeta)=(alpha)
where alpha is angular acceleration

And remember the only diffrence betweeen linear and angular SHM is

1. 'theeta' for 'x'
2. torque for Force
3. 'alpha' for 'a'
4. moment of inertia(I) in place of mass(M)

8. Apr 10, 2012

### darkxponent

Most phsics questions are not just formula based. You have to solve it.

Simple harmonc motion is that motion where accelaration is proportional to 'negative of displacement'. When we take angular displacement the anglular acceleration is taken.

Now the best way to solve a SHM question is to displace the body by a distance small 'x' or small 'theeta'(in angular SHM) and then show the forces on the body and hence find its accelation. You will get accelaration as a function of 'x' or 'theeta'(for angular SHM). And if you get accelaration as negative of displacement then only it is SHM.
And if you get so then just use the eqn

-(omega)^2*x=a

In case of angular SHM the eqn becomes

-(omega)^2*(theeta)=(alpha)
where alpha is angular acceleration

And remember the only diffrence betweeen linear and angular SHM is

1. 'theeta' for 'x'
2. torque for Force
3. 'alpha' for 'a'
4. moment of inertia(I) in place of mass(M)

9. Apr 10, 2012

### dat240

Thank you. I'm away from my notebook right now, but I will think about what you said and give it another (hopefully better) shot when I get home.

10. Apr 10, 2012

### dat240

Okay, so let me try again.

Tilting the rod a bit shows that the springs are always applying equivalent restorative forces. They are also applying equivalent torques since the forces are symmetrical about the point of rotation. The torques act in the same rotational direction.

So,

2τ = Iα

2kx(l/2) = Iα

kxl = Iα

From here, I wanted to plug in I = (1/12)ML^2 for the rod's moment of inertia, and α = (ω^2)∅ per your suggestion. While this does allow me a way to solve for ω, the x and ∅ variables don't cancel from what I can tell. And shouldn't ω be constant throughout?

11. Apr 11, 2012

### darkxponent

you are very near. Try to find a relation b/w 'x' and 'theeta'