# Determine the angular frequency of the system in SHM

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1. Apr 12, 2017

### richardz03

1. The problem statement, all variables and given/known data
Determine the angular frequency of the system in the image. The cable is ideal but the pulley is not. I will present the same solution but with different coordinate axes. For some reason they arent the same and neither of them are correct.

Given data: R is the radius of the pulley. K is the spring constant. m is the mass of the block. Icm is the moment of inertia of the pulley in form of a disk with respect to the center of mass of the pullet.

Variables: T is the tension, w is the angular frequency.

a and α will be translational acceleration and rotational acceleration.

2. Relevant equations

ΣF=ma
torque = (moment of inertia) * (angular acceleration)
Through the differential equation i can get the angular frequency w^2

3. The attempt at a solution

I have trouble with the signs. If I am not wrong, I should establish a system of coordinates and stay with it throughout the problem.
I will work on two different coordinate axes to show that I can't reach the solution and both are different. Something about my operation is wrong.

To get the angular frequency I decide to displace the block X meters lower than its position of equilibrium.

The magnitude of the spring force kx will be different in my view because if I put X axis pointing upward, then the displacement it would be positive and so |KX| should be KX. If I put X axis pointing downward then it would be |KX| = -KX

The correct answer is in the image and i cant seem to reach it. I hope i can get some help.

Last edited: Apr 12, 2017
2. Apr 12, 2017

### haruspex

Will the tension be the same both sides of the pulley?

3. Apr 12, 2017

### richardz03

My class consider the left side of the pulley has tension and the right side has instead of tension they just plug in the spring force of magnitude kx. Intuitively the spring force must be greater so there is a rotational acceleration that opposes to displacement.

4. Apr 12, 2017

### richardz03

Hi. Sorry to bother you. I updated the post to make it much more easier to understand. Hope you can help me.

5. Apr 13, 2017

### haruspex

In your first attempt, the problem is the T=ma substitution. You are measuring x, θ and α as positive for the mass descending. So a should be positive for the mass accelerating downwards. Which way does T act on the mass?

6. Apr 13, 2017

### richardz03

Hi. Dont coordinate axes determine their direction. The displacement X should be negative as I am moving the block down and let it do a SHM motion having the X axis pointing upward. In SHM at least in my class, they just dont matter alpha and acceleration signs. As long as it comes up like the solution

7. Apr 13, 2017

### Staff: Mentor

If you're having problems relating the directions of motions and forces and so on, consider laying out the problem linearly:

8. Apr 13, 2017

### richardz03

My problem is that in free body diagram i know exactly where they are pointing. Just that depending on axes, those forces will be negative or positive. Thats why my tension is ma in the first system and -ma the second system. What I am not sure is how i am applying second law of rotation, please tell me if my process in determining the signs of kx is correct. In the first system i put kx positive since the displacement of spring is positive. But in the second displacemet it would be -kx. This causes a change of sign and I am not sure why would it be wrong.

9. Apr 13, 2017

### haruspex

You have not actually stated a definition of x. In your first attempt you took the tension on the right as kx. That makes x the extension of the spring, i.e. an upwards movement onthe right is positive x. But an upwards movement on the right means a downwards movement for the mass, so for the mass x, and therefore a, are positive downwards: T=-ma.
Moral: write down clear definitions of all your variables.