Determine the angular frequency of the system in SHM

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The discussion focuses on determining the angular frequency of a system in simple harmonic motion (SHM) involving a pulley and a spring. Participants express confusion over the signs of forces and accelerations when establishing coordinate axes, which impacts their calculations. The tension in the system is debated, with questions about whether it remains consistent on both sides of the pulley. Clear definitions of variables and consistent application of the second law of motion are emphasized as crucial for solving the problem correctly. Ultimately, understanding the relationship between displacement, force direction, and coordinate systems is key to finding the correct angular frequency.
richardz03
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Homework Statement


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.

Capture.PNG

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.

Homework Equations



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

The Attempt at a Solution


[/B]
Capture2.PNG


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 can't seem to reach it. I hope i can get some help.
 
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Will the tension be the same both sides of the pulley?
 
haruspex said:
Will the tension be the same both sides of the pulley?
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.
 
haruspex said:
Will the tension be the same both sides of the pulley?
Hi. Sorry to bother you. I updated the post to make it much more easier to understand. Hope you can help me.
 
richardz03 said:
Hi. Sorry to bother you. I updated the post to make it much more easier to understand. Hope you can help me.
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?
 
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 don't matter alpha and acceleration signs. As long as it comes up like the solution
 
If you're having problems relating the directions of motions and forces and so on, consider laying out the problem linearly:

upload_2017-4-13_10-51-13.png
 
gneill said:
If you're having problems relating the directions of motions and forces and so on, consider laying out the problem linearly:

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.
 
richardz03 said:
The displacement X should be negative as I am moving the block down
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.
 

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