Small oscillation equation derivation

In summary: Fourth, the equation is not valid for arbitrary shapes or orientations of the bob.All of these problems make the derivation of the shear modulus very unreliable.
  • #1
unscientific
1,734
13
Hi guys, I have been trying to find the "floppy" resonant mode frequency of a simple oscillator. The displacement is in the order of nanometers, while the dimensions of the oscillator is in cm. I think small angle approximations apply here. I got to the point of the equation of motion, but I have no idea how to solve it.The equation for the shear modulus is defined. I used it to find the force.

I related force to torque.

I used maclaurin's expansion on cos θ and tan θ to approximate the equation in terms of θ.

oscillator1.png


Would appreciate any help.
 
Last edited:
Physics news on Phys.org
  • #2
hi unscientific! :smile:

you have θ'' = -C/θ

the standard trick to get a first integral is to multiply both sides by θ' to get …

θ'θ'' = -Cθ'/θ …

which integrates, and square-roots, to θ' = √(B - 2C*lnθ) :wink:

(how you get past that, I've no idea :redface:)
 
  • Like
Likes 1 person
  • #3
Thanks tim, I've tried wolframalpha but it gave a non-analytical solution..
 
  • #4
Expand C/θ in Taylor series around its equilibrium position and consider (by decree) that the higher order terms are negligible (That's what is usually meant by small oscillations)
 
  • #5
No, wait, you can't do that. There seems to be something wrong with your equation. Expand cosθ around its equilibrium position instead.
 
  • #6
Yes, there's definitely something wrong with your equation. The torque should be proportional to θ, not its inverse. Double check
 
  • #7
dauto said:
Yes, there's definitely something wrong with your equation. The torque should be proportional to θ, not its inverse. Double check

I can't see what's wrong with my equation..
 
  • #8
You should have sinθ in your equation instead of cosθ. The angle θ in your picture is not the angle between F and r.
 
  • #9
dauto said:
You should have sinθ in your equation instead of cosθ. The angle θ in your picture is not the angle between F and r.

the torque is r x F, which gives cosθ instead.
 
  • #10
Can you do a better (clearer) drawing? And what is the relationship between the two figures?
 
  • #11
nasu said:
Can you do a better (clearer) drawing? And what is the relationship between the two figures?

The figure on the right represents the part of the rod that's connected to the bob.
 
  • #12
So that' part of a rod? I am sorry, I cannot imagine your system from these drawings.
Maybe someone else.

Is there a real system you are trying to model?
 
  • #13
nasu said:
So that' part of a rod? I am sorry, I cannot imagine your system from these drawings.
Maybe someone else.

Is there a real system you are trying to model?

A force is applied on the disc, which is attached to the rod. As a result, the rod flexes.
 
  • #14
So what represent the right side of your drawing? The disk or the rod?
 
  • #15
nasu said:
So what represent the right side of your drawing? The disk or the rod?

the right side represents the flexing of the rod.
 
  • #16
I am sorry. It's even more confusing for me. It looks like a sort of shear deformation to me.
 
  • #17
nasu said:
I am sorry. It's even more confusing for me. It looks like a sort of shear deformation to me.

Ok. The picture on the right is simply the part of the rod that is connected to the bob at the top. When the bob rotates, the rod flexes.
 
  • #18
I cannot see any picture at all.
 
  • #19
voko said:
I cannot see any picture at all.

It's in the first post.
 
  • #20
unscientific said:
It's in the first post.

It does not work for me. Can you attach it here or use some other image hosting facility?
 
  • #21
voko said:
It does not work for me. Can you attach it here or use some other image hosting facility?


Does this work?

28rmvc6.png
 
  • #22
There are multiple problems with the derivation.

First, it is not correct to assume that the force is applied at some point and causes the rod to flex. Instead, the motion of the bob stresses the rod, and the restoring force (or torque) is generated. This torque is then applied to the bob, resulting in its oscillation.

Second, why are you dealing with the shear stress when the modes of vibrations you are after are flexural? My memory on the beam theory is rusty, but if I remember correctly, shear becomes important only with thick beams?

Third, can you really assume that all the flexing will be at the joint? It is much more reasonable to expect that the entire rod will be bending.
 

FAQ: Small oscillation equation derivation

1. What is a small oscillation equation?

A small oscillation equation is a mathematical model that describes the motion of a physical system when it is undergoing small, periodic oscillations around a stable equilibrium point. It takes into account the forces acting on the system and the system's natural frequency.

2. How is a small oscillation equation derived?

A small oscillation equation is typically derived by applying Newton's second law of motion and Hooke's law to the system. This involves setting up a system of differential equations and solving for the system's displacement over time using mathematical techniques such as the method of undetermined coefficients or Laplace transforms.

3. What are the assumptions made in deriving a small oscillation equation?

The main assumptions made in deriving a small oscillation equation include: the system is in stable equilibrium, the oscillations are small and remain within a linear regime, and there is no damping or external forces acting on the system.

4. How is a small oscillation equation used in science and engineering?

Small oscillation equations are used in various fields of science and engineering, including physics, mechanics, and electrical engineering. They provide a simplified model for understanding and predicting the behavior of physical systems, such as the motion of a pendulum or a mass-spring system.

5. Are there any limitations to the small oscillation equation?

Yes, there are some limitations to the small oscillation equation. It is only accurate for systems undergoing small, linear oscillations. It does not take into account damping or external forces, which may significantly affect the system's behavior in real-world situations. Additionally, it may not be applicable for systems with large displacements or when the system's equilibrium point is not stable.

Similar threads

Back
Top