I Thoughts about coupled harmonic oscillator system

AI Thread Summary
In a coupled harmonic oscillator system, the spring can still obey Hooke's law, but only within a linear strain region. When two oscillating modes are present, such as axial and transverse modes, they can behave independently unless a coupling mechanism is introduced. Hooke's law is an idealized approximation that holds true until the material is stretched beyond its elastic limit. Complex structures may exhibit regions where Hooke's law is applicable and others where it is not, depending on the stress applied. Understanding the relationship between local and global strains is crucial for analyzing the behavior of such systems.
phymath7
Messages
48
Reaction score
4
TL;DR Summary
In my physics lab,while mesuring the spring constant of a spiral spring,we were instructed to vibrate it in such a way that it only oscillates vertically.That means,suppose the spring vibrates along the 'y' axis,so it can't have any x or z component of oscillation.In simple term,it can't form a coupled harmonic osccilator system.
Same instruction was given while finding value of 'g' by a bar pendulum.
In the former case,does the spring obeys hooke's law while it forms a coupled harmonic oscillator system?Does the bar pendulum somehow breaks the simple harmonic motion(such that we can't apply the law for sumple harmonic motion)?
 
Physics news on Phys.org
phymath7 said:
does the spring obeys hooke's law while it forms a coupled harmonic oscillator system?
Sort of, yes.

So for clarity let's say you have two oscillating modes an axial mode and a single transverse mode (like p-waves and s-waves in earthquakes) in a simple bar of steel.

Hooke's law only applies in a linear region of strain (think of any very small region of your steel spring) where the material can stretch in one direction, let's say ##\hat{z}##, without a significant change in the spring constant either in that direction or in other orthogonal directions, like ##\hat{x}##. In this case the two oscillating modes can be viewed as uncoupled, they proceed to move in time independent of each other. If you hit it the right way you can excite either or both modes. But after that they will be essentially independent.

This is sort of by definition. Hooke's law applies when Hooke's law applies. If you stretch a spring too far, you can't use it anymore. How do you know what "too far" is? That's when Hooke's law doesn't work anymore. It's an idealized approximation which is simple and often true. But IRL you need to verify that it actually works that way.

But, you asked about coupled modes, where energy can move between modes. This requires some coupling mechanism which isn't defined, so I don't know what the effects are.

Complex structures may have some regions where Hooke's law works and others where it doesn't, it's best to think of this as an approximation of how a small region responds to stress and then figure out how that relates to the whole structure.

The structure may also create coupling between the modes. So, for example if you stretch a coil spring the entire structure will exhibit strain in tension (stretching) and torsion (twisting) because of the way it's constructed in relation to the stress direction you apply. But if you zoom into a small area inside the spring there will be only one strain direction which is a combination of the global tension and torsion directions. A solid metal bar won't couple global tension stress into a torsional strain because it is constructed differently. A tiny section of the coil spring can be viewed as a straight bar which is being pulled off-axis by the neighboring tiny sections it's attached to.
 
Thread 'Gauss' law seems to imply instantaneous electric field'
Imagine a charged sphere at the origin connected through an open switch to a vertical grounded wire. We wish to find an expression for the horizontal component of the electric field at a distance ##\mathbf{r}## from the sphere as it discharges. By using the Lorenz gauge condition: $$\nabla \cdot \mathbf{A} + \frac{1}{c^2}\frac{\partial \phi}{\partial t}=0\tag{1}$$ we find the following retarded solutions to the Maxwell equations If we assume that...
Hello! Let's say I have a cavity resonant at 10 GHz with a Q factor of 1000. Given the Lorentzian shape of the cavity, I can also drive the cavity at, say 100 MHz. Of course the response will be very very weak, but non-zero given that the Loretzian shape never really reaches zero. I am trying to understand how are the magnetic and electric field distributions of the field at 100 MHz relative to the ones at 10 GHz? In particular, if inside the cavity I have some structure, such as 2 plates...

Similar threads

Replies
131
Views
7K
Replies
8
Views
2K
Replies
17
Views
3K
Replies
11
Views
1K
Replies
18
Views
3K
Back
Top