spring - Better understanding natural resonance frequency and simple harmonic motion

Let me see if I'm getting this understood correctly. I'm trying to make sure my interpretation of simple harmonic motion is the right interpretation, including my take on resonant frequency.

Okay, so if something is going to oscillate simple harmonically, it needs the following conditions:

• It needs to be elastic - by that I mean an object is in some position where if it were to move in some given direction $+x$ let's say, the object's orientation would be such that there is a linear force opposing it trying to push it back to its original configuration. A spring has this property for example. When its compressed, it wants to decompress. When it's taught, it wants to relax back.


• Its net force on it must always oppose the direction of motion, equal to $F = -kx$ where $k$ is some constant. Thus, the net force on the object must vary with direction in time, but it must never go farther in one direction than the other, since we're considering normal simple harmonic motion, so its force will be highest at maximum displacement, but the values $F$ will take will always be the same in each direction. There's a symmetry to it.

Those two points probably imply eachother, but whatever.

If an object oscillates, this is implied:

$$F_{net} \propto -kx$$

$$\implies mx'' = -kx$$ $$\implies x'' = -\frac{k}{m} x \ \ \ \ \ \ \ \ \ \ \ \ \ (1)$$ This implies that the solution to this differential equation must satisfy the fact that taking its derivative twice returns the original function but with a negative.

The sine function satisfies this. Now, it needs to spit out $\sqrt\frac{k}{m}$ each time it's differentiated, so it needs to be a scalar on the argument of sine.

Define $\omega$ s.t. $\omega ^2 = \frac{k}{m}$

Then, our solution is of the form

$$sin(\omega t)$$

Questions I still have:

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  Why does everything have a natural resonant frequency, and not just things which can clearly oscillate like masses on springs?

  
  



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  This implies $\omega$ is a constant. So, for this spring, no matter what displacement I pull it at, it will oscillate at the same rate? This doesn't seem obvious, but seems like an implication of the system

  
  


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  Are there any holes in my interpretation?

  
  

Answer

Why does everything have a natural resonant frequency, and not just things which can clearly oscillate like masses on springs?

Because every structure that sits in equilibrium of some potential behaves, to the first order, like a SHO. This result is derived directly from the Taylor expansion of the potential

$$U(x)=U(x_{0})+\frac{1}{2}U^{\prime\prime}(x_0)\left(x-x_{0}\right)^{2}+\dots$$

Observe that the linear term vanishes because we expand the potential around a minima $x_{0}$. Thus Newton's law states that

$$m\ddot{x}=-\frac{{\rm d}U}{{\rm d}x}=-U^{\prime\prime}(x_0)\left(x-x_{0}\right)$$

so we can define an effective spring constant as $k\equiv U^{\prime\prime}(x_0)$.

This implies $\omega$ is a constant. So, for this spring, no matter what displacement I pull it at, it will oscillate at the same rate? This doesn't seem obvious, but seems like an implication of the system.

This statement is correct. In the case of a SHO, the frequency of oscillations is independent of the amplitude. This is a due to the linearity of the equation. We can write the equation of motion in the form

$$\ddot{x}=-\omega^{2}x$$

and as you can see, if $x$ is a solution, also $cx$ is a solution - with the same $\omega$.