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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:

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

  • 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

  • Are there any holes in my interpretation?

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  • $\begingroup$ Nonlinearity is a hole in your assumption that a system is linear. Almost all physical systems have some sort of nonlinearity. Viewed spectrally we talk about 'nonlinear distortion' in a harmonic. A spreading of sort of the natural frequency. $\endgroup$ – docscience Mar 7 '18 at 18:29
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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 $\omega$.

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  • $\begingroup$ Much appreciated. A few things: (a) I've never seen this explanation before. Brilliant! But, why does the potential around a minima $x_0$ mean? If we expand around some minima $x_0$ it must imply we know the image of $U(x)$, even though this is a general case. Also, why does that delete the linear term? (b) If I understand you correctly, you're saying for $cx = csin(\omega \ t)$, so $c$ is the amplitude, and when differentiating it twice, it becomes $-\omega ^2 csin(\omega \ t )$ so $c$ is also a valid solution to the equation but it doesn't alter the resonant frequency? $\endgroup$ – sangstar Nov 18 '17 at 18:59
  • $\begingroup$ Two more things, what is the resonant frequency for this case in $k \equiv U''(x_0)(x-x_0)$? And lastly, $m\ddot x = -U'$ should only apply if $m\ddot x$ is conservative. How do we know it is? $\endgroup$ – sangstar Nov 18 '17 at 19:02
  • $\begingroup$ (a) If you know the forces, you can (in the case of conservative forces) find the potential. So I assume the potential is given. The minimum of this potential is obtained by requiring its derivative (which is the force) to vanish. Thus when you expand the potential around a minima the linear term vanishes. (b) If you meant 'so $cx$ is also a valid solution...' then that's exactly what I'm saying. $\endgroup$ – eranreches Nov 18 '17 at 19:05
  • $\begingroup$ As for your new comment, this isn't $k$. With the $k=U^{\prime\prime}(x_{0})$ I gave, the resonant frequency is $\omega=\sqrt{\frac{U^{\prime\prime}(x_{0})}{m}}$. Regarding the conservative nature of forces, this is correct in most mechanical systems without dissipation (I think). When dissipation is introduced, it causes the amplitude of the oscillations to decrease in time. $\endgroup$ – eranreches Nov 18 '17 at 19:07
  • $\begingroup$ (b) Yes I did, oops! (a) The minimum of this potential, if it is a function of $x$, would be the case where $U'(x) = 0$. Is that what you mean by "requiring its derivative (which is the force) to vanish"? If that's the case taking the derivative of that point would definitely cause the linear term to vanish. Why is it expanded at a minimum then? $\endgroup$ – sangstar Nov 18 '17 at 19:15

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