Is there any simple explanation of why LC circuit period is equal to square root of LC?

Question

zyberg · Answer

@ganeshie8 maybe you would have any ideas?

osprey · Answer

I think that it's to do with the way the differential equation or the LC - or any other oscillatory circuit - is set up. in SHM, it's x double dot = - omega SQUARED x This leads to square roots in just about everything oscillatory - re the formula for a simple pendulum and it's periodic time, eg. http://perendis.webs.com

irishboy123 · Answer

that's the angular freq, not the period.....but ... really depends on what you mean by "why" you can just solve the DE: $\ddot I +\omega _{0}^{2}I =0$, and you're coming at it like you know all that. ....and so, then you can say that "that's that". but what is "why". https://www.youtube.com/watch?v=36GT2zI8lVA i can't see a physical analogue, even from a simple spring or pendulum system. i'm guessing that's what you're asking cos your question is so general. sorry can't be more helpful.

osprey · Answer

come to think of it, it's probably another "mathematical device", shall I say, to make things tidier/easier further down the chain. 
MAY be a bit like doing ict as the "fourth dimension" ... ct "converts" speed and time into "space-time" - distance, and the i (root of -1 ie imaginary) makes things look tidy in some invariant or other in spec rel.
And, after all, what is "i" but the square root of -1. Imaginary. Some people's imaginations ...

osprey · Answer

@IrishBoy123
I thought I saw a post about metre in poetry a day or so ago. It seems to have vanished both open and closed can't find it anywhere. Is this my imagination, or did I really see Iambic Pentameter ?????

radar · Answer

you could look at it like this:  Period (time) = 1/f 
frequency of  an LC circuit (resonance):$$f=1/(2\pi \sqrt{LC})$$You can now  write it for Period as:$$2\pi \sqrt{LC}$$
Why you may ask?  Check out the link.

osprey · Answer

@radar Why you may ask? .... looks like the answer is to make the maths elegant. After all, doesn't maths/physics delight in straight line graphs and any graph that can be "curve fitted" ? So, the practical people do the "knob twiddling", hand the results to the analysts who do the graphs and get the knob twiddles to match the maths and get the publications ?

irishboy123 · Answer

Trying out an anology ..... why is the period of a mass-spring system given by $2 \pi \sqrt {\dfrac{m}{k}} $ ??

We know from energy conservation that:

 $\frac{1}{2} k x^2 + \frac{1}{2} m \dot x ^2 = const \qquad \circ$.  


Why is energy conserved? Well, for present purposes, it just is.

Why is $U_s = \frac{1}{2} k x^2$? We have Hooke's Law: $\frac{dF}{dx} = k$, and we suppose that the intermolecular forces are inverse square but they linearise over short distances giving this linear relationship.  We then have all of Newton's ideas/definitions, especially that $W \equiv E \equiv F \cdot x$, and so we get that $\frac{1}{2} k x^2$ idea pretty easily.  Same for $T  = \frac{1}{2}m\dot x ^2$

But if we then plot the phase portrait $x$ vs $\dot x$ we have an ellipse ...... and that seems to be the key observation

Specifically, from $\circ$:

 $\left( \dfrac{x}{\sqrt{m}} \right)^2 + \left( \dfrac{\dot x}{\sqrt{k}} \right)^2  = const $, clearly an ellipse
 
 Or 
  $\left( \dfrac{x}{\gamma \sqrt{m}} \right)^2 + \left( \dfrac{\dot x}{\gamma \sqrt{k}} \right)^2  = 1 \qquad \triangle$
  
  .....where $\gamma$ is the interim fudge factor...
  
 We can parameterise any ellipse as $x = a \cos \omega t, y = b \sin \omega t $, ie:
 
  $\left( \dfrac{x}{a} \right)^2 + \left( \dfrac{y}{b} \right)^2  = 1 \qquad \star$
 
 and we also know from calculus that in our phase portrait with our particular ellipse that $y = \dot x$ and $y = \dot x = - a \omega \sin \omega t$, where $\omega$ can be any angular frequency
 
But we can pattern match $\star$ to:
 
 
   $\left( \dfrac{a \cos \omega t}{a} \right)^2 + \left( \dfrac{- a \omega \sin \omega t}{b} \right)^2  = 1 \qquad $
   
   ie $\omega =  \dfrac{b}{a}$
   
Then from $\triangle$ we have only one conclusion:

$\omega =  \dfrac{\gamma \sqrt{k}}{\gamma \sqrt{m}}    = \sqrt{ \dfrac{k}{m}  }$
 
So the answer in the case of a spring-mass system seems to be that the law of conservation of energy, Hooke's and Newton's Laws,  calculus and the geometry of an ellipse collectively demand that $ T = 2 \pi \sqrt{ \dfrac{m}{k}  }$


So, after all that faffing about, the smart answer I think is that the DE says it has to be so.  As it does, mutatis mutandis, in the case of  LC.

radar · Answer

@osprey Yes, not only in order to make the math elegant, but the speaker also. lol

osprey · Answer

@IrishBoy123
mutatis mutandis ... that looks suspiciously like Latin ? .....................................

irishboy123 · Answer

@osprey 

good spot !!!!


you do the French, i'll do the Latin ?!?!

osprey · Answer

@IrishBoy123 What does that latin mean ?

irishboy123 · Answer

literally, and Latin is beautifully concise, it means "all that needs to have been changed having been changed". 

so in an equation, you will have moved subscripts around.  

in plain English, i think the nearest idiom is "all things being equal".  but i don't really get that expression.