Question

LC oscillations analogy doubt!

Lc oscillations have excellent analogy correlation with Spring and mass.. So i like the following analogies

L = m
C = k
i = velocity
V (potential) = x
and everything is fine.. but i just don't get one thing

resonant frquency of spring mass = root(k/m) .. more k more frequency

but for LC its root (1/LC) .. more C less frequency.. that is something i am not able to digest.. any conceptual help?

Answer 1

@Vincent-Lyon.Fr

Answer 2

@ganeshie8

Answer 3

@Michio_kaku

Answer 4

Would you please explain what are you not getting? It does not seem rocket-science , what have you written

Answer 5

I think that you are assuming that the spring constant k is the analog of a capacitor. However this is not the case. The inverse spring constant is the mechanical analog of a capacitor.
The mechanical analog of a capacitor is the compliance of a spring, which is the quantity 1/k. Note that k is called the 'spring constant' or 'spring stiffness'.
The oscillation frequency increases for a stiffer spring (meaning a higher value of k) because the compliance of the spring reduces. If the stiffness of the spring is decreased, giving a lower value of k, the compliance of the spring increases (corresponding to an increase in capacitance) and the oscillation frequency will decrease.

Answer 6

Kropot is right,

F = kx is the equivalent of U = \(\frac 1C\) q

So electrical equivalent of stiffness k is 1/C

Answer 7

Why can't i think of it this way @Vincent-Lyon.Fr @kropot72

More the value of k, more is the work needed to stretch a spring to value x, and more energy U = 0.5 kx^2

Similarly more the value of C , more is the work needed to charge a capacitor to a particular value of v and more energy U = 0.5 CV^2 :D

so that gives me C being perfectly equivalent to k .. :D. (and not 1/k)

why is this logic wrong :(

Answer 8

Oh.. wait.. displacement x --> charge Q and not potential V ??

cause then i get U = 0.5 q^2/C

why did i inherently think of x as V ??

it makes sense .. conceptually, that more k .. more is the energy stored to stretch a spring..

but i don't get conceptually why more the C, Lesser the energy required to charge a capacitor to charge Q ?
oh.. is it like, if C is larger.. ITS EASIER to charge it up.. cause potential doesn't build up that easily? .. ahhh that seems to give me some intuition.. ! interesting!

Answer 9

so.. if i have a spring with INFINITE spring constant.. I need infinite energy to stretch, cause the spring wouldn't stretch at all

and if i have infinite capacitance, then i need zero energy to charge that baby up, cause potential will always be zero.. (somewhat like earth :D )


On the other hand.. zero k -- spring is veryyy springyy.. and can easily stretch with no work..
and if i have zero C, then its impossible to charge it up, thats the very meaning of ZERO CAPACITY to hold charges.. so inifinite energy would be required to charge it up..

OK OK.. this makes PERFECT sense now!!!.. thanks to ya guys!! :) :) :) :)

Answer 10

It is because energy should be written \(U_e=\dfrac 12 \dfrac{Q^2}{C}\).

Answer 11

yes yes.. i get it now :) :)!!!!