# AP Physics problem stumped me.

#### Brian Mcglinchey

Joined Feb 16, 2017
2

#### wayneh

Joined Sep 9, 2010
17,467
What's the frequency of an LC circuit?

#### RBR1317

Joined Nov 13, 2010
712
When an LC circuit is pulsed it will then ring (oscillate) at the resonance frequency determined by the LC values. Add a resistor and you will get a damped resonace as shown in your diagram. So see if you can determine the frequency of oscillation from the graph, then find what LC values will resonate at that frequency.

#### WBahn

Joined Mar 31, 2012
29,493
And keep in mind that the question only asks for the approximate inductance.

#### MrAl

Joined Jun 17, 2014
10,891
Hi,

I think we need to know if this is a parallel or series circuit first. A series circuit acts differently than a parallel circuit and so 1/sqrt(LC) does not always apply directly. I think we can assume a series circuit because they show the voltage across the resistor but it would be good to see the circuit. I think we also need to know the resistor value.

Since this is probably a series circuit, then 1/sqrt(LC) may not apply directly depending on the value of the resistor. Without giving too much away, the value of the resistor must not exceed a certain limit or else we cant get that waveshape, but even in the confines of that limit we can see a different apparent frequency in the ripples of the step response.

Again without giving too much away just yet, the actual apparent frequency is related to 1/sqrt(LC) by the factor:
sqrt(1-A^2)/sqrt(LC)

where A is a constant that varies from 0 to 1. If A is zero we get 1/sqrt(LC) but only then.

Readers may want to try this with a simulator with various L and C, making sure that R stays low enough to ensure we get that ripple type waveform with a step response. Compare the frequency measured graphically with the calculation of F=1/sqrt(LC). The higher the value of the resistor gets the farther the two frequencies will be away from each other, being careful not to go too high with R though which would void the whole experiment.

Of course this is all unless they allow the approximation 1/sqrt(LC) which may be the case too.

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#### WBahn

Joined Mar 31, 2012
29,493
The value of the resistor is not arbitrary -- it is fixed by the shape of the response and is determinable from the damping factor. Since this is a highly underdamped response, the resistance is low enough so that the undamped and damped natural frequencies are approximately the same and since the question only asks for an approximate value of the inductor....

#### MrAl

Joined Jun 17, 2014
10,891
The value of the resistor is not arbitrary -- it is fixed by the shape of the response and is determinable from the damping factor. Since this is a highly underdamped response, the resistance is low enough so that the undamped and damped natural frequencies are approximately the same and since the question only asks for an approximate value of the inductor....
Hi,

Since your post came right after mine but did not mention mine, i cant be sure if you are replying to my post or not. But to be clear, i was also stating the opposite: that the resistor value should be known too in order to get the response we see, and that although it is determinable also from a thorough analysis, it is not obvious from just sqrt(1/LC). That was the whole point I was not stating that the resistor could be anything, i was stating that it has to be less than some maximum but since we have a resistor it may alter the response from that of just 1/sqrt(LC), so we need to know the value. Yes we can compute the value i believe, but we can not compute it from sqrt(1/LC) without a little more.
Sometimes we can get away with sqrt(1/LC) though so i guess we could look into that by solving for the resistor and see if it makes a difference. Even though the resistor is limited in value, some values will alter the response even within that limit.
What i wasnt sure about was if they wanted to see a full analysis or just throw 1/LC at it anyway.

#### WBahn

Joined Mar 31, 2012
29,493
What more do we need? You want L, R, and C. You are given C, so you have two unknowns. You are given the response which shows two things -- the damped natural frequency and the damping factor. Two equations, two unknowns. It is not a matter of R having to be below some value in order to get the response that was given in the problem. -- R has to be a specific value, no more, no less, in order to the response that was given.

The question specifically only asks for an approximate value of L. The response is significantly underdamped -- the damping factor appears to be about 1/3 or so, so the damped natural frequency is about 95% of the natural frequency given by 1/sqrt(LC). So the actual value of L will be off by about 10%. Since the value of the capacitor is probably not known to that degree of certainty, this should qualify as a sufficient approximation and is almost certainly the intent of the person that wrote the problem.

#### wayneh

Joined Sep 9, 2010
17,467
Agreed. Which brings us back to #2. The TS has undoubtedly studied this and should be able to apply the formula, with data from the plot, to make an estimate. He may already 'know' everything he needs without realizing how to apply it.