Solving for Logarithmic Decrement in LCR Circuits

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Homework Statement



Basically, I have LQ''(t) + RQ'(t) + (1/C)Q(t)=0, and I'm supposed to

"Show that the ration of the charge Q between two successive maxima is given by exp(RTd/2L), where Td is the time between two successive maxima. The natural logarithm of this ration is called the logarithmic decrement.


Homework Equations



Dunno

The Attempt at a Solution



So I got a solution Q(t)=e(-Rt)/(2L) [ C1cos( (√(R2-4L/C) )/(2L)t) + C2sin( (√(R2-4L/C) )/(2L)t).

But I can't figure out how to find Td. I mean, I could always find t when dQ/dt=0; but then I'd have to plug two values of t back into Q(t) and find the difference, and ... So what's the right way to do this?
 
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First notice the sin & cos terms have the same argument & the choice of c1 & c2 will just choose an overall phase. So for this argument set c2 = 0.

Then the maxima will just be where cos is maximum and successive maxima will occur where the argument of cos has changed by 2pi
 
lanedance said:
First notice the sin & cos terms have the same argument & the choice of c1 & c2 will just choose an overall phase. So for this argument set c2 = 0.

Then the maxima will just be where cos is maximum and successive maxima will occur where the argument of cos has changed by 2pi

Not exactly. The max's don't agree with the max's of the cosine, but the right idea. To the OP, just look at e-btcos(at+c).
 
good pickup thanks - They will be pretty close when the natural frequency is much larger that the decay constant, but you do need to take the exponential into account
 
Last edited:
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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