Electrical circuit differential equation

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SUMMARY

The discussion centers on the differential equation for an electrical circuit, specifically q'' + 20q = 100 sin(ωt). Resonance occurs at ω = ±√20, confirming that both positive and negative angular frequencies are valid for resonance in this system. The general solution is expressed as q(t) = Acos(√20 t) + Bsin(√20 t) + t(Acos(√20 t) + Bsin(√20 t)), where A and B are arbitrary constants determined by initial conditions. The significance of negative frequencies lies in their equivalence to positive frequencies in terms of oscillation behavior.

PREREQUISITES
  • Understanding of differential equations, particularly second-order linear equations.
  • Familiarity with resonance concepts in electrical circuits.
  • Knowledge of trigonometric functions and their properties.
  • Basic skills in solving for constants in general solutions of differential equations.
NEXT STEPS
  • Study the implications of resonance in electrical circuits and its practical applications.
  • Learn about the characteristics of second-order linear differential equations.
  • Explore the role of initial conditions in determining constants A and B in solutions.
  • Investigate the effects of varying angular frequencies on circuit behavior.
USEFUL FOR

Electrical engineering students, circuit designers, and anyone interested in the mathematical modeling of electrical systems and resonance phenomena.

Polygon
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q''+ 20 q = 100 sin(ωt)

I have been asked to find all mathematically possible values of ω for which resonance will occur. From the homogeneous solution, q(t) = Acos(√20 t) +Bsin(√20 t), I can see that resonance occurs when ω=√20. My question is, should I also consider -√20? And if so, what is the significance of a negative angular frequency?

Thanks.
 
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You sure you have the full solution ?
What determines A and B ?
Is ##C\cos(-\sqrt{20} \, t) + D\sin(-\sqrt{20} \,t )\ ## much different from ##A\cos(\sqrt{20} \, t)+ B\sin(\sqrt{20} \,t ) \ ## ?

Likewise: what's the difference between ##100 \sin(\omega t)## and ##100 \sin(-\omega t)## ?
 
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BvU said:
You sure you have the full solution ?
What determines A and B ?
Is ##C\cos(-\sqrt{20} \, t) + D\sin(-\sqrt{20} \,t )\ ## much different from ##A\cos(\sqrt{20} \, t)+ B\sin(\sqrt{20} \,t ) \ ## ?

Likewise: what's the difference between ##100 \sin(\omega t)## and ##100 \sin(-\omega t)## ?

Thank you for your reply BvU. The only difference is that the sine term will have a negative in front, which doesn't really make a difference since A, B, C and D are arbitrary constants. So my particular solution in both cases (ω=+√20 and ω=-√20 ) will be of the form q(t) = t(acos(√20 t) ± bsin(√20 t)), which means the charge will oscillate with increasing amplitude and without bound as t increases. So these are the two (and only two) ω values for which resonance occurs in this system?
 
Polygon said:
$$q(t) = t(A \cos(√20 t) ± B\sin(√20 t))$$
(If that is what you mean) Doesn't look correct.
General solution = one particular solution + solution of homogeneous equation ,
the first without integration constants, the second with
is what I remember.
 
Sorry, that is not what I meant. I should not have written it like that. As you said, the general solution contains a particular solution and the homogeneous solution. So the general solution will be of the form q(t) = Acos(√20 t) +Bsin(√20 t) + t(acos(√20 t) + bsin(√20 t)).

So the values of ω that will give resonance are both +√20 and -√20, right? I just need some confirmation as I haven't been given a problem like this before and I have little knowledge of electrical circuits.
 
Polygon said:
q(t) = Acos(√20 t) +Bsin(√20 t) + t(acos(√20 t) + bsin(√20 t)).
Does that satisfy the differential equation ? I don't think it does...
 
BvU said:
Does that satisfy the differential equation ? I don't think it does...

It does after solving for the constants. Thanks for your time BvU. I think I've got it now :)
 
Also for ##\omega\ne \omega_0## ?

And: you're welcome.
 

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