Electrical circuit differential equation

In summary, ω=+√20 and -√20 are the only values of ω for which resonance occurs in this system.
  • #1
Polygon
4
0
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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  • #2
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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Likes Polygon
  • #3
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?
 
  • #4
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.
 
  • #5
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.
 
  • #6
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...
 
  • #7
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 :)
 
  • #8
Also for ##\omega\ne \omega_0## ?

And: you're welcome.
 

Related to Electrical circuit differential equation

What is an electrical circuit differential equation?

An electrical circuit differential equation is a mathematical representation of the behavior of an electrical circuit. It is used to describe the relationship between the voltage, current, and resistance in a circuit as well as how these variables change over time.

How is a differential equation applied in electrical circuits?

In electrical circuits, differential equations are used to analyze and predict the behavior of the circuit. They can be solved to determine the voltage, current, or resistance at a specific point in time, or to understand how changes in one variable affect the others.

What is the order of a circuit differential equation?

The order of a circuit differential equation is determined by the highest order derivative present in the equation. For example, a first-order circuit differential equation would have a first derivative, while a second-order equation would have a second derivative.

What are the initial conditions in an electrical circuit differential equation?

The initial conditions in an electrical circuit differential equation refer to the values of the variables at the beginning of the circuit's operation. These conditions are typically the starting voltage, current, and/or resistance, and are used to solve for the behavior of the circuit over time.

How are circuit differential equations solved?

Circuit differential equations can be solved using various methods, such as analytical solutions, numerical methods, or circuit simulation software. The method chosen depends on the complexity of the circuit and the desired level of accuracy.

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