Harmonic Motion Problem - Finding oscillation of charges in a circuit

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  • #1
JoeyBob
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Homework Statement:
See attached
Relevant Equations:
frequency=1/T T=2pi/w
So since V(cap) + V(ind)=0 then Q/C + L dI/dt=0

Now since I=dQ/dt, I can replace dI/dt with d^2Q/dt^2 resulting in Q/C + L d^2Q/dt^2 =0

Now L d^2Q/dt^2 looks like a harmonic motion thing I can solve, where w^2=L. This means I can find w. I get 0.0005385.

Now my issue is using this w gives the wrong frequency (the frequency is supposed to be 5.22 in MegaHertz). I obviously didnt use the first part of the equation with Q/C in any manner. I don't see how I can use Q/C in the harmonic motion equation with L d^2Q/dt^2 to find a different w.
 

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  • #2
kuruman
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Homework Statement: See attached
Attached where? Anyway, it looks like you dropped C from the equation. You should have $$\frac{d^2Q}{dt^2}+\frac{1}{LC}Q=0.$$What is the frequency?
 
  • #3
JoeyBob
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Attached where? Anyway, it looks like you dropped C from the equation. You should have $$\frac{d^2Q}{dt^2}+\frac{1}{LC}Q=0.$$What is the frequency?
My bad, fixed now. So frequency would be how often an oscillation occurs and period would be how long it takes an oscillation to occur in this context. Frequency can also be found using the equation 2(pi)/w, where w is the angular frequency.

In the case of the equation, I believe it ends up as $$\frac{Ld^2Q}{dt^2}+\frac{1}{C}Q=0.$$

$$\frac{Ld^2Q}{dt^2}$$ appears to be in a form where you can find the angular frequency, where w^2=L, but this isn't correct, for it leads to the wrong answer.

$$\frac{1}{C}Q$$ doesn't appear to be in a form where angular frequency can be determined.
 
  • #4
haruspex
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My bad, fixed now. So frequency would be how often an oscillation occurs and period would be how long it takes an oscillation to occur in this context. Frequency can also be found using the equation 2(pi)/w, where w is the angular frequency.

In the case of the equation, I believe it ends up as $$\frac{Ld^2Q}{dt^2}+\frac{1}{C}Q=0.$$

$$\frac{Ld^2Q}{dt^2}$$ appears to be in a form where you can find the angular frequency, where w^2=L, but this isn't correct, for it leads to the wrong answer.

$$\frac{1}{C}Q$$ doesn't appear to be in a form where angular frequency can be determined.
You are still ignoring the role of C. Look at @kuruman's form of the equation in post #2. L and C occur together as a product, and nowhere else, so you can treat that product as a single entity.
 
  • #5
JoeyBob
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You are still ignoring the role of C. Look at @kuruman's form of the equation in post #2. L and C occur together as a product, and nowhere else, so you can treat that product as a single entity.
Sorry, I don't understand how the L transfers over to the Q/C part of the equation. To me it appears as if L is part of V(ind) and not V(cap), I don't understand how you get it to be part of V(cap).

I think I understand how his equation can be used to find the frequency though.

$$\frac{d^2Q}{dt^2}+\frac{1}{LC}Q=0.$$ becomes $$\frac{d^2Q}{dt^2}=-\frac{1}{LC}Q.$$

Now w^2= Q/LC, and frequency can be found.
 
  • #6
haruspex
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Now w^2= Q/LC
No, Q is the dependent variable. The constants are L and C, or just "LC".
 
  • #7
JoeyBob
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No, Q is the dependent variable. The constants are L and C, or just "LC".
Yeah, my mistake, I knew that.
 
  • #8
DaveE
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You have the (a) correct version of the differential equation. But it seems like you are guessing at the solution.

If we just look at this mathematically, which function, when you take it's derivative twice is a negative constant times itself?

If you want a more physical interpretation, then we can look at the circuit elements.

Since the inductor has v=L⋅(di/dt), then for a given voltage (equal to the capacitor voltage), if L is large then di/dt must be small. That means that the current doesn't change as quickly as for a smaller L. That in turn determines the rate at which the capacitor is discharged. The inductor current is what causes the capacitor to discharge. So, the value of L matters, it must appear in your solution.

Likewise, since the capacitor has Q = ∫i⋅dt = Cv, or i = C⋅(dv/dt), then for a given current (the same as the inductor current), if C is large dv/dt must be small. That means that the capacitor voltage doesn't change as quickly as for a smaller C. That in turn determines the rate at which the inductor current changes. The capacitor voltage is what causes the inductor current to change. So, the value of C matters, it must appear in your solution.
 
  • #11
Delta2
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The equation of simple harmonic motion is $$\frac{d^2x}{dt^2}+\omega^2x=0$$. Compare it with the equation of the circuit $$\frac{d^2Q}{dt^2}+\frac{1}{LC}Q=0$$. What can you say about the respective quantities. What is ##x## and what is ##\omega## in this last equation.
 
  • #12
DaveE
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Honestly, there was some stuff I had to learn in school that I really didn't use. This problem isn't one of those things. Regardless of which STEM field you end up studying or working in you will need to really understand this problem. It comes up frequently and is fundamental to understanding how the world works. From mechanical engineering, EE, through quantum mechanics, you will need to become really familiar with this problem.
 
  • #13
JoeyBob
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So you have the answer now?
I mean yes but I still don't understand how L ends up being part of V(cap) when it was originally part of V(ind).
 
  • #14
Delta2
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I mean yes but I still don't understand how L ends up being part of V(cap) when it was originally part of V(ind).
This is a good old algebra trick. Multiply both sides of the equation by ##\frac{1}{L}##. I mean we have the right to multiply BOTH sides of any equation with the same quantity.
 
  • #15
DaveE
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I mean yes but I still don't understand how L ends up being part of V(cap) when it was originally part of V(ind).
Everything (voltages, currents, component values) in this circuit effects everything else. This circuit is an exact analogy to a mass hanging from the ceiling on a spring. If you put energy into the system, like pulling down the mass and letting it go, or putting some charge on the capacitor and connecting the circuit, everything changes in response to the other changes in the system. This is what oscillation really is. Whether it's a kid swinging on a swing set, a bell ringing, a guitar string vibrating, or your heart beating. They are all similar to this circuit.

In this case L determines how quickly the current changes in the inductor, the current in the inductor changes the capacitor voltage, which is also the inductor voltage.
 
  • #16
kuruman
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To @JoeyBob
Here is something that I have already posted in another thread. Simple harmonic motion is described by the generic form$$\frac{d^2\text{(something)}}{d\text{(something else)}^2}+\text{(constant)}*\text{(something)}=0.$$In this equation
something = the dependent variable
something else = the independent variable
constant = a constant characteristic of the physical situation that has units (something else)-2

The idea is that, once you bring your differential equation to this basic form (note that the second derivative has nothing other than 1 multiplying it), then the constant is a measure of how often the dependent variable oscillates as the independent variable is increased.

Examples
Spring-mass system (mechanics)
something = ##x## (position), something else = ##t## (time), constant = ##\dfrac{k}{m}.##
LC-circuit (E&M, this question)
something = ##Q## (charge on capacitor), something else = ##t## (time), constant = ##\dfrac{1}{LC}.##
Particle in a box (quantum mechanics)
something = ##\psi## (wavefunction), something else = ##x## (position), constant = ##{\dfrac{2mE}{\hbar^2}}.##

And the list goes on. Specifically, when the independent variable is time, the constant is the square of the oscillation frequency. Thus, when you are asked to find the "frequency of oscillations" in any situation involving harmonic motion, all you need to do is the algebra to bring the differential equation into the standard form and then write down the answer as the square root of "constant".
 
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  • #17
haruspex
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I mean yes but I still don't understand how L ends up being part of V(cap) when it was originally part of V(ind).
It's not that. It's that the physical behaviour depends on the relationship between the induction and capacitance, specifically, on the value of LC.
 
  • #18
JoeyBob
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This is a good old algebra trick. Multiply both sides of the equation by ##\frac{1}{L}##. I mean we have the right to multiply BOTH sides of any equation with the same quantity.
To @JoeyBob
Here is something that I have already posted in another thread. Simple harmonic motion is described by the generic form$$\frac{d^2\text{(something)}}{d\text{(something else)}^2}+\text{(constant)}*\text{(something)}=0.$$In this equation
something = the dependent variable
something else = the independent variable
constant = a constant characteristic of the physical situation that has units (something else)-2

The idea is that, once you bring your differential equation to this basic form (note that the second derivative has nothing other than 1 multiplying it), then the constant is a measure of how often the dependent variable oscillates as the independent variable is increased.

Examples
Spring-mass system (mechanics)
something = ##x## (position), something else = ##t## (time), constant = ##\dfrac{k}{m}.##
LC-circuit (E&M, this question)
something = ##Q## (charge on capacitor), something else = ##t## (time), constant = ##\dfrac{1}{LC}.##
Paticle in a box (quantum mechanics)
something = ##\psi## (wavefunction), something else = ##x## (position), constant = ##{\dfrac{2mE}{\hbar^2}}.##

And the list goes on. Specifically, when the independent variable is time, the constant is the square of the oscillation frequency. Thus, when you are asked to find the "frequency of oscillations" in any situation involving harmonic motion, all you need to do is the algebra to bring the differential equation into the standard form and then write down the answer as the square root of "constant".

Thanks guys.
 

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