Electrical analog and impedance

[Shackleford]

I'm a bit lost on this one. Also, what formula do I use for impedance?

http://i111.photobucket.com/albums/n149/camarolt4z28/2010-09-19202315.jpg?t=1284947003

http://i111.photobucket.com/albums/n149/camarolt4z28/2010-09-19202817.jpg?t=1284947014

 

[hikaru1221]

What are the differential equations for m1 and m2? Notice that now the damping force on m1 is $$-b_1(\dot{x}_1-\dot{x}_2)$$ (that is, it depends on the RELATIVE velocity between the masses).
After write down the 2 equations, relate it to an electrical circuit by using electromechanics analogy:
q ~ x
i ~ v (or dx/dt)
k ~ 1/C
L ~ m
b ~ R
Fcos(wt) ~ Vcos(wt)

The impedance is calculated from the circuit built

 

[Shackleford]

What are the differential equations for m1 and m2? Notice that now the damping force on m1 is $$-b_1(\dot{x}_1-\dot{x}_2)$$ (that is, it depends on the RELATIVE velocity between the masses).
After write down the 2 equations, relate it to an electrical circuit by using electromechanics analogy:
q ~ x
i ~ v (or dx/dt)
k ~ 1/C
L ~ m
b ~ R
Fcos(wt) ~ Vcos(wt)

The impedance is calculated from the circuit built

That makes sense. The problem doesn't imply they move with the same velocity. That's why I was adding the masses.

Actually, the damping force on m2 is $$-b_2(\dot{x}_2)$$. I'll write out the DEs tomorrow. I'm tired. Thanks for the help.

 

[Shackleford]

Are my equations correct? If so, do I then start substituting the electrical analogs? If so and assuming I get that correct, I'm not sure what to do after that. I suppose I could find my electrical solutions and then plug those into the impedance formula (whatever that is).

http://i111.photobucket.com/albums/n149/camarolt4z28/2010-09-22220323.jpg?t=1285211103

 

[hikaru1221]

It's not correct, because you forgot the elastic force F=-kx
Well, the next thing to do is to convert these into an equivalent circuit Just replace F with V, k with 1/C, etc, and what you get is 2 voltage equations. You will then know what to do.

 

[Shackleford]

It's not correct, because you forgot the elastic force F=-kx
Well, the next thing to do is to convert these into an equivalent circuit Just replace F with V, k with 1/C, etc, and what you get is 2 voltage equations. You will then know what to do.

Crap. I knew I was forgetting something. I was pretty tired last night. Let me re-write this. It's absolutely important that I firstly understand the situation physically then construe it mathematically.

However, I have a question. The mass m1 has the spring restoring force, but the mass m2 does not. M2 has the frictional force from the floor based on its velocity and then again the force of friction from m1.

 

[hikaru1221]

The mass m1 has the spring restoring force, but the mass m2 does not. M2 has the frictional force from the floor based on its velocity and then again the force of friction from m1.

Yes, you're right
But... is that a question?

 

[Shackleford]

Yes, you're right
But... is that a question?

Haha! Oops.

Okay, I'll give you a question now. Is this correct?

http://i111.photobucket.com/albums/n149/camarolt4z28/2010-09-23191532.jpg?t=1285288944

 

[hikaru1221]

Your equations are correct.

 

[Shackleford]

Your equations are correct.

Hey, that's great. Now, I can attempt my electrical analogs. However, should I write it in terms of I or q? And, isn't impedance equal to the square root of the sum of the squares of R_L and R_C. I vaguely remember the equation.

 

[hikaru1221]

Just do as I suggested earlier. Convert all the mechanical quantities into the equivalent electrical quantities, and you have 2 voltage equations. Each term of the voltage equations corresponds to a device (L, R, C, source). From the equation, you can draw the equivalent circuit and then from the circuit, calculate the impedance.

 

[Shackleford]

Just do as I suggested earlier. Convert all the mechanical quantities into the equivalent electrical quantities, and you have 2 voltage equations. Each term of the voltage equations corresponds to a device (L, R, C, source). From the equation, you can draw the equivalent circuit and then from the circuit, calculate the impedance.

Here are my equations. One of my classmates was saying something about the delta-q being the current through the R1 resistor or something like that. Each of these terms has to be voltage.

http://i111.photobucket.com/albums/n149/camarolt4z28/2010-09-25172307.jpg?t=1285453923

 

[hikaru1221]

You're on the right track. Take your friend's hint, replace $$\ddot{q}$$ with $$\frac{di}{dt}$$ and $$\dot{q}$$ with $$i$$; you will see it clearer. And yes, each term corresponds to voltage across some electrical device.

P.S.: I saw the term corresponding to $$kx_1$$ is a bit incorrect.

 

[Shackleford]

You're on the right track. Take your friend's hint, replace $$\ddot{q}$$ with $$\frac{di}{dt}$$ and $$\dot{q}$$ with $$i$$; you will see it clearer. And yes, each term corresponds to voltage across some electrical device.

P.S.: I saw the term corresponding to $$kx_1$$ is a bit incorrect.

Ah. You're right. I accidentally put the dot over the q. Let me see what I can come up with. I took Physics II two years ago. That's the last time I really dealt with impedance, circuits, and so forth.

http://i111.photobucket.com/albums/n149/camarolt4z28/2010-09-25233241.jpg?t=1285475708

 

[hikaru1221]

The last term of 2nd equation is a bit incorrect. Anyway, from here, can you build a corresponding circuit?

 

[Shackleford]

The last term of 2nd equation is a bit incorrect. Anyway, from here, can you build a corresponding circuit?

Crap. I put the dot somewhere but not where it belonged!

The circuit has two inductors, two resistors, and on capacitor. Does it have a parallel branch?

 

[hikaru1221]

Yes. That's why I told you to replace q' with i There should be some parallel branches. Hint: You can rewrite the 2nd equation as $$R_1(i_1-i_2)=L_2\frac{di_2}{dt}+R_2i_2$$ so that you can see a common term of both equations. That shows something about the parallel branches.

 

[Shackleford]

Yes. That's why I told you to replace q' with i There should be some parallel branches. Hint: You can rewrite the 2nd equation as $$R_1(i_1-i_2)=L_2\frac{di_2}{dt}+R_2i_2$$ so that you can see a common term of both equations. That shows something about the parallel branches.

Unfortunately, I don't have a damn idea what to do or how to calculate the impedance. I suppose I could insert the common term into equation one, but I don't see how that helps me. I guess I first need to know how I can calculate the impedance then try to see how my equations could get me there.

 

[hikaru1221]

Well you must first build the equivalent circuit There is no way around I suppose.
You don't have to substitute the common term into the 1st equation. Let me explain the circuit for you.
_ From the 1st equation: $$L_1\frac{di_1}{dt}+R_1(i_1-i_2)+\frac{q_1}{C}=V$$

That is, there is a source V, connected in series with an inductor L1, a resistor R1, a capacitor C. If you look at the current accompanying each term, you can see that L1 is accompanied by I1 and C is accompanied by Q1 which is directly related to I1. However, R1 is accompanied by (I1 - I2), i.e. the current through R1 is only a part of I1. Therefore there must be something connected in parallel to R1.

_ Indeed, the 2nd equation shows so: $$R_1(i_1-i_2)=L_2\frac{di_2}{dt}+R_2i_2$$

We can see that, sharing the same voltage as R1, there is a branch consisting of L2 and R2, both of which are accompanied by I2.

So now can you draw the equivalent circuit? From the circuit, you can easily find the impedance.

 

[Shackleford]

That makes sense.

This is probably wrong, but it kind of makes sense since this in this electrical analog the voltage source would provide AC.

Edited for embarrassment.

 

[hikaru1221]

Sorry, it's really wrong
Remember that the branch (L2 and R2) is parallel to R1. Besides the source has 2 terminals.

 

[Shackleford]

Sorry, it's really wrong
Remember that the branch (L2 and R2) is parallel to R1. Besides the source has 2 terminals.

What the heck did I just write. That's not a circuit. I'm multitasking here, so let me redo that. -_-

Okay, here it is.

http://i111.photobucket.com/albums/n149/camarolt4z28/2010-09-26162147.jpg?t=1285536201

 

[hikaru1221]

What you drew is that the branch (L1 + C), not R1, is parallel to (L2 + R2)!

 

[Shackleford]

What you drew is that the branch (L1 + C), not R1, is parallel to (L2 + R2)!

I thought the difference in current indicated a parallel circuit going into R1.

 

[Shackleford]

http://i111.photobucket.com/albums/n149/camarolt4z28/2010-09-26164217.jpg?t=1285537508

Wait. I see what you're saying now. Hold on.

 

[hikaru1221]

Now you got it. Then calculate the impedance from the circuit.

 

[Shackleford]

Now you got it. Then calculate the impedance from the circuit.

So, my terminals are correct?

If I'm looking at this the right way, there's one charge going "directly" through R1 and then the "other" charge going through L2 and R2. Those two charges constitute the I1 current through the R1, L1, and C series.

 

[hikaru1221]

What's with "charges"? You mean "current"?

 

[Shackleford]

What's with "charges"? You mean "current"?

Yes. Current is simply the movement of the "charges" with respect to time.

 

[hikaru1221]

Let me redraw the circuit in the standard way. See the picture attached.
Now can you calculate the impedance from the circuit?

 

[Shackleford]

Let me redraw the circuit in the standard way. See the picture attached.
Now can you calculate the impedance from the circuit?

The answer the back of the book has is very long and nasty. I don't know how to get their answer.

 

[hikaru1221]

Well if you don't remember how to calculate impedance, you'd better review your knowledge Hint: Use the complex forms of impedance:
$$Z_R = R$$
$$Z_L = jL\omega$$
$$Z_C = -\frac{j}{C\omega}$$

 

[Shackleford]

Well if you don't remember how to calculate impedance, you'd better review your knowledge Hint: Use the complex forms of impedance:
$$Z_R = R$$
$$Z_L = jL\omega$$
$$Z_C = -\frac{j}{C\omega}$$

Well, I put an equation for impedance in a previous post. The only difference is it's not in complex form. Why does it have to be in complex form?

http://i111.photobucket.com/albums/n149/camarolt4z28/2010-09-25233241.jpg?t=1285475708

 

[hikaru1221]

That's not the right one. That formula is only true for the case that there are only 3 elements R, L, C in series with the source (RLC circuit). It doesn't apply to this case where the circuit is much more complicated.

 

[Shackleford]

That's not the right one. That formula is only true for the case that there are only 3 elements R, L, C in series with the source (RLC circuit). It doesn't apply to this case where the circuit is much more complicated.

Okay. Let me see what I can dig up in my Physics II book.