Calling all Oxbridge Physicists!

Yeah that sounds about right - output voltage lags the input by

\frac{\pi}{4}

(woo, latex - havent used it for ages :biggrin:

)

Reply 102

well i looked at it like: Vout/Vin = T. We can pick Vin to be whatever phase we want at any given moment cos we can set Vin=Ve^(jwt) for example
So the phase shift will just be the argument of the T = pi/4

I dont understand what you've done in your derivation nor how your tutor arrived at arctan(1/WRC)....I just do argument of T, which is you do a but of jiggery pokery comes out to be arctan(1) whatever W is.......which makes me think I'M WRONG about something. But i got the right answer

Reply 103

Hoofbeats worked out the phase of each voltage using

Unparseable latex formula:

\font{4}\mathrm{phase}=\arctan\left(\frac{\Im (V)}{\Re (V)}\right)

then subtracted one from the other

Reply 104

are those supposed to be Imaginary part and Real Part of V.....in which case what are you taking to be V....
I worked out the Im and Re parts of T, which are both exactly the same (RWC/1+(RWC)^2) so what is she taking as Vin and Vout then?

Reply 105

Willa

are those supposed to be Imaginary part and Real Part of V.....in which case what are you taking to be V....
I worked out the Im and Re parts of T, which are both exactly the same (RWC/1+(RWC)^2) so what is she taking as Vin and Vout then?

Yeah, for some reason it only gives you R and I instead of Re and Im. She was taking Vout = jwRC and Vin = 1 + jwRC

Reply 106

well then yes her method is valid because the question allows Vin to be taken as anything you wish....if you choose Vin to be 1+jwRC then Vout is correctly Vout. My method just happened to use that Vin = Some unspecified value, and hence Vout=TVin, so the phase shift is just arg T...which comes out to arctan(1). The problem that arises is that is the phase shift really dependant on W?

Reply 107

shiny

12

Willa

I worked out the Im and Re parts of T, which are both exactly the same (RWC/1+(RWC)^2) so what is she taking as Vin and Vout then?

T
= jRWC / (1 + jWRC)
= jRWC (1-jRWC) / (1+jRWC)(1-jRWC)
= (RWC)^2 + jRWC / (1 + (RWC)^2)

Reply 108

shiny

T
= jRWC / (1 + jWRC)
= jRWC (1-jRWC) / (1+jRWC)(1-jRWC)
= (RWC)^2 + jRWC / (1 + (RWC)^2)

oops i take everything back, i forgot to square it!

in which case phase shift is: arctan(Im/Re) = arctan(RWC/RWC^2) = arctan(1/RWC) which is what the tutor has scribbled. And in the specific case RWC = 1, so the phase shift is pi/4.

sorry, this is exactly what i do all the time: make stupid mistakes!

since we're on circuits..has anyone got an answer for me on my circuit question on the last page...the purpose of the large resistor...i'm still stumped and i need to understand for my report! thanks!

Reply 109

OP

Sorry, still don't have a clue about that large resistor! I'm sure it's something to do with a practical we did in Electronics lab - Nick/Andy/Phil? any ideas?

Also, another question (well I had two, but I emailed my tutor about the other!) I have a circuit (see attachment) and have an emf is applied which is zero for all times before t=0, then steps up abruptly to E at t=0, and remains at E for all subsequent times. I need to find the current I(t).

Right, branch equations:

E - V1 = L.dI/dt
V1 - 0 (I - I2)R = L.dI2/dt + I2.R

so I = (L/R)*dI2/dt + 2I2

differentiating we get dI/dt = (L/R)*(d^2I2/dt^2) + 2.dI2dt

Then somehow I got the following expression...

E - L.dI2/dt - I2.R = L.dI/dt

...but I'm not sure where it's come from! Can someone explain? Thanks. Until I get that, I can't go onto the next bit!

(Apologies for my terribly diagram and the fact I can't use latex - how do I get latex?!)

Prelim CV2i).JPG10.3KB

Reply 110

OP

Never mind! Just being thick as usual, have spotted where it comes from (although I included a rather odd step last time! I've just said that V1=L.dI2/dt + I2R and substituted in! Thanks anyways

Reply 111

latex is built into the board...use [ tex ] tags. To learn how to use latex just google "latex" and you can find the documentation...or you can play around with typing stuff between [ tex ] [ /tex ] tags (obviously without the spaces)

Reply 112

out of pure curiosity...how do you finish that question to get I(t)

where do you go from: E - L.dI2/dt - I2.R = L.dI/dt ?????

Reply 113

OP

well you can either substitute in the dI/dt = (L/R)*(d^2I2/dt^2) + 2.dI2dt and then solve the ODE for I2 and then plug that back into get I1 or rearrange differently to begin with and form an ODE for I1 and solve. To find the boundary conditions, we know that just before the step up, I=0 and immediately after I is still zero and t=0. We also know that because the current flowing afterwards is initially v.small, most of the voltage drop occurs over the inductor so we can estimae E=L.dI/dt at t=0.

Reply 114

OP

Another quick question...we (another Lincoln physicist and I) were wondering...is it right to say that at resonance there will be no phase shift between V-out and V-in?!

Reply 115

Hoofbeat

Another quick question...we (another Lincoln physicist and I) were wondering...is it right to say that at resonance there will be no phase shift between V-out and V-in?!

Yeah, I think so - ring's a bell anyway!

Reply 116

OP

Hey guys, a wave question for you (Andy & Nick, it's Q2a from the problem sheet Waves 2).

We have a string of uniform linear density p which is stretched to a tension T and it's ends are fixed at x=0 and x=L. Using separation of variables, you end up with the general solution:

y(x,t) = (Acos.kx + Bsin.kx)(Ccos.kct + Dsin.kct)
But since y(0,t)=0 and y(L,t)=0, we get A=0 and kL=r.pi

thus y(x,t) = Bsin(rx.pi/L)[Ccos(rct.pi/L)+ Dsin(rct.pi/L)]

I know have to show that the following are both solutions, obeying the boundary conditions:

y(x,t) = Esin(rx.pi/L)sin(rct.pi/L)
y(x,t) = Fsin(rx.pi/L)cos(rct.pi/L)

I can get the second one by using the fact that it's initially at rest, so dy/dt=0, thus in my earlier solution, D=0 and F=BC. How do I obtain the first solution however? I clearly have to use different boundary conditions, but I can't see what they should be. Suggestions?

Reply 117

OP

Bezza

Yeah, I think so - ring's a bell anyway!

It definitely is, by the way (in case anyone's wondering) - I checked with my tutor :smile:

Reply 118

OP

Also, another wave question guys; If we're dealing with a relativistic particle (ie. in quantum mechanics) then I know the group velocity = dw/dk, but how do we know that

Vg = dw/dk = dE/dp

where E=relativistic energy and p=relativistic momentum

Thanks :smile:

Reply 119