Recent content by Miles123K

Since the membrane doesn't break, the wave is continuous at ##x=0## such that ##\psi_{-}(0,y,t) = \psi_{+}(0,y,t)## ##A e^{i(k \cos(\theta)x + k \sin(\theta)y - \omega t)} = A e^{i(k' \sin(\theta ') y- \omega t)}## Which is only true when ## k' \sin(\theta ') = k \sin(\theta) ##. From the...

Right. The electrodes didn't look any different but I will attempt another trial to check if it's actually graphite exfoliation.

That makes a lot of sense! However, I also did the same procedure with other electrolytes. Also Chlorides. And those electrolytes didn't show the same thing. Does the presence of barium make the exfoliation easier or something? Thanks for your answer!

I did an electrolysis experiment with 1.6M Barium Chloride solution and the electrolyte turned black after the electrolysis. I am fairly positive that Chlorine was produced at the anode from the smell. However, Ba(OH)2 is colorless. I used graphite electrodes for this experiment. Does anyone...

Oh okay, the new dispersion relations I obtained are as follows: ##\omega^2=\frac {K} {m} (2 - e^{ka} - e^{-ka}) + \frac {g} {l}## which simplifies to: ##\omega^2=\frac {2K} {m} (1 - cosh(ka)) + \frac {g} {l}## If I plug in the ##ka## I got above I get the answer?

First I worked out the dispersion relations, which is pretty easy: ##M \ddot x_j = K x_{j-1} + K x_{j+1} - 2K x_j -mg \frac {x_j} {l} ## (All t-derivatives) We know ##x_j## will be in the form ##Ae^{ijka}e^{-i\omega t}## so the above becomes: ## -\omega^2M = K (e^{-ika}+e^{ika}-2)-\frac {g}...

Okay. I kind of get the idea now. Thanks a lot. :biggrin:

Why should we take ##V_6 = 0## instead of treating the last capacitor as a "free end" though? I know in this case that would be correct but is there a prompt for doing so?

Wow thank you so much! I am curious about how you got the normal mode frequency though, would you mind explaining that? Also, what software did you use to plot this graph?

What about part C where it talks about the resonance of this system? I assume that means the local maximums of the amplitude? Do I attempt to put the denominator in terms of ##\omega##? The argument in chapter five referred to should be this:

Because of the complexity of the algebra I won't upload the entire thing. The voltage at ##C_1## would be: ##\tilde V_1(t) = V_0 \frac 1 {32cos^5(ka) - 32cos^3(ka) + 6cos(ka)} e^{-i\omega t }## where ##cos(ka)## are to be substituted with ##(1-\frac {LC} 2 \omega^2) - \frac {RC} 2 \omega i ##...

I seem to be able to get the coefficient as ##\frac 1 {32cos^5(ka) - 32cos^3(ka) + 6cos(ka)}##, which would be complex in this case. Is that correct? If so, I will upload my result.

Oh okay. In that case: ##cos(ka)=(1-\frac {LC} 2 \omega^2) - \frac {RC} 2 \omega i## I don't really get the part of expressing the coefficient of ##e^{-i\omega t}## though. Use trig identity to make the function in terms of ##cos(ka)##? EDIT: Ok I seem to get it. I will upload my attempted...

Sorry but I don't know " finite-difference equations" or "z-transform":frown: I am not a college student, just learning for my interest as a high school student. The mathematical methods I have learned are first year or maybe second year content of Multi-variable calculus, differential equations...

Okay thanks. Now I have some clue but I am not sure if I am going down the correct path. The forced oscillation should produce a mode to the positive x-direction in the form: ##\psi(x,t) = A ((\frac {e^{i(k_r+ik_i)x}-e^{-i(k_r+ik_i)x}} {e^{i(k_r+ik_i)l}-e^{-i(k_r+ik_i)l}}))e^{-i\omega t}## To...