Recent content by lepton5

No, I think you can't do that..since in your formula for Idisk still using \lambda as mass distribution in length, while your formula is related to area. you should use \sigma as mass distribution per area and start over.

for normal component, using the fact \nabla\cdot A = 0, or write it in integral form, \oint \vec{A}\cdot d\vec{a} = 0, we will get A_{above}^{\perp} = A_{below}^{\perp}. or we can write it in cartesian coordinate, A_{2z} = A_{1z}. for tangential component, Griffith also said that A is...

sory my bad, mix it up with kinematics. you should you use a (acceleration) for SHM, that is a = \omega ^2 A

well, actually it comes from \Delta y = y_1 - y_2 = 0. it was based on your information that other wave move oppositely from other one, so the substraction is there.

time t (period / 4) to reach max velocity from min velocity.

you can find acceleration just using simple kinematic eqn, that relate Vmax, Vmin, a, t.

do you know the relation between \omega, k and m ?

it's more simple if you use energy relation. for maximum velocity, you can equate ME = KE or \frac{1}{2}k A^2 = \frac{1}{2}m v^2. use relation \omega ^2 = \frac{k}{m} and you will get your velocity to about 140m/s

maybe this way, just equate y1 and y2 to find condition where A = 0 like below, e^{-(x_0 - 2t + 4)^2} = e^{-(x_0 + 2t - 2)^2}, if initially x0 same for both pulse, then you get t = 1.5 s

I will try to sharp two eqn above based on common textbook, 1) W = q . [V(a) - V(b)] this eqn tells us for the work done to move a charge from point (a) to point (b). V here means potential difference between points (a) and (b). 2) W = \frac{1}{2}\Sigma^{n}_{i=1} q_i V(r_i) this eqn...

Hi, Griffith's fan too here, you see from \mu_o(\vec{K}\times \hat{n}) = \mu_o K(-\hat{y}) automatically x and y component from eqn 1 will dissapear. then, eqn 1 will become eqn 2 as below: -\hat{y}\left[\frac{\partial A_{2z} }{\partial x}-\frac{\partial A_{2x} }{\partial z}-...

May i help you to solved the problem with another approach (only use V = I R rather than Kirchoof rule). Like this: with R23 = 18.75 you can find voltage here V23 = I1 . R23 = 4.84 V then back to your original circuit diagram. I2 that flow through R2 is I_2 = \frac{V_{23}}{R_2} = about...

Note: this relation n \simeq \sqrt{K_{\epsilon}} (Maxwell Relation) only holds for simple gases (air, Helium, Hydrogen). For water, this relation doesn't work well because K_{\epsilon} and then n are actually frequency-dependent, known as 'dispersion'. You can consult to your Optics book for...

use this one, \frac{H - h}{H} = \frac{r}{R}. you are wrong when you use h as denominator at left side, since you have equate it with R (radius of cone) you must also use H (height of cone). can you see : big triangle vs little triangle (from top of cone).

Well, actually you are evaluating the existence of limit by continuity test. maybe if you want to more detail, you can make table for 0^+ (like 1/5, 1/10, 1/100, ...) vs result of limit and for 0^- (like -1/5, -1/10, ...). and check it for left and right - limit (or you can make graph from...