316 reputation
17
bio website twitter.com/MJalalvand
location Iran
age 22
visits member for 1 year, 8 months
seen May 18 '13 at 21:37

I'm a Physics student at Univesity of Tehran.


Apr
28
awarded  Yearling
May
3
comment How we approach RLC circult from RLGC model?
Note that to get the impedance for RLGC model, the only math we use is the loop and junction law That's not true. The formula you're given for characteristic impedance is derived from telegrapher equation assuming infinite number of infinitesimal components (integration).
May
3
comment Calculating the coefficient of thermal expansion in liquid
No, this one works because you get V=1 again so mass is equal to density. For volumes other than unity it's incorrect.
May
2
answered How we approach RLC circult from RLGC model?
May
2
answered Calculating the coefficient of thermal expansion in liquid
May
1
comment Volume of gas at which relative fluctuation of gas density occurs
@UnkleRhaukus By the way what \left or \right do? Sorry, I should have told you before undoing them.
May
1
comment Volume of gas at which relative fluctuation of gas density occurs
@UnkleRhaukus Yes, Thanks for your corrections but the \left/\right made angle brackets appear lower than where they should be.
Apr
30
revised Volume of gas at which relative fluctuation of gas density occurs
deleted 33 characters in body
Apr
30
answered Electric Field in Dieletric
Apr
30
comment Volume of gas at which relative fluctuation of gas density occurs
Well, $k$ is the Boltzmann constant $k = 1.38 \times 10^{-23} J/K$ , $T = 293.15 K$ , $P=10^5 Pa$ , $\delta = 0.01$. Just put them in the last formula.
Apr
30
comment Volume of gas at which relative fluctuation of gas density occurs
@Andrew123321 May I ask what grade are you in and where you see the question?
Apr
30
comment Volume of gas at which relative fluctuation of gas density occurs
${(V - V_0) \over V} = 0.1$ So approximately $\langle {(V - V_0)^2 \over V^2} \rangle = 0.01$
Apr
30
answered Volume of gas at which relative fluctuation of gas density occurs
Apr
30
comment A Type of Pendulum
Also except the sign case your Lagrangian is correct.
Apr
30
comment A Type of Pendulum
Both terms in $\dot y$ should be positive. If you correct the sign there all $\phi + \omega t$ will become $\phi - \omega t$. Except this your ${\partial L \over \partial \dot \phi}$ is right. The second term in ${d \over dt}{\partial L \over \partial \dot \phi}$ should be $m r l \omega (\dot \phi - \omega) cos (\phi - \omega t)$. First term in ${\partial L \over \partial \phi}$ should be $+m r l \omega \dot \phi cos(\phi - \omega t)$.
Apr
30
answered Damped oscilator - logarithmic decrement of damping
Apr
30
comment A Type of Pendulum
Just solve this $\frac{d}{dt} \frac{\partial L}{\partial \dot \phi} - \frac{\partial L}{\partial \phi} = 0$
Apr
30
comment A Type of Pendulum
what are mistakes? You shouldn't deviate relative to $\omega$
Apr
30
comment A Type of Pendulum
Yes, it should only be $\phi$ and I think your only $q_i$ is $\phi$.
Apr
30
comment A Type of Pendulum
If you take $\omega$ to be constant you get only one equation involving one variable and its derivatives $\phi$. But if you take $\omega$ as a variable as well (meaning that the disk could oscillate in addition to pendulum) you'll get two equations each of which containing both variable and its derivatives. This is a coupled non-linear ODE system that you can only find the answer with linear approximation or numerical methods. You can't decouple it analytically.