Re: LC tank, voltage
It is a classic second order differential equation. There are no losses in the circuit so there will be sustained oscillations.
Mark voltages \[v_C\] and \[v_L\] for the inductor and capacitor from the top node to ground and mark currents \[i_C\] and \[i_L\] for the inductor and capacitor flowing from the top node to ground.
Just write the differential equations for the circuit.
\[i(t) = C \frac{d v_C}{dt} + i_L\] ----(1)
\[v_L = L \frac{d i_L}{dt}\] -----(2)
note also that \[v_C=L \frac{d i_L}{dt}\] (since \[v_C=v_L\]). Now taking derivative on both sides of \[v_C=L \frac{d i_L}{dt}\] gives \[ \frac{d v_C}{dt} = L \frac{d^2 i_L}{dt^2}\]. Now substitute this into (1) and you will get
\[i(t) = LC \frac{d^2 i_L}{dt^2} + i_L\]
This is a classic 2nd order differential equation. Given that \[i(t)=I_m u(t)\] is a step function of amplitude \[ I_m\] and that initial inductor current and capacitor voltage is zero \[i_L(0)=0\], \[v_C(0)=0\], the solution is
\[ i_L(t)= I_m - I_m \cos( \frac{t}{\sqrt{LC}}) = I_m - I_m \cos(\omega_0 t)\]
\[ v_C(t)= I_m Z_0 \sin( \frac{t}{\sqrt{LC}}) = I_m Z_0 \sin(\omega_0 t)\]
where \[Z_0=\sqrt{\frac{L}{C}}\] is called characteristic impedance of the tank circuit. In the example that you had \[I_m=1\], \[Z_0=1\] and \[\omega_0=1\], so your solutions were
\[ i_L(t)= 1 - \cos(t)\]
\[ v_C(t)= \sin(t)\]
Best regards,
\[v_c\]
Added after 1 hours 45 minutes:
Sorry to reply to my own post but I had to correct some typos and clean up the equations and put them in TeX mode. I think they should be OK now.
\[v_C\]