Transfer Function: Trying to understand the Laplace analysis of this circuit
May I suggest that you start by simplifying the 2 resistors into one resistor. I'm suggesting this because it makes the math a lot easier and I think you have a mistake in your final formula. So, the voltage source reduces from 10 volts to 9.09 volts in series with a resistance of 90.90 ohms due to standard circuit theorums.

This then has a standard formula: -

$$H(s) = \dfrac{\omega_n^2}{s^2 +2\zeta\omega_n s+ \omega_n^2}$$
Where \$\omega_n = \dfrac{1}{\sqrt{LC}}\$ 

And \$\zeta= \dfrac{R}{2}\sqrt{\dfrac{C}{L}}\$

The value for R is the newly recalculated value of 90.9 ohms. To offer a little more help, \$\omega_n\$ is the natural resonant frequency of the filter and \$\zeta\$ is called the damping ratio (also equal to the reciprocal of 2Q).

Now if you want to see what this looks like you can use this on-line calculator and plug-in the values: -



Resonance occurs at about 503 Hz and there is a peak in the response of about 11 dB. Circuit Q is about 3.5.

But don't forget that there is an overall attenuation (due to me simplying the two resistors into one resistor) so the formulas are multiplied by 0.909.

I haven't used your equations because I can see that there is an error in the final formula you derived.