Yes, if you take the inductor as the output. Now you can do the Bode plot. If you take the capacitor as output, you get the same thing except the numerator is 1:
1
G(s)=-------------
LC*s^2+1
A clarification must be made:
In circuit theory, the resonant frequency is the same as natural frequency in control systems i.e. ωn
The resonant frequency is not the natural frequency, ωr=ωn√1-2ξ2
When ξ=0, both are equal.
Let see it better with an example.
Series RLC circuit, with output in the capacitor (because of easier transfer function).
ωn=1/√LC (this is the natural frequency but in circuit theory is called resonant frequency)
ωr=√(2L-R2C)/2L2C (this is the full expresion for the resonant frequency for a RLC circuit)
Why in circuit theory the natural frequency is called the resonant frequency and they never calculate the real resonant frequency ?
Because natural frequency "ωn" is easier to calculate and approaches to the "ωr" (resonant frequency)
Let us use a RLC circuit like the following: R=0.1 Ω, L=90 mH and C= 50 µF
fn=75.02635968 Hz
fr=75.02625548 Hz
And the voltage gain at fr is = 424.26436 V/V or 52.55273 dB
If you want to check this, go ahead and do the Bode plot in LTspice and see that the peak is reached at the resonant frequency "fr" NOT at the natural frequency (in circuit theory called resonant frequency because it is easier to calculate and it is a good aproximation).
PS: I have made the Bode plot in OrCAD and I needed 500 000 points in order to get enough decimals of the peak frequency in order to check calculations.