Then it's a second-order L-section filter, not a 3rd order PI-section filter.
At a rough approximation, it would give 20dB of attenuation about an octave and a half above the resonant frequency.
That would give a cutoff frequency of about 35Hz, and so L would be 1/(4π^2f^2C) = 2H
Of course, the load impedance would have an effect as well.

Hi,

Yes, that's what it appears to be at first glance. However, the problem looks like there will be two frequencies on the output not just the one we want. That would be considered distortion, and it could be a lot.
The AC signal analysis shows something similar to your result but the time domain transform clearly shows two different frequencies similar to this (this is not exact just representative):
sin(w*t)+sin(t/sqrt(LC))
which we could write:
sin(w1*t)+sin(w2*t)
where
w1 is the input frequency and w2 is the frequency caused by the undamped LC combination.

What you think?

 

Hello again,

I looked at this a little mor and found that with an input resistance one of the sin/cos pairs can be converted to sinh/cosh pair which means just one sin/cos pair remains which means just one frequency. It's tricky though as the exponential decay that kills one of the sin/cos pairs can be very long not something we would want in an actual real life design, which would mean a very distorted wave for some time until those exponentials damped out.
There is a possibility that with any input resistance at all (even 1uOhm) that the second sin/cos pair can be killed, but it's not clear if the exponential that causes that will take too long or not.