The nodal points for a metal or wooden bar vibrating in the free-free
transverse mode are 22.42% in from each end, or to put it another way,
at fractional points 0.2242 and 0.7758 along the length.  These are
therefore the points where you would mount the bar to its support
rails.

Also of interest is that the fundamental frequency is proportional to
the inverse of the square of the length, assuming that the width and
thickness are constant between bars.  Halving the bar's length will
therefore raise the frequency by a factor of four, or two octaves.
This is different to the organ pipe where halving the length raises
the pitch by one octave.

Once you have determined the length of the lowest note bar it is
a simple matter to calculate all the other bar lengths.  Each bars
length is the (24th root of a half) of the one before, since 24 equal
ratios (12 notes in an octave) give the two octave halving of length.

I used these equations successfully many years ago when I built a
glockenspiel for an orchestrion.  Over two and a half octaves I found
that this length and pitch relationship was slightly in error in that
the top bar was a semitone flat.  It was the a simple matter to finally
tune the bars to their correct pitch.

I have no reason to believe that wood behaves any differently to metal in
the respect of transverse vibrations.  Maybe others could enlighten us.

Best regards,
Nicholas Simons, GB