A sword that resonate at \(7.489\,Hz\),
given that the speed of sound in steel is \(5000\,ms^{-1}\).
Unfortunately...
\(f.\lambda=v\)
\(l_s=\cfrac{\lambda}{2}=\cfrac{v}{2f}\)
\(l_s=\cfrac{5000}{2*7.489}\)
\(l_s=333.82\,m\)
over three hundred and thirty meters!
If however, we fold the molten smelt repeatedly during the forging process, such that many half wavelengths spread over the length of the sword we set at,
\(l_s=0.72\,m\)
\(l_s=n.\cfrac{\lambda}{2}=n.\cfrac{v_n}{2f}\)
\(n=l_s*\cfrac{2f}{v_n}\)
we assume further that the speed reduced by a factor of \(\cfrac{1}{n^2}\)
\(v_n=\cfrac{v}{n^2}\)
\(n=\cfrac{v_n}{2fl_s}\)
\(n=\cfrac{333.82}{0.72}\)
In practice, the number of folds possible is binary,
\(n_b=\left\lceil\cfrac{log(n)}{log(2)}\right\rceil=\left\lceil 8.856 \right\rceil=9\)
ie,
\(n_p=2^9=512\)
This would be,
\(p=\cfrac{n_p}{n}=\cfrac{512}{333.82}*{0.72}=1.104\)
times the intended \(l_s=0.72\). So, with nine folds of the smelt, we turn it such that the folds run perpendicularly along the sword and elongate the smelt to a length of,
\(l_b=p*0.72=0.795\)
and file this blade down to \(0.72\). This way the sword at \(0.72\,m\) will have the intended \(n\) number of folds along its length.
The speed of sound along the blade is an approximated at \(5000\,ms^{-1}\) and we have assumed that the folds on the blade slow the speed of sound along it by a factor of \(\cfrac{1}{n^2}\).
If both these assumptions are true, this sword should be light as a feather.
