where $T = $ weight of part of rope hanging below the point under consideration
$ = \left( {\frac{M}{L}} \right)xg$
==> $v = \sqrt {\frac{{\left( {\frac{M}{L}} \right)xg}}{{\left( {\frac{M}{L}} \right)}}} = \sqrt {xg} $.

Statement$-2:$ Intensity of waves of given frequency in same medium is proportional to square of amplitude only.
