\(v \propto \frac{1}{{\sqrt \rho }}\)
==>\(\frac{{{v_1}}}{{{v_2}}} = \sqrt {\frac{{{\rho _2}}}{{{\rho _1}}}} = \sqrt {\frac{4}{1}} = 2:1\)
$(a)$ $\left(x^2-v t\right)^2$
$(b)$ $\log \left[\frac{(x+v t)}{x_0}\right]$
$(c)$ $e^{\left\{-\frac{(x+v t)}{x_0}\right\}^2}$
$(d)$ $\frac{1}{x+v t}$
(ગુણકારનો અચળાંક $1$ લો)