$tan \theta \,\, = \,\,\frac{F}{{Mg}}\,\,$ ($\theta $ નાનો છે)

$\theta \,\, = \,\,\frac{F}{{Mg}}\,\, \Rightarrow \,\,F\,\, = \,\,Mg\theta $

$\frac{{K{Q^2}}}{{{x^2}}}\, = \,\,Mg\,\,\frac{x}{\ell }$

${Q^2}\,\, = \,\,\frac{{Mg}}{{K\ell }}\,{x^3}\,\, \Rightarrow \,\,Q\,\, = \,\,\sqrt {\frac{{Mg}}{{K\ell }}} \,{x^{\frac{3}{2}}}$

$\left( {\frac{{dQ}}{{dt}}} \right)\,\, = \,\,\sqrt {\frac{{Mg}}{{K\ell }}} \,\,\left( {\frac{3}{2}\,{x^{\frac{1}{2}}}} \right)\,\,\left( {\frac{{dx}}{{dt}}} \right)$

$ = \,\,\left( {\frac{{dQ}}{{dt}}} \right)\,$ અચળ છે તેથી $\frac{{dx}}{{dt}}\, \propto \,{x^{ – \frac{1}{2}}}$ અથવા $V\,\, \propto \,\,{x^{ – \frac{1}{2}}}$