\(v_{n}=\frac{m v}{2 L}\) where \(n=1,2,3, \ldots .\)
The difference between two consecutive resonant frequencies is
\(\Delta v_{n}=v_{n+1}-v_{n}=\frac{(n+1) v}{2 L}-\frac{n v}{2 L}=\frac{v}{2 L}\)
which is also the lowest resonant frequency \((n=1)\)
Thus the lowest resonant frequency for the given string
\(=420 \mathrm{Hz}-315 \mathrm{Hz}=105 \mathrm{Hz}\)
(હવામાં ધ્વનિની ઝડપ $=330\, ms^{-1}$)