\(\Rightarrow \frac{n+1}{n}=\frac{420}{315} \Rightarrow n=3\)
Hence \(3 \times \frac{\mathrm{v}}{2 \ell}=315 \Rightarrow \frac{\mathrm{v}}{2 \ell}=105 \mathrm{Hz}\)
The lowest resonant frequency is when
\(\mathrm{n}=1\)
Therefore lowest resonant frequency
\(=105 \mathrm{Hz}\)