A string is stretched between two rigid supports separated by \(75 \mathrm{~cm}\). There are no resonant frequencies between \(420 \mathrm{~Hz}\) and \(315 \mathrm{~Hz}\). The lowest resonant frequency for the string is

As there is no resonant frequency between \(315 \mathrm{~Hz}\) and \(420 \mathrm{~Hz}\), let \(315 \mathrm{~Hz}\) be \(\mathrm{n}^{\text {th }}\) overtone and \(420 \mathrm{~Hz}\) be \((\mathrm{n}+1)^{\text {th }}\) overtone.
Now, \(v=\frac{\text { nv }}{2 l}\)
\(
\therefore \quad 315=\frac{\mathrm{nv}}{2 l} \text { and } 420=\frac{(\mathrm{n}+1) \mathrm{v}}{2 l}
\)
Taking the ratio,
\(
\begin{array}{ll}
& \frac{315}{420}=\frac{n}{n+1} \\
\therefore \quad & 315 n+315=420 n \\
\therefore \quad & n=3
\end{array}
\)

The resonant frequency is \(v_0=\frac{\mathrm{v}}{2 l}\)
Therefore, from equation (i) we get, \(v_0=\frac{v}{n}=\frac{315}{3}=105 \mathrm{~Hz}\)