Physics - Wave on String Question with Solution | TestHub

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, $v=\frac{\text{ nv }}{2l}$ $$\therefore 315=\frac{\mathrm{nv}}{2l}\text{ and }420=\frac{(\mathrm{n}+1)\mathrm{v}}{2l}$$ Taking the ratio, $$\begin{array}{ll}& \frac{315}{420}=\frac{n}{n+1}\\ \therefore & 315n+315=420n\\ \therefore & n=3\end{array}$$ The resonant frequency is $v_0=\frac{\mathrm{v}}{2l}$ Therefore, from equation (i) we get, $v_0=\frac{v}{n}=\frac{315}{3}=105\mathrm{Hz}$