A sonometer wire resonates with a given tuning fork forming standing waves with five antinodes between the two bridges when a mass of 9 kg is suspended from the wire. When this mass is replaced by a mass M, the wire resonates with the same tuning fork forming three antinodes for the same positions of the bridges. The value of M is

Let L be the length of the wire between the bridges and let m be the mass per unit length of the wire.

Five antinodes on a length L implies that $\mathrm{L}=\frac{5}{2}{\mathrm{\lambda }}_1$ or ${\mathrm{\lambda }}_1=\frac{2\mathrm{L}}{5}$. Thus in this case we have

${\mathrm{v}}_1=\frac{V_1}{{\mathrm{\lambda }}_1}=\frac{5}{2\mathrm{L}}\sqrt{\frac{{\mathrm{T}}_1}{\mathrm{m}}}$, where ${\mathrm{T}}_1={\mathrm{M}}_1\mathrm{g}$

Three antinodes on a length L implies that $\mathrm{L}=\frac{3}{2}{\mathrm{\lambda }}_2$ or ${\mathrm{\lambda }}_2=\frac{2\mathrm{L}}{3}$. In this case, we have ${\mathrm{v}}_2=\frac{V_2}{{\mathrm{\lambda }}_2}=\frac{3}{2\mathrm{L}}\sqrt{\frac{{\mathrm{T}}_2}{\mathrm{m}}}$, where ${\mathrm{T}}_2={\mathrm{M}}_2\mathrm{g}$

Given ${\mathrm{v}}_1={\mathrm{v}}_2,{\mathrm{M}}_1=9\mathrm{kg}\text{ and }{\mathrm{M}}_2=\mathrm{M}$. Therefore, we have

 $\frac{5}{2\mathrm{L}}\sqrt{\frac{9\mathrm{g}}{\mathrm{m}}}=\frac{3}{2\mathrm{L}}\sqrt{\frac{\mathrm{Mg}}{\mathrm{m}}}$

which gives M = 25 kg.