If two anti-nodes are observed on either side of C, the frequency of the mode in which the rod is vibrating will be ($\mathrm{y}=2\times {10}^{11}\mathrm{N}/{\mathrm{m}}^2, \mathrm{d}=8\times {10}^3\mathrm{kg}/{\mathrm{m}}^3$) length of the rod is 2.5 m

 Velocity of the longitudinal waves in the rod $=\mathrm{v}=\sqrt{\mathrm{Y}/\mathrm{d}}=\sqrt{2\times {10}^{11}/8000}=5000 \mathrm{m}/\mathrm{s}$
The wavelength of the wave for the mode of vibration in which 2 anti-nodes occur, is  $\mathrm{\lambda }=\frac{1.25}{\left(3/4\right)}=\frac{5}{3}\mathrm{m}$
Hence frequency of vibration  $=\mathrm{n}=\frac{\mathrm{V}}{\mathrm{\lambda }}=\frac{5000}{5/3}\mathrm{Hz}=3000\mathrm{Hz}$