JEE Main · 2018 · 08 Apr 2018 · Offline · Waves & Sound

The speed of a longitudinal wave in the rod is set by its stiffness and density, $v=\sqrt{\frac{Y}{\rho }}=\sqrt{\frac{9.27\times 10^{10}}{2.7\times 10^3}}\approx 5.86\times 10^3$ m/s. A rod clamped at its middle cannot move there, so the centre is a displacement node while the two free ends are antinodes; the fundamental therefore fits half a wavelength into the whole length, $\lambda =2L=1.2$ m. The fundamental frequency is then $f=\frac{v}{\lambda }=\frac{5.86\times 10^3}{1.2}\approx 4.9$ kHz, essentially $5$ kHz.