Question

A steel rod having a length of $1m$ is fastened at its middle. Assuming young's modulus to be $2\times {10}^{11}Pa$, and density to be $8gm/c{m}^3$ if the fundamental frequency of the longitudinal vibration is $\frac{x}{2}kHz$. Find $x$ :

Answer 1

Given: $L=1mY=2\times {10}^{11}Pa$

$\rho =8gm/c{m}^3=8000\frac{kg}{m^3}$

Velocity of sound wave in steel rod, $v=\sqrt{\frac{Y}{\rho }}$

$v=\sqrt{\frac{2\times {10}^{11}}{8000}}=\frac{{10}^4}{2}m/s$

Frequency of fundamental mode i.e $n=1$, ${\nu }_1=\frac{v}{2L}$

${\nu }_1=\frac{\frac{{10}^4}{2}}{2\times 1}⟹{\nu }_1=\frac{5}{2}kHz$

Thus $x=5$