Question

Two rods $P$, $Q$ are considered such that the young's modulus of elasticity of rod $P$ is twice that of the other while the density of rod $Q$ is eighteen times that of rod $P$. If a sound wave is allowed to transverse $2m$ distance through each rod, then in which case will the sound take lesser time and by how many times?

Answer 1

${\rho }_Q=18{\rho }_P$

$E_P=2E_Q$

$d=2m$

$\frac{t_P}{t_Q}$ OR $\frac{t_Q}{t_P}$

We know that $t=\frac{d}{v}$

$\therefore t_P=\frac{d}{v_P}$ and $t_Q=\frac{d}{v_Q}$

$\therefore \frac{t_P}{t_Q}=\frac{v_Q}{v_P}$

We know that $v=\sqrt{\frac{\gamma E}{\rho }}$

$\therefore v_P=\sqrt{\frac{\gamma E_P}{{\rho }_P}}$ and $v_Q=\sqrt{\frac{\gamma E_Q}{{\rho }_Q}}$

$\therefore \frac{v_Q}{v_P}=\sqrt{\frac{\gamma \times E_Q\times {\rho }_P}{\gamma \times E_P\times {\rho }_Q}}=\sqrt{\frac{E_Q\times {\rho }_P}{2E_Q\times 18{\rho }_P}}=\sqrt{\frac{1}{36}}=\frac{1}{6}$

$\therefore \frac{t_P}{t_Q}=\frac{1}{6}$

This shows that $t_P$ is 6 times smaller than $t_Q.$

The sound will take lesser time in rod P and by 6 times lesser than that in Q.