Chapter 11, Problem 40P

To determine

To Find: The speed of the longitudinal wave and the factor of their highness.

Answer to Problem 40P

The rod which is less dense is having the higher speed of longitudinal waves and the factor of their highness is $\text{1}\text{.41}$ .

Explanation of Solution

Given:

Elastic modulus of rod $1$ is equal to Elastic modulus of rod $2$ .

  $\text{So, }{E}_1=E_2$

Density of rod $1$$={\rho }_1$

Density of rod $2$$={\rho }_2=2\times {\rho }_1$

Velocity of rod $1$$\text{ = }{v}_1$

Velocity of rod $2$$\text{ = }{v}_2$

Formula used:

The speed of the longitudinal waves in solid rod is obtained by the following formula,

  $\begin{array}{l}v\text{ =}\sqrt{\frac{\text{Elastic modulus}}{\text{Density}}}\\ =\sqrt{\frac{E}{\rho }}\end{array}$

Here,

  $v$ is speed of longitudinal wave,

  $E$ is elastic modulus,

  $\rho$ is density.

Calculation:

Substitute the given values in the above equation,

  $\begin{array}{l}{v}_1=\sqrt{\frac{E_1}{{\rho }_1}}\\ v_2=\sqrt{\frac{E_2}{{\rho }_2}}=\sqrt{\frac{E_1}{2\times {\rho }_1}}\\ \frac{v_1}{v_2}=\frac{\sqrt{\frac{E_1}{{\rho }_1}}}{\sqrt{\frac{E_1}{2\times {\rho }_1}}}=\sqrt{2}=1.41\end{array}$

Conclusion:

Thus, the rod which is less dense is having the higher speed of longitudinal waves compared to the denser rod and by the factor of $1.41$ .