91Ó°ÊÓ

The given data is listed below as-

• The interatomic spring stiffness of element A is three times larger than the interatomic spring stiffness for element B i.e,$k_A=3k_B$
• The mass of an atom of element A is three times greater than the mass of an atom of element B i.e.$m_A=3m_B$,
• The atomic diameters are approximately the same for A and B. i.e.$d_A=d_B$,

The speed of sound depends upon stiffness of interatomic bond , mass of the atom and atomic diameter of the solids.

The concept of the interatomic bond gives the magnitude of the speed of sound.

The equation of the magnitude of the speed of sound is expressed as,

$V=\frac{k}{m}â‹\dots d$

Here,k is interatomic spring stiffness,m mass of an atom of element, dis the atomic diameter.

The ratiothe speed of sound in rod A to the speed of sound in rod B.

$\frac{V_A}{V_B}=\frac{\sqrt{{\displaystyle \frac{k_A}{m_A}}}.d}{\sqrt{{\displaystyle \frac{k_B}{m_B}}}.d}$

For,$k_A=3k_B$ , $m_A=3m_B$ and $d_A=d_B$

$$\begin{gathered} \frac{V_A}{V_B}=\frac{\sqrt{{\displaystyle \frac{3k_B}{3m_B}}}.d}{\sqrt{{\displaystyle \frac{k_B}{m_B}}}.d} \\ \frac{V_A}{V_B}=1 \end{gathered}$$

Thus, the ratio of the speed of sound in rod A to the speed of sound in rod B is 1.