Question

Two metal rods are made of different elements. The interatomic spring stiffness of element A is three times larger than the interatomic spring stiffness for element B. The mass of an atom of element A is three times greater than the mass of an atom of element B. The atomic diameters are approximately the same for A and B. What is the ratio of the speed of sound in rod A to the speed of sound in rod B?

Short answer

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

Step-by-step solution

Identification of the given data

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.$d_A=d_B$,

Significance of the speed of sound

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.

Determination of the ratio of the speed of sound in rod A to the speed of sound in rod B.

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

$V=\frac{k}{m}\cdot d$

Here,k is interatomic spring stiffness,m mass of an atom of element, d is 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.