One mole of Nickel $\left(6.02\times {10}^{23}\mathrm{atoms}\right)$has a mass of , and its density is $\mathbf{8}\mathbf{.}\mathbf{9}\mathbf{g}\mathbf{/}{\mathbf{cm}}^{\mathbf{3}}$. You have a bar of nickel 2.5 m long, with a cross-section of 2mm on a side. You hang the rod vertically and attach a 40 kg mass to the bottom, and you observe that the bar becomes 1.2 mm longer. Next, you remove the 40 kg mass, place the rod horizontally, and strike one end with a hammer. How much time will elapse before a microphone at the other end of the bar will detect a disturbance?

The given data is listed below as-

• The mass is,$m=40\mathrm{kg}$
• The length the bar is,$L=2.5\mathrm{m}$
• The area is,$A={\left(2\times {10}^{-3}\mathrm{m}\right)}^2=4\times {10}^{-6}{\mathrm{m}}^2$
• The stretched length is,$∆L=1.2\mathrm{mm}=0.0012\mathrm{m}$ .
• Density of nickel is$\mathrm{\rho }=8.9\mathrm{g}/{\mathrm{cm}}^3=8.9\times {10}^3\mathrm{kg}/{\mathrm{m}}^3$
• The mass of one mole of nickel is$m_I=59\mathrm{g}$

The speed of sound will give the required time. Use the spring model to get the equation for the speed of sound.

The speed of sound in a solid object depends upon the mass of the atom and stiffness of the metal.

The mass of the atom is expressed as,

$m_a=\frac{m_I}{N_A}$

Here,$m_I$is the mass of one mole,$N_A$is the Avagadro’s Number.

Substitute 0.059kg for$m_I$ and$6.02\times {10}^{23}\mathrm{atom}$ for$N_A$ in the above equation.

$$\begin{gathered} m_a=\frac{m_I}{N_A} \\ =\frac{0.059\mathrm{kg}}{6.02\times {10}^{23}\mathrm{atom}} \\ =9.79\times {10}^{-26}\mathrm{kg}/\mathrm{atom} \end{gathered}$$

The density is given by,

$$\begin{gathered} \rho =\frac{m_a}{d^3} \\ d=\sqrt[3]{\frac{m_a}{\rho }} \end{gathered}$$

Substitute$9.79\times {10}^{-26}\mathrm{kg}/\mathrm{atom}$ for$m_a$ and$8.9\times {10}^3\mathrm{kg}/{\mathrm{m}}^3$ for$\rho$ in the above equation.

$$\begin{gathered} d=\sqrt[3]{\frac{m_a}{\rho }} \\ =\sqrt[3]{\frac{9.79\times {10}^{26}kg}{8.9\times {10}^{3}kg/m^3}} \\ =2.2\times {10}^{-10}\mathrm{m} \end{gathered}$$

The equation of young’s modulus is expressed as,

$$\begin{gathered} Y=\frac{stress}{strain} \\ Y=\frac{F/A}{∆L/L} \end{gathered}$$

Here, F is the force exerted due to weight of the ball,A is the cross sectional area of the rod,$∆L$ is the stretched length of the rod and L is the length of the rod.

Substitute mg for F in the above equation,

$$\begin{gathered} Y=\frac{F/A}{∆L/L} \\ =\frac{{\displaystyle \frac{mg}{A}}}{{\displaystyle \frac{∆L}{L}}} \end{gathered}$$

Substitute 40kg for m,$9.8m/s^2$ for g,$4\times {10}^{-6}{m}^2$ for A,0.0012m for$∆L$ and 2.5m forL in the above equation.

$$\begin{gathered} E=\frac{{\displaystyle \frac{\left(40kg\right)\left(9.81m/s^2\right)}{4\times {10}^{-6}{m}^2}}}{0.0012\mathrm{m}/2.5\mathrm{m}} \\ =204\times {10}^9\mathrm{N}/{\mathrm{m}}^2 \end{gathered}$$

Thus, the value of the young’s modulus is$204\times {10}^9\mathrm{N}/{\mathrm{m}}^2$.

Now, young’s modulus is related to interatomic stiffness by

role="math" localid="1658900470516" $k_{s,i}=Yd$

Here, is the young’s modulus, and is the diameter of atom

Find the value of interatomic stiffness using the formula below as:

$$\begin{gathered} k_{s,i}=Yd \\ =\left(204\times {10}^9\mathrm{N}/{\mathrm{m}}^2\right)\left(2.22\times {10}^{-10}\mathrm{m}\right) \\ =45.29\mathrm{N}/\mathrm{m} \end{gathered}$$

Thus, the magnitude of the inter-atomic stiffness is 45/29N/m .

The equation of speed of sound is expressed as

$v=\sqrt{\frac{k_{s,i}}{m_a}d}$

Here,$k_{s,i}$is the inter-atomic bond stiffness of the metal,$m_a$is the mass of the atom and $d$is the diameter of the atom.

Substitute$45.29\mathrm{N}/\mathrm{m}\mathrm{for}{\mathrm{k}}_{\mathrm{s},\mathrm{i}},9.79\times {10}^{-26}\mathrm{kg}\mathrm{for}{\mathrm{m}}_{\mathrm{a}}\mathrm{and}2.22\times {10}^{-10}\mathrm{m}$ ford in the above equation.

$$\begin{gathered} v=\sqrt{\frac{k_{s,i}}{m_a}}d \\ =\sqrt{\frac{45.29{\displaystyle \frac{N}{m}}}{9.79\times {10}^{-26}kg}}\left(2.22\times {10}^{-10}\mathrm{m}\right) \\ =4774\mathrm{m}/\mathrm{s} \end{gathered}$$

Length of the rod is considered as distance travelled by the sound wave.

Obtain the travel time as ratio of distance travelled by sound wave and speed of sound

$$\begin{gathered} t=\frac{L}{v} \\ t=\frac{2.5\mathrm{m}}{4475{\displaystyle \frac{\mathrm{m}}{\mathrm{s}}}} \\ t=0.523\mathrm{ms} \end{gathered}$$

Thus, the timeT that will elapse before a microphone at the other end of the bar will detect a disturbance is 0.523 ms .