A hanging titanium wire with diameter$\mathbf{2}\mathbf{}\mathbf{mm}$($\mathbf{2}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathbf{}\mathbf{m}$) is initially $\mathbf{3}\mathbf{}\mathbf{m}$long. When a $\mathbf{5}\mathbf{}\mathbf{kg}$mass is hung from it, the wire stretches an amount $\mathbf{0}\mathbf{.}\mathbf{4035}\mathbf{}\mathbf{mm}$, and when a $\mathbf{10}\mathbf{}\mathbf{kg}$mass is hung from it, the wire stretches an amount $\mathbf{0}\mathbf{.}\mathbf{807}\mathbf{}\mathbf{mm}$. A mole of titanium has a mass of $\mathbf{48}\mathbf{g}$, and its density is $\mathbf{4}\mathbf{.}\mathbf{51}\mathbf{g}\mathbf{/}{\mathbf{cm}}^{\mathbf{3}}$. Find the approximate value of the effective spring stiffness of the interatomic force, and explain your analysis .

The given data is listed below as,

$$\begin{gathered} \mathrm{The}\mathrm{given}\mathrm{data}\mathrm{is}\mathrm{listed}\mathrm{below}\mathrm{as}, \\ •\mathrm{The}\mathrm{diameter}\mathrm{of}\mathrm{the}\mathrm{titanium}\mathrm{wire}\mathrm{is},{\mathrm{a}}_1=2\mathrm{mm} \\ •\mathrm{The}\mathrm{length}\mathrm{of}\mathrm{the}\mathrm{titanium}\mathrm{wire}\mathrm{is}\mathrm{initially},{\mathrm{a}}_2=3\mathrm{m} \\ •\mathrm{The}\mathrm{first}\mathrm{mass}\mathrm{hung}\mathrm{from}\mathrm{the}\mathrm{titanium}\mathrm{wire}\mathrm{is},{\mathrm{m}}_1=5\mathrm{kg} \\ •\mathrm{The}\mathrm{stretching}\mathrm{length}\mathrm{of}\mathrm{the}\mathrm{wire}\mathrm{initially}\mathrm{is}{\mathrm{x}}_1=0.4035\mathrm{mm} \\ •\mathrm{The}\mathrm{second}\mathrm{mass}\mathrm{hung}\mathrm{from}\mathrm{the}\mathrm{titanium}\mathrm{wire}\mathrm{is},{\mathrm{m}}_2=10\mathrm{kg} \\ •\mathrm{The}\mathrm{stretching}\mathrm{length}\mathrm{of}\mathrm{the}\mathrm{wire}\mathrm{finally}\mathrm{is}{\mathrm{x}}_2=0.807\mathrm{mm} \\ •\mathrm{The}\mathrm{mass}\mathrm{of}\mathrm{one}\mathrm{mole}\mathrm{of}\mathrm{titanium}\mathrm{is},\mathrm{m}=48\mathrm{g} \\ •\mathrm{The}\mathrm{density}\mathrm{of}\mathrm{one}\mathrm{mole}\mathrm{of}\mathrm{titanium}\mathrm{is},\mathrm{d}=4.51\mathrm{g}/{\mathrm{cm}}^3 \end{gathered}$$

The stiffness of the spring is directly proportional to the force exerted and inversely proportional to the displacement of the wire.

The equation of the stiffness of the spring gives the effective spring stiffness of the titanium wire.

For the 5 kg mass,

The equation of the force is expressed as:

$\mathrm{F}={\mathrm{m}}_1\mathrm{g}$

Here, ${\mathrm{m}}_1$is the mass of the first object and g is the acceleration due to gravity and .

$\mathrm{g}=9.81\mathrm{m}/{\mathrm{s}}^2$

Substitute all the values in the above expression.

$$\begin{gathered} \mathrm{F}=5\mathrm{kg}\times 9.8\mathrm{m}/{\mathrm{s}}^2 \\ =49\mathrm{kg}.\mathrm{m}/{\mathrm{s}}^2 \\ =49\mathrm{kg}.\mathrm{m}/{\mathrm{s}}^2\times \frac{1\mathrm{N}}{1\mathrm{kg}.\mathrm{m}/{\mathrm{s}}^2} \\ =49\mathrm{N} \end{gathered}$$

For the 10 Kg mass,

The equation of the force is expressed as:

$\mathrm{F}={\mathrm{m}}_2\mathrm{g}$

Here, ${\mathrm{m}}_2$is the mass of the second object and g is the acceleration due to gravity

Substitute all the values in the above expression.

$$\begin{gathered} \mathrm{F}=10\mathrm{kg}\times 9.8\mathrm{m}/{\mathrm{s}}^2 \\ =98\mathrm{kg}.\mathrm{m}/{\mathrm{s}}^2 \\ =98\mathrm{kg}.\mathrm{m}/{\mathrm{s}}^2\times \frac{1\mathrm{N}}{1\mathrm{kg}.\mathrm{m}/{\mathrm{s}}^2} \\ =98\mathrm{N} \end{gathered}$$

For the 5 kg mass, the equation of the stiffness of the spring can be expressed as:

$\mathrm{k}=\frac{\mathrm{F}}{{\mathrm{x}}_1}$

Substituting the values in the above equation.

$$\begin{gathered} \mathrm{k}=\frac{49\mathrm{N}}{0.4035\mathrm{mm}\times \left({\displaystyle \frac{{10}^{-3}\mathrm{m}}{1\mathrm{m}\mathrm{m}}}\right)} \\ =\frac{49\mathrm{N}}{4.035\times {10}^{-3}\mathrm{m}} \\ =12143.74\mathrm{N}/\mathrm{m}\times \left(\frac{{10}^{-3}\mathrm{kN}}{1\mathrm{N}}\right) \\ =12.14\mathrm{kN}/\mathrm{m} \end{gathered}$$

For the 10 mass, the equation of the stiffness of the spring can be expressed as:

$k=\frac{F}{x_2}$

Here, F is the force exerted and $x_2$is the displacement of the particle by the second mass.

Substituting the values in the above equation.

$$\begin{gathered} \mathrm{k}=\frac{98\mathrm{N}}{0.807\mathrm{m}\mathrm{m}\times \left({\displaystyle \frac{{10}^{-3}\mathrm{m}}{1\mathrm{m}\mathrm{m}}}\right)} \\ =\frac{98\mathrm{N}}{8.07\times {10}^{-4}\mathrm{m}} \\ =121437.4\mathrm{N}/\mathrm{m}\times \left(\frac{{10}^{-3}\mathrm{kN}}{1\mathrm{N}}\right) \\ =121.43\mathrm{k}\mathrm{N}/\mathrm{m} \end{gathered}$$

Thus, the value of the effective spring stiffness of the interatomic force of the 5 kg and the 10 kg mass are $12.14\mathrm{kN}/\mathrm{m}\mathrm{and}121.43\mathrm{kN}/\mathrm{m}$respectively.