(a) What is the fastest transverse wave that can be sent along a steel wire? For safety reasons, the maximum tensile stress to which steel wires should be subjected is $\mathbf{7}\mathbf{.}\mathbf{00}\mathit{x}{\mathbf{10}}^{\mathbf{18}\mathbf{}}\mathit{N}\mathbf{/}{\mathit{m}}^{\mathbf{2}}$.The density of steel is$\mathbf{7800}\mathbf{}\mathit{k}\mathit{g}\mathbf{/}{\mathit{m}}^{\mathbf{3}}$. (b) Does your answer depend on the diameter of the wire?

Tensile stress is$T_{stress}=7.00\times {10}^{8}N/m^2$

Density of steel is$\mathrm{\rho }=7800\mathrm{kg}/{\mathrm{m}}^3$

We use the concept of velocity of a transverse wave. We use the relation between density, area, and linear mass density. Then plugging the values in the equation of wave velocity we can find the maximum velocity that can be sent through the steel.

Formulae:

The velocity of the string, $v=\sqrt{\frac{T}{\mu }}.......\left(1\right)$

The linear density of the string, role="math" localid="1660983986609" $\mu =\rho A.......\left(2\right)$

We know tensile stress is tension per unit area that is given as:

$$\begin{gathered} T_{stress}=\frac{T}{A} \\ T=T_{stress}A \end{gathered}$$

Substituting the value of tension and linear density from equation (2) in equation (1), we get the maximum velocity as given:

$$\begin{gathered} v_{max}=\sqrt{\frac{T_{stress}A}{\rho A}} \\ =\sqrt{\frac{T_{stress}}{\rho }} \\ =\sqrt{\frac{\left(7.00\times {10}^8\right)}{7800}} \\ =300m/s \end{gathered}$$

Hence, the value of maximum velocity is 300 m/s

The above equation is independent of the area, so the answer doesn’t depend on the diameter of the wire.