Chapter 11, Problem 42P

(a)

To determine

To Find: The speed of the pulse.

Answer to Problem 42P

The speed of the pulse is $77.5\text{ m/s}$

Explanation of Solution

Given:

Length of the steel cable, $L=620\text{ m }$

Distance, $\Delta x=2\times L$

Diameter of steel cable, $d=1.5\text{ cm }=0.015\text{ m}$

Time taken to return, $\Delta t=16\text{ s}$

Formula used:

Speed of the wave pulse is obtained by the following formula

  $v=\frac{\Delta x}{\Delta t}$

Here,

  $v$ is speed of wave pulse.

  $\Delta x$ is the distance travelled by the pulse.

and $\Delta t$ is the time taken by the pulse to return.

Calculation:

It is observed that the wave pulse travels twice the length of the cable. Substitute the given values in the above equation,

  $v=\frac{2\times L}{\Delta t}=\frac{2\times 620}{16}=77.5\text{ m/s }$

Conclusion:

Thus, the speed of the pulse is $77.5\text{ m/s}$ .

(b)

To determine

To Find: The tension in the cable.

Answer to Problem 42P

The tension in the cable is $8.3\times {10}^3\text{ N}$

Explanation of Solution

Given:

Length of the steel cable, $L=620\text{ m }$

Diameter of steel cable, $d=1.5\text{ cm }=0.015\text{ m}$

Speed of the wave pulse, $v=77.5\text{ m/s}$

Density of steel, $\rho =7.8\times {10}^3{\text{kg/m}}^{\text{3}}$

Formula used:

The tension in the cable is calculated by the following formula,

  $v=\sqrt{\frac{T}{\left(\frac{m}{L}\right)}}$

Here,

  $T$ is tension in the cord,

  $m$ is the mass of cord,

  $L$ is the length of cord.

The expression for the density.

  $\rho =\frac{\text{mass}}{\text{Volume}}=\frac{m}{A\times L}=\frac{m}{L}\times \frac{1}{A}$

The expression for the Area of cross-section of steel cable.

Area, $A=\frac{\pi }{4}\times d^2$

Calculation:

Calculate the area of cross-section of the cable.

  $\begin{array}{c}\text{Area }A=\frac{\pi }{4}\times d^2\\ =\frac{\pi }{4}\times {0.015}^2\\ =1.77\times {10}^{-4}{\text{ m}}^{\text{2}}\end{array}$

Calculate the ratio of $\frac{m}{L}$ using the density equation.

  $\begin{array}{l}\rho =\frac{m}{L}\times \frac{1}{1.77\times {10}^{-4}}\\ \frac{m}{L}=7.8\times {10}^3\times 1.77\times {10}^{-4}=1.38\text{ kg/m}\end{array}$

Calculate the value of T using the velocity equation.

  $\begin{array}{l}v=\sqrt{\frac{T}{1.38}}\\ $ T=1.38\times v^2=1.38\times {(77.5)}^2=8.3\times {10}^3\text{ N}$\end{array}$

Conclusion:

Thus, the tension in the cable is $8.3\times {10}^3\text{ N}$ .