Chapter 17, Problem 59AP

(a)

To determine

The speed of the compressional wave.

Answer to Problem 59AP

The speed of the compressional wave is $\underset{\_}{5.04\times {10}^3\text{m}/\text{s}}$.

Explanation of Solution

Write the expression for compressional wave.

  $\upsilon =\sqrt{\frac{Y}{\rho }}$                                                                                                                     (I)

Here, $\upsilon$ is the speed of the compressional wave, $Y$ is the Young’s modulus, and $\rho$ is the density of the medium.

Conclusion:

Substitute, $20.0\times {10}^{10}\text{N}/{\text{m}}^2$ for $Y$, and $7860\text{kg}/{\text{m}}^3$ for $\rho$ in equation (I).

  $\begin{array}{c}\upsilon =\sqrt{\frac{20.0\times {10}^{10}\text{N}/{\text{m}}^2}{7860\text{kg}/{\text{m}}^3}}\\ =5.04\times {10}^3\text{m}/\text{s}\end{array}$

Therefore, the speed of the compressional wave is $\underset{\_}{5.04\times {10}^3\text{m}/\text{s}}$.

(b)

To determine

The time taken by the back end of the rod to come to stop its motion.

Answer to Problem 59AP

The time taken by the back end of the rod to come to stop its motion is $\underset{\_}{1.59\times {10}^{-4}\text{s}}$.

Explanation of Solution

Write the expression for the time taken by the signal to stop to reach at the back end.

  $\Delta t=\frac{L}{\upsilon }$                                                                                                                     (II)

Here, $L$ is the length of the rod.

Conclusion:

Substitute, $0.800\text{m}$ for $L$, and $5.04\times {10}^3\text{m}/\text{s}$ for $\upsilon$ in equation (II).

  $\begin{array}{c}\Delta t=\frac{0.800\text{m}}{5.04\times {10}^3\text{m}/\text{s}}\\ =1.59\times {10}^{-4}\text{s}\end{array}$

Therefore, the time taken by the back end of the rod to come to stop its motion is $\underset{\_}{1.59\times {10}^{-4}\text{s}}$.

(c)

To determine

The distance moved by the back end of the rod at time $\Delta t$.

Answer to Problem 59AP

The distance moved by the back end of the rod at time $\Delta t$ is $\underset{\_}{1.90\text{mm}}$.

Explanation of Solution

Let the velocity with which the back end of the rod moving be ${\upsilon }_i$ and it is equal to $12.0\text{m}/\text{s}$.

Write the equation for distance moved by the back end of the rod.

  $\Delta L={\upsilon }_i\Delta t$                                                                                                               (III)

Conclusion:

Substitute, $12.0\text{m}/\text{s}$ for ${\upsilon }_i$, and $1.59\times {10}^{-4}\text{s}$ for $\Delta t$ in equation (III).

  $\begin{array}{c}\Delta L=\left(12.0\text{m}/\text{s}\right)\left(1.59\times {10}^{-4}\text{s}\right)\\ =1.90\text{mm}\end{array}$

Therefore, the distance moved by the back end of the rod at time $\Delta t$ is $\underset{\_}{1.90\text{mm}}$.

(d)

To determine

The strain of the rod.

Answer to Problem 59AP

The strain of the rod is $\underset{\_}{2.38\times {10}^{-3}}$.

Explanation of Solution

Strain defined as the change in dimension by original dimension.

Write the expression for strain.

  $s=\frac{\Delta L}{L}$                                                                                                                    (IV)

Here, $\Delta L$ is the change in length, and $L$ is the original length of the rod.

Conclusion:

Substitute, $1.90\text{mm}$ for $\Delta L$, and $0.800\text{m}$ for $L$ in equation (IV).

  $\begin{array}{c}s=\frac{1.90\text{mm}\times \frac{{10}^{-3}\text{m}}{\text{1}\text{mm}}}{0.800\text{m}}\\ =2.38\times {10}^{-3}\end{array}$

Therefore, the strain of the rod is $\underset{\_}{2.38\times {10}^{-3}}$.

(e)

To determine

The stress of the rod.

Answer to Problem 59AP

The stress of the rod is $\underset{\_}{4.76\times {10}^8\text{N}/{\text{m}}^2}$.

Explanation of Solution

Young’s modulus is the ratio of stress by strain. From the known values of young’s modulus and strain, stress can be determined.

Write the expression for the stress of the rod.

  $\sigma =Y\left(\frac{\Delta L}{L}\right)$                                                                                                             (V)

Conclusion:

Substitute, $20.0\times {10}^{10}\text{N}/{\text{m}}^2$ for $Y$, and $2.38\times {10}^{-3}$ for $\frac{\Delta L}{L}$ in equation (V).

  $\begin{array}{c}\sigma =20.0\times {10}^{10}\text{N}/{\text{m}}^2\left(2.38\times {10}^{-3}\right)\\ =4.76\times {10}^8\text{N}/{\text{m}}^2\end{array}$

Therefore, the stress of the rod is $\underset{\_}{4.76\times {10}^8\text{N}/{\text{m}}^2}$.

(f)

To determine

The maximum impact speed of the rod.

Answer to Problem 59AP

The maximum impact speed of the rod is $\underset{\_}{\frac{{\sigma }_y}{\sqrt{\rho Y}}}$.

Explanation of Solution

The expression for the speed of the wave is.

  $\upsilon =\sqrt{\frac{Y}{\rho }}$                                                                                                                  (VI)

Even if the front end strikes on wall, the back end will be in motion, and the time taken for the forward motion is.

  $\Delta t=\frac{L}{\upsilon }$                                                                                                                  (VII)

Substitute equation (VI) in (VII).

  $\Delta t=L\sqrt{\frac{\rho }{Y}}$                                                                                                            (VIII)

The distance traveled at time $\Delta t$ is.

  $\Delta L={\upsilon }_i\Delta t$                                                                                                               (IX)

The strain of the rod is.

  $s=\frac{\Delta L}{L}$                                                                                                                   (IX)

Substitute, equation (VIII) in (IX).

  $s=\frac{{\upsilon }_i\Delta t}{L}$                                                                                                                    (X)

Substitute, equation (VIII) in (X).

  $\begin{array}{c}s=\frac{\Delta L}{L}\\ ={\upsilon }_i\sqrt{\frac{\rho }{Y}}\end{array}$                                                                                                                (XI)

The stress of the rod is.

  $\sigma =Y\left(\frac{\Delta L}{L}\right)$                                                                                                          (XII)

Substitute, equation (XI) in (XII).

  $\begin{array}{c}\sigma =Y\left({\upsilon }_i\sqrt{\frac{\rho }{Y}}\right)\\ ={\upsilon }_i\sqrt{\rho Y}\end{array}$                                                                                                      (XIII)

From equation (XIII) the expression for maximum speed, if the above stress is less than the yield stress is.

  ${\upsilon }_{\text{max}}=\frac{{\sigma }_y}{\sqrt{\rho Y}}$

Conclusion:

Therefore, the maximum impact speed of the rod is $\underset{\_}{\frac{{\sigma }_y}{\sqrt{\rho Y}}}$.