Chapter 17, Problem 17.59AP

(a)

To determine

The speed of the one dimensional compression wave.

Answer to Problem 17.59AP

The speed of the one dimensional compression wave is $5.04\times {10}^3\text{m}/\text{s}$.

Explanation of Solution

Given info: The young’s modulus of steel is $20\times {10}^{10}\text{N}/{\text{m}}^{\text{2}}$, the density of the steel is $7.86\times {10}^3\text{kg}/{\text{m}}^{\text{3}}$, the yield strength of the steel is $400\text{MPa}$, the length of the rod is $80.0\text{cm}$ and the speed of the rod is $12.0\text{m}/\text{s}$.

Write the expression to calculate the speed of the one dimensional compression wave.

$v=\sqrt{\frac{Y}{\rho }}$

Here,

$Y$ is the young’s modulus of steel.

$\rho$ is the density of the steel.

Substitute $20\times {10}^{10}\text{N}/{\text{m}}^{\text{2}}$ for $Y$ and $7.86\times {10}^3\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$ in above expression.

$\begin{array}{c}v=\sqrt{\frac{20\times {10}^{10}\text{N}/{\text{m}}^{\text{2}}}{7.86\times {10}^3\text{kg}/{\text{m}}^{\text{3}}}}\\ =5.04\times {10}^3\text{m}/\text{s}\end{array}$

Conclusion:

Therefore the speed of the one dimensional compression wave is $5.04\times {10}^3\text{m}/\text{s}$.

(b)

To determine

The time interval of the wave.

Answer to Problem 17.59AP

The time interval of the wave is $1.59\times {10}^{-4}\text{s}$.

Explanation of Solution

Given info: The young’s modulus of steel is $20\times {10}^{10}\text{N}/{\text{m}}^{\text{2}}$, the density of the steel is $7.86\times {10}^3\text{kg}/{\text{m}}^{\text{3}}$, the yield strength of the steel is $400\text{MPa}$, the length of the rod is $80.0\text{cm}$ and the speed of the rod is $12.0\text{m}/\text{s}$.

Write the expression to calculate time interval of the wave.

$t=\frac{L}{v}$

Here,

$L$ is the length of the rod.

Substitute $5.04\times {10}^3\text{m}/\text{s}$ for $v$ and $80.0\text{cm}$ for $L$ in above expression.

$\begin{array}{c}t=\frac{80.0\text{cm}\times \frac{1\text{m}}{100\text{cm}}}{5.04\times {10}^3\text{m}/\text{s}}\\ =1.59\times {10}^{-4}\text{s}\end{array}$

Conclusion:

Therefore the time interval of the wave is $1.59\times {10}^{-4}\text{s}$.

(c)

To determine

The distance of the travel by the back end of the rod.

Answer to Problem 17.59AP

The distance of the travel by the back end of the rod is $1.90\times {10}^{-3}\text{m}$.

Explanation of Solution

Given info: The young’s modulus of steel is $20\times {10}^{10}\text{N}/{\text{m}}^{\text{2}}$, the density of the steel is $7.86\times {10}^3\text{kg}/{\text{m}}^{\text{3}}$, the yield strength of the steel is $400\text{MPa}$, the length of the rod is $80.0\text{cm}$ and the speed of the rod is $12.0\text{m}/\text{s}$.

The expression for the distance of the travel by the back end of the rod.

$d=v_{\text{r}}t$

Here,

$v_{\text{r}}$ is the speed of the rod.

Substitute $12.0\text{m}/\text{s}$ for $v_{\text{r}}$ and $1.59\times {10}^{-4}\text{s}$ for $t$ in above expression.

$\begin{array}{c}d=12.0\text{m}/\text{s}\times 1.59\times {10}^{-4}\text{s}\\ =1.90\times {10}^{-3}\text{m}\end{array}$

Conclusion:

Therefore the distance of the travel by the back end of the rod is $1.90\times {10}^{-3}\text{m}$.

(d)

To determine

The strain in the rod.

Answer to Problem 17.59AP

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

Explanation of Solution

Given info: The young’s modulus of steel is $20\times {10}^{10}\text{N}/{\text{m}}^{\text{2}}$, the density of the steel is $7.86\times {10}^3\text{kg}/{\text{m}}^{\text{3}}$, the yield strength of the steel is $400\text{MPa}$, the length of the rod is $80.0\text{cm}$ and the speed of the rod is $12.0\text{m}/\text{s}$.

The expression for the strain in the rod.

$\epsilon =\frac{\Delta L}{L}$

Here,

$\Delta L$ is the change in the length.

Substitute $1.90\times {10}^{-3}\text{m}$ for $\Delta L$ and $80.0\text{cm}$ for $L$ in above expression.

$\begin{array}{c}\epsilon =\frac{1.90\times {10}^{-3}\text{m}}{80.0\text{cm}\times \frac{1\text{m}}{100\text{cm}}}\\ =2.38\times {10}^{-3}\end{array}$

Conclusion:

Therefore the strain in the rod is $2.38\times {10}^{-3}$.

(e)

To determine

The stress in the rod.

Answer to Problem 17.59AP

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

Explanation of Solution

Given info: The young’s modulus of steel is $20\times {10}^{10}\text{N}/{\text{m}}^{\text{2}}$, the density of the steel is $7.86\times {10}^3\text{kg}/{\text{m}}^{\text{3}}$, the yield strength of the steel is $400\text{MPa}$, the length of the rod is $80.0\text{cm}$ and the speed of the rod is $12.0\text{m}/\text{s}$.

The expression for the stress in the rod is

$\sigma =Y\epsilon$

Substitute $20\times {10}^{10}\text{N}/{\text{m}}^{\text{2}}$ for $Y$ and $2.38\times {10}^{-3}$ for $\epsilon$ in above expression.

$\begin{array}{c}\sigma =20\times {10}^{10}\text{N}/{\text{m}}^{\text{2}}\times 2.38\times {10}^{-3}\\ =4.76\times {10}^8\text{N}/{\text{m}}^{\text{2}}\end{array}$

Conclusion:

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

(f)

To determine

The maximum impact speed of the rod in terms of ${\sigma }_{\text{y}}$, $Y$ and $\rho$.

Answer to Problem 17.59AP

The maximum impact speed of the rod in terms of ${\sigma }_{\text{y}}$, $Y$ and $\rho$ is $\frac{{\sigma }_{\text{y}}}{\sqrt{\rho Y}}$.

Explanation of Solution

Given info: The young’s modulus of steel is $20\times {10}^{10}\text{N}/{\text{m}}^{\text{2}}$, the density of the steel is $7.86\times {10}^3\text{kg}/{\text{m}}^{\text{3}}$, the yield strength of the steel is $400\text{MPa}$, the length of the rod is $80.0\text{cm}$ and the speed of the rod is $12.0\text{m}/\text{s}$.

The expression for time is,

$t=\frac{L}{v}$

Substitute $\sqrt{\frac{Y}{\rho }}$ for $v$ in above expression.

$t=\frac{L}{\sqrt{\frac{Y}{\rho }}}$

Thus the time is $\frac{L}{\sqrt{\frac{Y}{\rho }}}$.

The expression for change in length is,

$\Delta L=\frac{{\sigma }_{\text{y}}L}{Y}$

The expression for the maximum impact speed of the rod is,

$v_{{\max }_{\text{i}}}=\frac{\Delta L}{t}$

Substitute $\frac{{\sigma }_{\text{y}}L}{Y}$ for $\Delta L$ and $\frac{L}{\sqrt{\frac{Y}{\rho }}}$ for $t$ in above expression.

$\begin{array}{c}{v}_{{\max }_{\text{i}}}=\frac{\frac{{\sigma }_{\text{y}}L}{Y}}{\frac{L}{\sqrt{\frac{Y}{\rho }}}}\\ =\frac{{\sigma }_{\text{y}}}{\sqrt{\rho Y}}\end{array}$

Conclusion:

Therefore the maximum impact speed of the rod in terms of ${\sigma }_{\text{y}}$, $Y$ and $\rho$ is $\frac{{\sigma }_{\text{y}}}{\sqrt{\rho Y}}$.