Chapter 17, Problem 68AP

To determine

The ratio of $\frac{L_1}{L_2}$ for the wave to travel the length of rod 1 and 2, in the same time interval required to travel through rod 3.

Answer to Problem 68AP

The ratio of $\frac{L_1}{L_2}$ for the wave is the $\underset{\_}{6.45}$.

Explanation of Solution

The time taken by the wave to travel a length $L$ is.

  $t=\frac{L}{\upsilon }$                                                                                                                          (I)

Here, $L$ is the length of the rod, and $\upsilon$ is the speed of the wave.

Substitute, $\sqrt{Y/\rho }$ for $\upsilon$ in equation (I).

  $t=\frac{L}{\sqrt{Y/\rho }}$                                                                                                                 (II)

Use the above equation (II) to find the time taken to travel through each rod.

The time taken to travel through rod 1.

  $t_1=\frac{L_1}{\sqrt{Y_1/{\rho }_1}}$                                                                                                              (III)

The time taken to travel through rod 2.

  $t_2=\frac{L_2}{\sqrt{Y_2/{\rho }_2}}$                                                                                                           (IV)

Substitute. $L_3-L_1$ for $L_2$ in equation (IV).

  $t_2=\frac{L_3-L_1}{\sqrt{Y_2/{\rho }_2}}$                                                                                                            (V)

The time taken to travel through rod 3.

  $t_3=\frac{L_3}{\sqrt{Y_3/{\rho }_3}}$                                                                                                           (VI)

The required time is.

  $t_1+t_2=t_3$                                                                                                               (VII)

Conclusion:

Substitute, $2.70\times {10}^3\text{kg}/{\text{m}}^3$ for ${\rho }_1$, and $7.00\times {10}^{10}\text{N}/{\text{m}}^2$ for $Y_1$ in equation (III).

  $\begin{array}{c}{t}_1=\frac{L_1}{\sqrt{7.00\times {10}^{10}\text{N}/{\text{m}}^2/2.70\times {10}^3\text{kg}/{\text{m}}^3}}\\ =L_1\left(1.96\times {10}^{-4}\text{s}/\text{m}\right)\end{array}$

Substitute, $\left(1.50\text{m}-L_1\right)$ for $L_2$, $11.3\times {10}^3\text{kg}/{\text{m}}^3$ for ${\rho }_2$, and $1.60\times {10}^{10}\text{N}/{\text{m}}^2$ for $Y_2$ in equation (III).

  $\begin{array}{c}{t}_2=\frac{\left(1.50\text{m}-L_1\right)}{\sqrt{\left(11.3\times {10}^3\text{N}/{\text{m}}^2/1.60\times {10}^{10}\text{kg}/{\text{m}}^3\right)}}\\ =1.26\times {10}^{-3}-L_1\left(8.40\times {10}^{-4}\text{s}/\text{m}\right)\end{array}$

Substitute, $8.80\times {10}^3\text{kg}/{\text{m}}^3$ for ${\rho }_2$, and $11.0\times {10}^{10}\text{N}/{\text{m}}^2$ for $Y_2$ in equation (III).

  $\begin{array}{c}{t}_3=\frac{1.50\text{m}}{\sqrt{11.0\times {10}^{10}\text{N}/{\text{m}}^2/8.80\times {10}^3\text{kg}/{\text{m}}^3}}\\ =4.24\times {10}^{-4}\text{s}\end{array}$

Substitute, $L_1\left(1.96\times {10}^{-4}\text{s}/\text{m}\right)$ for $t_1$, $1.26\times {10}^{-3}-L_1\left(8.40\times {10}^{-4}\text{s}/\text{m}\right)$, and $4.24\times {10}^{-4}\text{s}$ for $t_3$ in equation (VII).

  $\begin{array}{c}{L}_1\left(1.96\times {10}^{-4}\text{s}/\text{m}\right)+1.26\times {10}^{-3}-L_1\left(8.40\times {10}^{-4}\text{s}/\text{m}\right)=4.24\times {10}^{-4}\text{s}\\ L_1=1.30\text{m}\end{array}$

Since, $L_2$ is equal to $L_3-L_1$.

  $\begin{array}{c}{L}_2=1.50\text{m}-1.30\text{m}\\ \text{=0}\text{.20}\text{m}\end{array}$

Take the ratio of length’s.

  $\begin{array}{c}\frac{L_1}{L_2}=\frac{1.30\text{m}}{0.201\text{m}}\\ =6.45\end{array}$

Therefore, the ratio of $\frac{L_1}{L_2}$ for the wave is the $\underset{\_}{6.45}$.