Chapter 17, Problem 66AP

(a)

To determine

The sound which arrives first.

Answer to Problem 66AP

The sound through the metal arrives first.

Explanation of Solution

When a hammer is used at one end, the sound can travel through both metal rod and air. The speed of the wave through metal rod is $3560\text{m}/\text{s}$, and that of speed of the sound in air is $343\text{m}/\text{s}$.

Since, speed of the sound in metal is greater than the sped of the sound in air, the sound through metal arrives first.

Conclusion:

Therefore, the sound through the metal arrives first.

(b)

To determine

The length of the rod as a function of $\Delta t$.

Answer to Problem 66AP

The length of the rod as a function of $\Delta t$ is $\underset{\_}{380\Delta t}$.

Explanation of Solution

Write the expression for the time taken by the sound to travel.

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

Here, $L$ is the length of the rod, $\upsilon$ is the speed, and $t$ is the time taken.

The delay between the arrival of pulses through copper and air is.

  $\begin{array}{c}\Delta t=L\left(\frac{1}{{\upsilon }_{\text{air}}}-\frac{1}{{\upsilon }_{\text{cu}}}\right)\\ =L\frac{{\upsilon }_{\text{cu}}-{\upsilon }_{\text{air}}}{{\upsilon }_{\text{air}}{\upsilon }_{\text{cu}}}\end{array}$                                                                                                   (II)

Here, ${\upsilon }_{\text{air}}$ is the speed of the sound in air, ${\upsilon }_{\text{cu}}$ is the speed of the sound in copper rod.

Rearrange equation (II) to obtain an expression for $L$.

  $L=\frac{{\upsilon }_{\text{air}}{\upsilon }_{\text{cu}}}{{\upsilon }_{\text{cu}}-{\upsilon }_{\text{air}}}\Delta t$                                                                                                      (III)

Conclusion:

Substitute, $343\text{m}/\text{s}$ for ${\upsilon }_{\text{air}}$, and $3560\text{m}/\text{s}$ for ${\upsilon }_{\text{cu}}$ in equation (III).

  $\begin{array}{c}L=\frac{\left(343\text{m}/\text{s}\right)\left(3560\text{m}/\text{s}\right)}{\left(3560\text{m}/\text{s}-343\text{m}/\text{s}\right)}\Delta t\\ =380\Delta t\end{array}$

Therefore, length of the rod as a function of $\Delta t$ is $\underset{\_}{380\Delta t}$.

(c)

To determine

The length of the rod if $\Delta t=127\text{ms}$.

Answer to Problem 66AP

The length of the rod if $\Delta t=127\text{ms}$ is $\underset{\_}{48.2\text{m}}$.

Explanation of Solution

Write the expression for the length of the rod.

  $L=380\Delta t$                                                                                                              (IV)

Conclusion:

Substitute, $127\text{ms}$ for $\Delta t$ in equation (IV).

  $\begin{array}{c}L=\left(380\text{m}/\text{s}\right)\left(127\text{ms}\times \frac{{10}^{-3}\text{s}}{1\text{ms}}\right)\\ =48.2\text{m}\end{array}$

Therefore, the length of the rod if $\Delta t=127\text{ms}$ is $\underset{\_}{48.2\text{m}}$.

(d)

To determine

The length of the rod in terms of $t$ and ${\upsilon }_r$.

Answer to Problem 66AP

The length of the rod in terms of $t$ and ${\upsilon }_r$ is $\underset{\_}{L=\frac{\Delta t}{\frac{1}{343\text{m}/\text{s}}-\frac{1}{{\upsilon }_r}}}$.

Explanation of Solution

The speed of the sound in rod is given as ${\upsilon }_r$. The speed of the sound in air is $343\text{m}/\text{s}$.

Substitute, ${\upsilon }_r$ for ${\upsilon }_{\text{cu}}$, $343\text{m}/\text{s}$ for ${\upsilon }_{\text{air}}$ in equation (II) and rewrite to obtain an expression for $L$.

  $\begin{array}{c}\Delta t=L\left(\frac{1}{343\text{m}/\text{s}}-\frac{1}{{\upsilon }_{\text{r}}}\right)\\ L=\frac{\Delta t}{\left(\frac{1}{343\text{m}/\text{s}}-\frac{1}{{\upsilon }_{\text{r}}}\right)}\end{array}$

Conclusion:

Therefore, the length of the rod in terms of $t$ and ${\upsilon }_r$ is $\underset{\_}{L=\frac{\Delta t}{\frac{1}{343\text{m}/\text{s}}-\frac{1}{{\upsilon }_r}}}$.

(d)

To determine

The length of the rod when speed of the sound goes to infinity.

Answer to Problem 66AP

The length of the rod when speed of the sound goes to infinity is $\underset{\_}{\Delta t\left(343\text{m}/\text{s}\right)}$.

Explanation of Solution

Write the expression for the length of the rod.

  $L=\frac{\Delta t}{\left(\frac{1}{343\text{m}/\text{s}}-\frac{1}{{\upsilon }_{\text{r}}}\right)}$                                                                                                 (V)

Conclusion:

Substitute, $\propto$ for ${\upsilon }_{\text{r}}$ in equation (V).

  $\begin{array}{c}L=\frac{\Delta t}{\left(\frac{1}{343\text{m}/\text{s}}-\frac{1}{\propto }\right)}\\ =\frac{\Delta t}{\left(\frac{1}{343\text{m}/\text{s}}\right)}\\ =\Delta t\left(343\text{m}/\text{s}\right)\end{array}$

Therefore, the length of the rod when speed of the sound goes to infinity is $\underset{\_}{\Delta t\left(343\text{m}/\text{s}\right)}$.