Chapter 15, Problem 87A

(a)

To determine

To graph:The wavelength versus the frequency graphfor the given data andinterpret their relationship from the graph.

Answer to Problem 87A

The graph shows an inverse relationship between frequency and wavelength.

Explanation of Solution

Given:

The wavelengths of sound waves produced by a set of tuning forks:

Calculation:

As per the problem,

Plot the graph with frequencies of the tuning fork along x-axis and wavelengths along y-axis

Take $1\text{ division}=10\text{ Hz}$ along x-axis and $1\text{ division}=0.02\text{ m}$ along y-axis.

Conclusion: The graph shows an inverse relationship between frequency and wavelength.

(b)

To determine

To graph:The wavelength versus the inverse of frequency ( $\frac{1}{f}$)graph for the givendata and interpret their relationship from the graph.

To determine the speed of sound from the graph.

Answer to Problem 87A

The graph obtained is straight line graph.

The speed of sound is $341.9\text{ m/s}$.

Explanation of Solution

Given:

The table:

Formula used:

Slope of a straight line:

  $\text{slope=}\frac{y_2-y_1}{x_2-x_1}$

Where $(x_1,y_1)\text{ and (}{x}_2,y_2)$ are any two points on the line.

Speed of sound is given by:

  $\begin{array}{l}f=\frac{v}{\lambda }\\ v=\frac{\lambda }{(1/f)}\end{array}$

Where $f$ is the frequency of sound, $v$ is the speed of sound and $\lambda$ is the wavelength of sound.

Calculation:

As per problem,

To plot wavelength versus inverse of frequency, calculate inverse of each frequency given in the table.

For $f=262\text{ Hz}$

  $\frac{1}{f}=\frac{1}{\text{262 Hz}}=3.82\times {10}^{-3}\text{ s}$

For $f=294\text{ Hz}$

  $\frac{1}{f}=\frac{1}{\text{294 Hz}}=3.40\times {10}^{-3}\text{ s}$

For $f=330\text{ Hz}$

  $\frac{1}{f}=\frac{1}{\text{330 Hz}}=3.03\times {10}^{-3}\text{ s}$

For $f=349\text{ Hz}$

  $\frac{1}{f}=\frac{1}{\text{349 Hz}}=2.86\times {10}^{-3}\text{ s}$

For $f=392\text{ Hz}$

$\frac{1}{f}=\frac{1}{\text{392 Hz}}=2.55\times {10}^{-3}\text{ s}$

For $f=440\text{ Hz}$

  $\frac{1}{f}=\frac{1}{440\text{ Hz}}=2.27\times {10}^{-3}\text{ s}$

Tabulate the above calculated data along with corresponding wavelength:

Plot the graph of wavelength with the corresponding inverse of frequency.

Take $1\text{ division}=0.04\text{ ms}$ along x-axis and $1\text{ division}=0.02\text{ m}$ along y-axis.

The graph represents a straight line.

The slope of the graph can calculate speed of sound.

To calculate speed of sound use formula

  $\begin{array}{l}f=\frac{v}{\lambda }\\ v=\frac{\lambda }{(1/f)}\end{array}$

Consider any two end pointsA $\text{(2}\text{.27,0}\text{.78)}$ andG $\text{(3}\text{.82,1}\text{.31)}$ from the graph

  $\begin{array}{l}\text{slope}=v=\frac{(1.31\text{ m}-0.78\text{ m})}{(3.82\text{ ms}-2.27\text{ ms}\text{})}\\ v=341.9\text{ m/s}\end{array}$

Conclusion:

The graph obtained is straight line graph.

The speed of sound is $341.9\text{ m/s}$.