Chapter 12, Problem 48P

To determine

To calculate: The number of beats per second that can be heard.

Answer to Problem 48P

  $f_1=129.92{\text{ s}}^{-1}$ and $f_2=124.27{\text{ s}}^{-1}$

Explanation of Solution

Given:

Wavelength $\left({\lambda }_1\right)$ $=2.64\text{ m},$

Wavelength $\left({\lambda }_2\right)$ $=2.76\text{ m},$

Temperature $\left(T_2\right)$ $=20°\text{C}$

Formula used:

Write the expression for the frequency of the first wavelength.

  $f_1=\frac{v}{{\lambda }_1}\text{ Hz}$

  $f_2=\frac{v}{{\lambda }_2}\text{ Hz}$

  $f_1=$ Frequency of the first wavelength, in Hz

  $f_2$ = Frequency of the second wavelength, in Hz

Calculation:

To find the speed of the wave $\left(v\right)$ ,

At temperature $20°\text{C}$

  $\begin{array}{l}v=331+(0.6\times 20)\text{ m/s}\\ \text{ = 343 m/s}\text{.}\end{array}$

Now, find the number of beats at each wavelength

  ${\lambda }_1=2.64\text{ m},$

  $f_1=\frac{v}{{\lambda }_1}\text{ Hz}$

  $f_1=\frac{343\text{ m/s}}{2.64\text{ m}}\text{ Hz}\text{.}$

  $f_1=129.92{\text{ s}}^{-1}$

Now, for ${\lambda }_2=2.76\text{ m},$

  $f_2=\frac{v}{{\lambda }_2}\text{ Hz}\text{.}$

  $f_2=\frac{343\text{ m/s}}{2.76\text{ m}}\text{ Hz}\text{.}$

  $f_2=124.27{\text{ s}}^{-1}$

Conclusion:

The number of beats for ${\lambda }_2=2.76\text{ m}$ is $f_2=124.27{\text{ s}}^{-1}$ and for ${\lambda }_1=2.64\text{ m}$ is, $f_1=129.92{\text{ s}}^{-1}$ .