Chapter SH, Problem 9.4APP

To determine

To calculate: Difference in frequency of the two waves.

Answer to Problem 9.4APP

Difference in frequency of the two waves is $\underset{\_}{1632\text{ Hz}}$ .

Explanation of Solution

Given:

Speed of both waves = $\text{340 m/s}$

Wavelength of first wave = $\text{5}\text{.0 m}$

Wavelength of second wave = $0.2\text{ m}$

Formula used:

Speed is given by,

  $v=f\lambda$

where f is frequency and $\lambda$ is wavelength.

Calculation:

Frequency of first wave is calculated as:

  $\begin{array}{l}v=f_1{\lambda }_1\\ f_1=\frac{v}{{\lambda }_1}\\ f_1=\frac{340\text{ m/s}}{5.0\text{ m}}\\ f_1=68\text{ Hz}\end{array}$

Frequency of second wave is calculated as:

  $\begin{array}{l}v=f_2{\lambda }_2\\ f_2=\frac{v}{{\lambda }_2}\\ f_2=\frac{340\text{ m/s}}{\text{0}\text{.2 m}}\\ f_2=1700\text{ Hz}\end{array}$

Difference in frequency is calculated as:

  $\begin{array}{l}\text{ 1700 Hz}-68\text{ Hz}\\ \text{= }1632\text{ Hz}\end{array}$

Conclusion:

Difference in frequency of the two waves is $\underset{\_}{1632\text{ Hz}}$ .