Chapter 22, Problem 29P

To determine

The longer wavelength and its factor.

Answer to Problem 29P

The AM radio frequency is larger by $315.957411m$ .

Explanation of Solution

Given info:

First frequency, $f_1=940kHz$

Second frequency, $f_2=94MHz$

Factor, $c=3.00\times {10}^{8}m/s$

Formula used:

The relation between frequency and wavelength: $f=\frac{c}{\lambda }$

Where,

  $f$ is frequency

  $\lambda$ is wavelength

  $c$ is a factor

Calculation:

Rearrange the relation between frequency and wavelength: $f=\frac{c}{\lambda }$ . So, $\lambda =\frac{c}{f}$

According to the question, $940$ on the AM radio is $940kHz$ . This is equals to $940,000Hz$ . By using the formula: ${\lambda }_1=\frac{c}{f_1}$

  ${\lambda }_1=\frac{300,000,000m/s}{940,000Hz}=319.1489m$

According to the question, $94$ on the FM radio is $94MHz$ . This becomes $94,000,000Hz$ .By using the formula: ${\lambda }_2=\frac{c}{f_2}$

  ${\lambda }_2=\frac{300,000,000m/s}{94,000,000Hz}=3.191489m$

By comparing the value of ${\lambda }_1$ and ${\lambda }_2$ :

  $319.1489m-3.191489m=315.957411m$

So, the AM radio frequency has the larger wavelength and also, it is larger by $315.957411m$ .

Conclusion: AM radio frequency has a longer wavelength.