Chapter 5.1, Problem 4PP

Interpretation Introduction

Interpretation:

Wavelength needs to be calculated having frequency of FM radio station $94.7MHz$ and AM station $820kHz$. Also, the drawing which represents the FM station and AM station needs to be determined.

Concept introduction:

The distance between two nearest crests or nearest troughs is called the wavelength. It is denoted by $\lambda$. If wavelength is more, the distance between two crests and troughs will be more.

Explanation of Solution

Given Information:

Frequency $94.7MHz=94.7\times {10}^{6}Hz$

Frequency $820kHz=820\times {10}^{3}Hz$

The wavelength of wave can be calculated from frequency as follows:

$\lambda =\frac{c}{v}$

v = Frequency

c = velocity of light $\left(3\times {10}^{8}m{s}^{-1}\right)$

$\lambda$ = wavelength

Wavelength of radio waves having Frequency of $94.7MHz=94.7\times {10}^{6}Hz$ can be calculated as follows:

$\lambda =\frac{3\times {10}^{8}m/s}{94.7\times {10}^{6}Hz}$

$\lambda =3.16\times {10}^{12}m$

Thus, wavelength of FM station is $3.16\times {10}^{12}m$.

Similarly, wavelength of radio waves having Frequency of $820kHz=820\times {10}^{3}Hz$

$\lambda =\frac{3\times {10}^{8}m/s}{820\times {10}^{3}Hz}$

$\lambda =3.65\times {10}^{8}m$

Thus, wavelength of AM station is $3.65\times {10}^{8}m$.

Since, the wavelength of AM station is less than that of FM, the distance between two crests and troughs will be less in case of AM station drawing than FM.

Thus, drawing (a) represents FM station and drawing (b) represents AM station.