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(A) The wavelength $\lambda$ of a signal is given by $\lambda = c/f$,where $c$ is the speed of light $(3 	imes 10^8 \ m/s)$ and $f$ is the frequency.

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longer antenna for the $AM$ channel and shorter for the $FM$

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(A) The wavelength $\lambda$ of a signal is given by $\lambda = c/f$,where $c$ is the speed of light $(3 	imes 10^8 \ m/s)$ and $f$ is the frequency.For $AM$ at $1020 \ kHz$ $(1.02 	imes 10^6 \ Hz)$,the wavelength is $\lambda_{AM} = (3 	imes 10^8) / (1.02 	imes 10^6) \approx 294 \ m$.For $FM$ at $89.5 \ MHz$ $(89.5 	imes 10^6 \ Hz)$,the wavelength is $\lambda_{FM} = (3 	imes 10^8) / (89.5 	imes 10^6) \approx 3.35 \ m$.For efficient radiation and reception,the height of the antenna should be comparable to a quarter wavelength $(\lambda/4)$ of the signal.Since $\lambda_{AM} > \lambda_{FM}$,the $AM$ channel requires a much longer antenna than the $FM$ channel.

$A$ radio station has two channels. One is $AM$ at $1020 \ kHz$ and the other is $FM$ at $89.5 \ MHz$. For good results,you will use:

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