Question

compare 930 on the am dial to 92.9 on the fm. which has the longer wavelength?
the fm station.
the am station.
both stations have the same wavelength.
previous answers
correct
part b
by what factor is it larger?
express your answer using three significant figures.

Explanation:

Step1: Recall the relationship between frequency and wavelength

The speed of light \(c\) is related to frequency \(f\) and wavelength \(\lambda\) by the formula \(c = f\lambda\), so \(\lambda=\frac{c}{f}\). For AM, the frequency is in \(kHz\) (e.g., \(f_{AM}=930\space kHz = 930\times10^{3}\space Hz\)), and for FM, the frequency is in \(MHz\) (e.g., \(f_{FM}=92.9\space MHz=92.9\times 10^{6}\space Hz\)).

Step2: Calculate the ratio of wavelengths

\(\frac{\lambda_{AM}}{\lambda_{FM}}=\frac{\frac{c}{f_{AM}}}{\frac{c}{f_{FM}}}=\frac{f_{FM}}{f_{AM}}\)
Substitute \(f_{AM} = 930\times10^{3}\space Hz\) and \(f_{FM}=92.9\times 10^{6}\space Hz\) into the formula:
\(\frac{\lambda_{AM}}{\lambda_{FM}}=\frac{92.9\times 10^{6}}{930\times10^{3}}\)
\(\frac{\lambda_{AM}}{\lambda_{FM}}=\frac{92900}{930}\approx99.9\)

Answer:

\(99.9\)