# Question 0a254

Apr 24, 2017

#### Explanation:

The wavelength of $550 \text{ kHz}$ is:

(3.0xx10^8" m/s")/(5.5xx10^5" Hz")= 545.5" m"

The wavelength of $1600 \text{ kHz}$ is:

(3.0xx10^8" m/s")/(1.6xx10^6" Hz")= 187.5" m"

The wavelength of 88 MHz is:

(3.0xx10^8" m/s")/(8.8xx10^7" Hz")= 3.4" m"

The wavelength of 108 MHz is:

(3.0xx10^8" m/s")/(1.08xx10^8" Hz")= 2.8" m"

Apr 24, 2017

AM signals: $188 \text{m"-545"m}$
FM signals: $2.8 \text{m"-3.4"m}$

#### Explanation:

From the chemistry reference sheet, $C = \lambda \nu$ where $C$ is the speed of the wave, $\lambda$ is the wavelength, and $\nu$ is the frequency. Solving this formula for wavelength, $\lambda = \frac{C}{\nu}$.

For AM signals, $C = 3.0 \times {10}^{8} \text{m/s}$ and $\nu$ ranges from $550 \times {10}^{3} \text{Hz}$ to $1600 \times {10}^{3} \text{Hz}$ (since $1 \text{kHz"=10^3"Hz}$). Therefore, the wavelength ranges from lambda=(3.0times10^8"m/s")/(550times10^3"Hz")=545"m" to lambda=(3.0times10^8"m/s")/(1600times10^3"Hz")=188"m".

For FM signals, $C = 3.0 \times {10}^{8} \text{m/s}$ and $\nu$ ranges from $88 \times {10}^{6} \text{Hz}$ to $108 \times {10}^{6} \text{Hz}$ (since $1 \text{MHz"=10^6"Hz}$). Therefore, the wavelength ranges from lambda=(3.0times10^8"m/s")/(88times10^6"Hz")=3.4"m" to lambda=(3.0times10^8"m/s")/(108times10^6"Hz")=2.8"m"#.