Problem 27
Question
FM radio stations broadcast at different frequencies. Calculate the wavelengths corresponding to the broadcast frequencies of the following college radio stations: (a) KCSU-FM (Fort Collins, CO), \(90.5 \mathrm{MHz} ;\) (b) WVUD (Newark, DE), \(91.3 \mathrm{MHz} ;\) (c) KUCR (Riverside, CA), \(88.3 \mathrm{MHz}\)
Step-by-Step Solution
Verified Answer
Question: What are the wavelengths corresponding to the broadcast frequencies of the following college radio stations:
(a) KCSU-FM (Fort Collins, CO): 90.5 MHz
(b) WVUD (Newark, DE): 91.3 MHz
(c) KUCR (Riverside, CA): 88.3 MHz
Answer: The wavelengths corresponding to the broadcast frequencies of the college radio stations are:
(a) KCSU-FM (Fort Collins, CO): λ ≈ 3.31 m
(b) WVUD (Newark, DE): λ ≈ 3.29 m
(c) KUCR (Riverside, CA): λ ≈ 3.40 m
1Step 1: Write down given frequencies
The frequencies of the radio stations are given as:
(a) KCSU-FM (Fort Collins, CO): \(90.5 \mathrm{MHz}\)
(b) WVUD (Newark, DE): \(91.3 \mathrm{MHz}\)
(c) KUCR (Riverside, CA): \(88.3 \mathrm{MHz}\)
2Step 2: Convert MHz to Hz
Before we can calculate the wavelengths, we need to convert the frequencies from MHz to Hz by multiplying by \(10^6\):
(a) \(90.5 \mathrm{MHz} = 90.5 \times 10^6 \mathrm{Hz}\)
(b) \(91.3 \mathrm{MHz} = 91.3 \times 10^6 \mathrm{Hz}\)
(c) \(88.3 \mathrm{MHz} = 88.3 \times 10^6 \mathrm{Hz}\)
3Step 3: Calculate the wavelengths
Using the formula \(\lambda = \frac{c}{f}\) and the speed of light \(c = 3 \times 10^8 \mathrm{m/s}\), we can calculate the wavelengths of the radio stations:
(a) \(\lambda_{\mathrm{KCSU-FM}} = \frac{3 \times 10^8 \mathrm{m/s}}{90.5 \times 10^6 \mathrm{Hz}} \approx 3.31 \mathrm{m}\)
(b) \(\lambda_{\mathrm{WVUD}} = \frac{3 \times 10^8 \mathrm{m/s}}{91.3 \times 10^6 \mathrm{Hz}} \approx 3.29 \mathrm{m}\)
(c) \(\lambda_{\mathrm{KUCR}} = \frac{3 \times 10^8 \mathrm{m/s}}{88.3 \times 10^6 \mathrm{Hz}} \approx 3.40 \mathrm{m}\)
4Step 4: Present the final results
The wavelengths corresponding to the broadcast frequencies of the college radio stations are:
(a) KCSU-FM (Fort Collins, CO): \(\lambda \approx 3.31 \mathrm{m}\)
(b) WVUD (Newark, DE): \(\lambda \approx 3.29 \mathrm{m}\)
(c) KUCR (Riverside, CA): \(\lambda \approx 3.40 \mathrm{m}\)
Key Concepts
Frequency ConversionSpeed of LightWavelength Formula
Frequency Conversion
Frequency conversion is an essential step when dealing with radio waves, especially when the initial data is expressed in different units such as Megahertz (MHz).
To perform frequency conversion, particularly from MHz to Hertz (Hz), you need to remember that 1 MHz is equal to 1 million Hz (\(10^6\)).
Hence, the conversion from MHz to Hz involves multiplying the given frequency by \(10^6\).
To perform frequency conversion, particularly from MHz to Hertz (Hz), you need to remember that 1 MHz is equal to 1 million Hz (\(10^6\)).
Hence, the conversion from MHz to Hz involves multiplying the given frequency by \(10^6\).
- If the radio station broadcasts at 90.5 MHz, you calculate it in Hz as 90.5 \(\times 10^6\) Hz.
- Similarly, 91.3 MHz becomes 91.3 \(\times 10^6\) Hz, and 88.3 MHz turns into 88.3 \(\times 10^6\) Hz.
Speed of Light
The speed of light, denoted as \(c\), is a fundamental constant in physics integral for many calculations related to light and electromagnetic waves.
For calculations of radio wavelengths, we use the speed of light value of \(3 \times 10^8\) meters per second (\(\text{m/s}\)).
This constant helps connect the speed, frequency, and wavelength of waves. Because radio waves, like light, travel at the speed of light in a vacuum, this value is crucial.
For calculations of radio wavelengths, we use the speed of light value of \(3 \times 10^8\) meters per second (\(\text{m/s}\)).
This constant helps connect the speed, frequency, and wavelength of waves. Because radio waves, like light, travel at the speed of light in a vacuum, this value is crucial.
- It ensures that we can calculate the wavelength by relating it to frequency.
- Additionally, understanding this value and its application solidifies our grasp of how electromagnetic waves propagate.
Wavelength Formula
The wavelength formula illustrates the relationship between frequency, speed, and wavelength of a wave. The equation is \(\lambda = \frac{c}{f}\), where \(\lambda\) represents the wavelength in meters, \(c\) is the speed of light (\(3\times 10^8\) \(\text{m/s}\)), and \(f\) is the frequency in Hz.
To use this formula effectively:
For KCSU-FM at 90.5 MHz (or \(90.5 \times 10^6\) Hz), the wavelength \(\lambda\) is \(\frac{3 \times 10^8}{90.5 \times 10^6} \approx 3.31\) meters.
This same method applies to the other frequencies, ensuring a consistent approach for calculating wavelengths.
To use this formula effectively:
- Ensure that the frequency is converted to Hz if not initially given in this unit.
- Substitute the values into the equation, dividing the speed of light by the frequency.
- Calculate to find the wavelength corresponding to the given frequency.
For KCSU-FM at 90.5 MHz (or \(90.5 \times 10^6\) Hz), the wavelength \(\lambda\) is \(\frac{3 \times 10^8}{90.5 \times 10^6} \approx 3.31\) meters.
This same method applies to the other frequencies, ensuring a consistent approach for calculating wavelengths.
Other exercises in this chapter
Problem 25
A neon light emits radiation of \(\lambda=616 \mathrm{nm} .\) What is the frequency of this radiation?
View solution Problem 26
In the 1990 s the Russian and American navies developed extremely low- frequency communications networks to send messages to submerged submarines. The frequency
View solution Problem 28
Which radiation has the longer wavelength: (a) radio waves from an AM radio station broadcasting at \(680 \mathrm{kHz}\) or (b) infrared radiation emitted by th
View solution Problem 29
Which radiation has the lower frequency: (a) radio waves from an AM radio station broadcasting at \(1030 \mathrm{kHz}\) or (b) the red light \((\lambda=633 \mat
View solution