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Chapter 16: Problem 120

A radio station broadcasts radio waves at a frequency of \(101.7 \mathrm{MHz} .\) The radio waves move through the air at approximately the speed of light in a vacuum. What is the wavelength of the radio waves?

Short Answer

Expert verified
The wavelength of the radio waves broadcasted at a frequency of \(101.7\,\mathrm{MHz}\) is approximately \(2.95\,\mathrm{m}\).

Step by step solution

01

1. Write the data given in the problem

We are given: - Frequency, \(f = 101.7\,\mathrm{MHz} = 101.7\,\times10^6 \,\mathrm{Hz}\) - Speed of light in vacuum, \(c = 3\times10^8\,\mathrm{m/s}\) Our goal is to find the wavelength, \(\lambda\).
02

2. Use the wave speed formula

The formula for the speed of a wave is given by: \(v=f\lambda\). Since the speed of the radio waves is approximately equal to the speed of light, we can substitute \(c\) for \(v\): \[c = f\lambda\]
03

3. Find the wavelength

Now, we solve for the wavelength, \(\lambda\), by dividing both sides of the equation by the frequency, \(f\): \[\lambda = \frac{c}{f}\] Plug in the given values for \(c\) and \(f\): \[\lambda = \frac{3\times10^8\,\mathrm{m/s}}{101.7\times10^6\,\mathrm{Hz}}\]
04

4. Calculate the wavelength

Perform the division to find the value of \(\lambda\): \[\lambda = \frac{3\times10^8\,\mathrm{m/s}}{101.7\times10^6\,\mathrm{Hz}} \approx 2.95\,\mathrm{m}\] The wavelength of the radio waves is approximately 2.95 meters.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Wave Speed Formula
Understanding the wave speed formula is crucial when studying physics, especially in the context of waves. In its basic form, the wave speed formula is expressed as \( v = f\lambda \), where \( v \) is the velocity or speed of the wave, \( f \) is the frequency with which the wave oscillates, and \( \lambda \) (lambda) represents the wavelength, or the distance between two consecutive peaks or troughs of the wave.

This formula tells us that the speed of a wave is directly proportional to both its frequency and its wavelength. In other words, if you know two of these properties, you can calculate the third. It's particularly useful in determining properties of waves that travel through different media, although in the case of radio waves in our exercise, they move through air at a speed close to a very special constant, the speed of light in a vacuum.
Frequency
Frequency is a fundamental concept in both physics and the entire field of wave mechanics. It refers to the number of times a wave repeats its cycle within a second and is measured in hertz (Hz). 1 Hz equates to one cycle per second. For example, a frequency of 101.7 MHz, or megahertz, is \( 101.7 \times 10^6 \) Hz, which means 101.7 million cycles per second. High-frequency waves have short wavelengths, and low-frequency waves have long wavelengths.

Understanding frequency allows us to grasp how different types of waves, such as radio waves, microwaves, and even light waves, interact with materials and communicate information. In telecommunications, different frequencies are assigned for different services to avoid interference between channels.
Speed of Light
The speed of light is a physical constant essential in many areas of physics, including optics and relativity. In a vacuum, it is precisely \( 299,792,458 \) meters per second (approximately \( 3 \times 10^8 \) m/s). When light travels through other mediums, like air or glass, it slows down slightly due to interaction with the material's particles. However, for many practical calculations such as our exercise, we approximate the speed of light in air to its value in a vacuum.

This speed limit also sets the stage for understanding space, time, and causality in the universe. It's fascinating that all observers, regardless of their relative motion, will measure the speed of light in a vacuum to be the same. In the context of our exercise, using the speed of light allowed us to find the wavelength of radio waves being broadcast, which is an important aspect of understanding their transmission and reception.

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Most popular questions from this chapter

An electromagnetic wave, such as light, does not require a medium. Can you think of an example that would support this claim?

Consider a stretched spring, such as a slinky. The stretched spring can support longitudinal waves and transverse waves. How can you produce transverse waves on the spring? How can you produce longitudinal waves on the spring?

Your ear is capable of differentiating sounds that arrive at each ear just 0.34 ms apart, which is useful in determining where low frequency sound is originating from. (a) Suppose a low-frequency sound source is placed to the right of a person, whose ears are approximately 18 \(\mathrm{cm}\) apart, and the speed of sound generated is \(340 \mathrm{m} / \mathrm{s}\). How long is the interval between when the sound arrives at the right ear and the sound arrives at the left ear? (b) Assume the same person was scuba diving and a low- frequency sound source was to the right of the scuba diver. How long is the interval between when the sound arrives at the right ear and the sound arrives at the left ear, if the speed of sound in water is \(1500 \mathrm{m} / \mathrm{s}\) ? (c) What is significant about the time interval of the two situations?

Two sinusoidal waves are moving through a medium in the same direction, both having amplitudes of \(3.00 \mathrm{cm}, \mathrm{a}\) wavelength of \(5.20 \mathrm{m},\) and a period of \(6.52 \mathrm{s}\), but one has a phase shift of an angle \(\phi\). What is the phase shift if the resultant wave has an amplitude of \(5.00 \mathrm{cm} ?[\)Hint: Use the trig identity \(\sin u+\sin v=2 \sin \left(\frac{u+v}{2}\right) \cos \left(\frac{u-v}{2}\right)\).

If the tension in a string were increased by a factor of four, by what factor would the wave speed of a wave on the string increase?
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