/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 20 Radio station WJR broadcasts at ... [FREE SOLUTION] | 91Ó°ÊÓ

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Radio station WJR broadcasts at \(760 \mathrm{kHz}\). The speed of radio waves is \(3.00 \times 10^{8} \mathrm{~m} / \mathrm{s}\). What is the wavelength of WJR's waves?

Short Answer

Expert verified
The wavelength of WJR's waves is approximately 394.74 meters.

Step by step solution

01

Identify the Given Values

The problem provides the frequency of the radio station WJR as \(760\,\mathrm{kHz}\) and the speed of radio waves as \(3.00 \times 10^{8}\,\mathrm{m/s}\).
02

Convert Frequency to Hz

Since the frequency is given in kilohertz (kHz), convert it to hertz (Hz) by multiplying by 1,000: \(760 \times 10^3 = 760,000\,\mathrm{Hz}\).
03

Recall the Wave Equation

Use the equation for wave speed: \(v = f \lambda\), where \(v\) is the speed of the wave, \(f\) is the frequency, and \(\lambda\) is the wavelength. We want to find \(\lambda\).
04

Solve for Wavelength

Rearrange the wave equation to solve for wavelength: \(\lambda = \frac{v}{f}\).
05

Calculate the Wavelength

Substitute the given values into the formula: \(\lambda = \frac{3.00 \times 10^8 \, \mathrm{m/s}}{760,000 \, \mathrm{Hz}}\).
06

Complete the Calculation

Perform the division to find the wavelength: \(\lambda = 394.74 \, \mathrm{m}\).

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Key Concepts

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

Wavelength
The concept of wavelength is crucial in understanding how waves propagate through different mediums. In the context of radio waves, the wavelength is the distance between two identical points on consecutive waves, such as crest to crest or trough to trough. This helps us visualize how long each 'wave' of the radio signal is as it moves through space.

Wavelength can be calculated using the wave equation, which we'll discuss in another section, but keep in mind:

  • In electromagnetic waves, like radio waves, the longer the wavelength, the lower the frequency, and vice versa.
  • Wavelength is usually measured in meters.
  • Knowing the wavelength is important for understanding things like antenna design and signal propagation.
With radio station WJR, we calculated the wavelength to be approximately 394.74 meters. This gives us a physical grasp on the "size" of the waves emitted by the radio station.
Frequency Conversion
Conversion of frequency from one unit to another is a common task, particularly moving from kilohertz (kHz) to hertz (Hz). Frequency is a measure of how many times a wave oscillates per second, and Hertz (Hz) is the standard unit of frequency, equal to one cycle per second.

When converting frequency:

  • Recognize the need to move from kHz to Hz by multiplying by 1,000, as there are 1,000 Hz in a kHz.
  • A frequency given as 760 kHz would thus be converted to 760,000 Hz.
This process ensures accuracy when plugging values into equations since most scientific formulas, including the wave equation, use Hz. Such conversion is essential to consistent results and calculations across different problems and fields.
Radio Waves
Radio waves are part of the electromagnetic spectrum, characterized by longer wavelengths than visible light. Due to their long wavelengths, they can travel long distances and diffract around obstacles, making them perfect for communication systems like radio broadcasting.

Key points about radio waves include:

  • They travel at the speed of light, which is approximately 3.00 x 10^8 meters per second in a vacuum.
  • They are used not only in radio broadcasting but also in television, mobile phones, and even radar.
  • Radio frequencies vary from very low frequencies (VLF) to extremely high frequencies (EHF), all impacting their propagation and usage.
For WJR, which broadcasts at 760 kHz, the radio waves transmitted are capable of traveling across great distances, sometimes even influenced by atmospheric conditions.
Wave Equation
The wave equation is fundamental to understanding the relationship between the speed, frequency, and wavelength of a wave. It is given by the formula: \[ v = f \lambda \]where:
  • \(v\) is the wave speed, typically given in meters per second (m/s).
  • \(f\) is the frequency, given in hertz (Hz).
  • \(\lambda\) is the wavelength, given in meters (m).
The equation essentially tells us that the speed of a wave is the product of its frequency and its wavelength. This relationship is extremely useful in calculating one variable if the other two are known.

For WJR's broadcast, we used the wave equation to calculate the wavelength by rearranging it to \(\lambda = \frac{v}{f}\), and substituting in the known values to find a wavelength of 394.74 meters. This understanding helps in multiple applications, from designing communication equipment to exploring wave phenomena in physics.

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Most popular questions from this chapter

Determine the length of the shortest air column in a cylindrical jar that will strongly reinforce the sound of a tuning fork having a vibration rate of \(512 \mathrm{~Hz}\). Use \(v=340 \mathrm{~m} / \mathrm{s}\) for the speed of sound in air.

A uniform flexible cable is \(20 \mathrm{~m}\) long and has a mass of \(5.0\) kg. It hangs vertically under its own weight and is vibrated (perpendicularly) from its upper end with a frequency of \(7.0 \mathrm{~Hz}\). (a) Find the speed of a transverse wave on the cable at its midpoint. (b) What are the frequency and wavelength at the midpoint? (b) Because wave crests do not pile up along a string or cable, the number passing one point must be the same as that for any other point. Therefore, the frequency, \(7.0 \mathrm{~Hz}\), is the same at all points. To find the wavelength at the midpoint, we must use the speed we found for that point, \(9.9 \mathrm{~m} / \mathrm{s}\). That gives us $$ \lambda=\frac{v}{f}=\frac{9.9 \mathrm{~m} / \mathrm{s}}{7.0 \mathrm{~Hz}}=1.4 \mathrm{~m} $$

A copper wire \(2.4 \mathrm{~mm}\) in diameter is \(3.0 \mathrm{~m}\) long and is used to suspend a 2.0-kg mass from a beam. If a transverse disturbance is sent along the wire by striking it lightly with a pencil, how fast will the disturbance travel? The density of copper is \(8920 \mathrm{~kg} / \mathrm{m}^{3}\).

A wire under tension vibrates with a fundamental frequency of 256 Hz. What would be the fundamental frequency if the wire were half as long, twice as thick, and under one-fourth the tension?

A string vibrates in five segments at a frequency of \(460 \mathrm{~Hz} .(a)\) What is its fundamental frequency? (b) What frequency will cause it to vibrate in three segments? If the string is \(n\) segments long, then fromwe have \(n\left(\frac{1}{2} \lambda\right)\) \(=L\). But \(\lambda=v / f_{n}\), so \(L=n\left(\mathrm{u} / 2 f_{n}\right)\). Solving for \(f_{n}\) provides $$ f_{n}=n\left(\frac{v}{2 L}\right) $$ We are told that \(f_{5}=460 \mathrm{~Hz}\), and so $$ 460 \mathrm{~Hz}=5\left(\frac{v}{2 L}\right) \quad \text { or } \quad \frac{v}{2 L}=92.0 \mathrm{~Hz} $$ Substituting this in the above relation gives $$ f_{n}=(n)(92.0 \mathrm{~Hz}) $$ (a) \(f_{1}=92.0 \mathrm{~Hz}\). (b) \(f_{3}=(3)(92 \mathrm{~Hz})=276 \mathrm{~Hz}\) Alternative Method Recall that for a string held at both ends, \(f_{n}=n_{f 1}\). Knowing that \(f_{5}\) \(=460 \mathrm{~Hz}\), it follows that \(f_{1}=92.0 \mathrm{~Hz}\) and \(f_{3}=276 \mathrm{~Hz}\).

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