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Chapter 22: Problem 21

Radar waves with \(3.4 \mathrm{~cm}\) wavelength are sent out from a transmitter. Their speed is \(3.00 \times 10^{8} \mathrm{~m} / \mathrm{s}\). What is their frequency?

Short Answer

Expert verified
The frequency of the radar waves is approximately 8.82 GHz.

Step by step solution

01

Understand the Relationship

To find the frequency of a wave, use the wave equation: \[ v = f \cdot \lambda \]where \( v \) is the speed of the wave, \( f \) is the frequency, and \( \lambda \) is the wavelength.
02

Convert Units

First, ensure all units are compatible. The wavelength must be in meters. Convert the given wavelength from centimeters to meters:\[ 3.4 \text{ cm} = 0.034 \text{ m} \]
03

Solve for Frequency

Rearrange the wave equation to solve for frequency \( f \):\[ f = \frac{v}{\lambda} \]Substitute the given values of speed \( v = 3.00 \times 10^8 \text{ m/s} \) and wavelength \( \lambda = 0.034 \text{ m} \):\[ f = \frac{3.00 \times 10^8}{0.034} \]
04

Calculate Frequency

Calculate the frequency:\[ f = 8.82 \times 10^9 \text{ Hz} \]Thus, the frequency of the radar waves is approximately \( 8.82 \text{ GHz} \).

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Key Concepts

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

Wave Speed
Wave speed is a fundamental concept in physics, describing how fast a wave propagates through a medium. This speed, denoted as \( v \), is calculated using the distance a wave travels over a period of time. In our context, it's given as \( 3.00 \times 10^8 \text{ m/s} \). This is the speed of light, which radar waves follow since they are electromagnetic. Understanding wave speed allows you to determine how quickly a signal is transmitted.
Key characteristics of wave speed include:
  • It varies across different media - light travels slower in water than in air.
  • It's determined by the medium鈥檚 properties - density and elasticity affect sound wave speed.
  • It's crucial for calculating other wave properties, such as frequency and wavelength.
Remember, when dealing with wave equations, consistency in units is important to ensure accurate calculations.
Wave Equation
The wave equation is a powerful tool in physics, linking wave speed, frequency, and wavelength through the simple formula: \( v = f \cdot \lambda \). It says the speed of a wave \( v \) is the product of its frequency \( f \) and its wavelength \( \lambda \). Using this equation, you can solve for any of the three components if you have the other two.
Here鈥檚 a breakdown of the components:
  • Frequency \( f \): Measured in hertz (Hz), it reflects the number of wave cycles per second.
  • Wavelength \( \lambda \): The distance between successive peaks (or troughs) of a wave, in meters.
  • Wave Speed \( v \): Describes how fast a wave carries energy through space or a medium.
To solve for an unknown, rearrange the equation appropriately. For example, if you know \( v \) and \( \lambda \), solve for \( f \) using \( f = \frac{v}{\lambda} \). This relationship is key when converting radar wavelengths to frequencies.
Wavelength Conversion
Wavelength conversion involves adjusting wavelength measurements to match standard units before using them in calculations. When you find a wavelength in centimeters, like \( 3.4 \text{ cm} \), it's essential to convert it into meters to use in equations that require SI units.
The conversion process:
  • Identify the initial unit - in this case, centimeters.
  • Convert to meters by dividing by 100, as \( 1 \text{ cm} = 0.01 \text{ m} \).
  • So \( 3.4 \text{ cm} = 3.4 / 100 = 0.034 \text{ m} \).
This conversion is critical for solving problems using the wave equation, ensuring your units are consistent throughout the calculations. This simple step aids in achieving precise results and avoids common mistakes in wave-related computations.

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Most popular questions from this chapter

A string has both its total mass and length doubled; all else kept constant, what happens to the speed of transverse waves that can be set up on the string?

Suppose that depicts standing waves on a metal string under a tension of \(88.2 \mathrm{~N}\). Its length is \(50.0 \mathrm{~cm}\) and its mass is \(0.500\) g. ( \(a\) ) Compute \(v\) for transverse waves on the string. (b) Determine the frequencies of its fundamental, first overtone, and second overtone. (a) \(v=\sqrt{\frac{\text { Tension }}{\sqrt{\text { Mass per t unit length }}}}=\sqrt{\frac{88.2 \mathrm{~N}}{15.00 \times 10^{-4} \mathrm{~kg} /(0.500 \mathrm{~m})}}=297 \mathrm{~m} / \mathrm{s}\) (b) We recall that the length of the segment is \(\lambda / 2\) and we use \(\lambda=\) \(\mathrm{u} / \mathrm{f}\). For the fundamental: $$ \lambda=1.00 \mathrm{~m} \quad \text { and } \quad f=\frac{297 \mathrm{~m} / \mathrm{s}}{1.00 \mathrm{~m}}=297 \mathrm{~Hz} $$ For the first overtone: $$ \lambda=0.500 \mathrm{~m} \quad \text { and } \quad f=\frac{297 \mathrm{~m} / \mathrm{s}}{0.500 \mathrm{~m}}=594 \mathrm{~Hz} $$ For the second overtone: $$ \lambda=0.333 \mathrm{~m} \quad \text { and } \quad f=\frac{297 \mathrm{~m} / \mathrm{s}}{0.333 \mathrm{~m}}=891 \mathrm{~Hz} $$

An organ pipe closed at one end is \(61.0 \mathrm{~cm}\) long. What are the frequencies of the first three overtones if \(u\) for sound is \(342 \mathrm{~m} / \mathrm{s}\) ?

What must be the length of an iron rod that has the fundamental frequency \(320 \mathrm{~Hz}\) when clamped at its center? Assume longitudinal vibration at a speed of \(5.00 \mathrm{~km} / \mathrm{s}\).

A string \(2.0 \mathrm{~m}\) long is driven by a 240 -Hz vibrator at its end. The string resonates in four segments forming a standing wave pattern. What would be the speed of a transverse wave on such a string? Let's first determine the wavelength of the wave from part (d) of Since each segment is \(\lambda / 2\) long, $$ 4\left(\frac{\lambda}{2}\right)=L \quad \text { or } \quad \lambda=\frac{L}{2}=\frac{2.0 \mathrm{~m}}{2}=1.0 \mathrm{~m} $$ Then, using \(\lambda=u T=v / f\), $$ \mathrm{u}=f \lambda=\left(240 \mathrm{~s}^{-1}\right)(1.0 \mathrm{~m})=0.24 \mathrm{~km} / \mathrm{s} $$
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