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Doppler Radar. A giant thunderstorm is moving toward a weather station at \(45.0 \mathrm{mi} / \mathrm{h}(20.1 \mathrm{m} / \mathrm{s}) .\) If the station sends a radar beam of frequency 200.0 \(\mathrm{MHz}\) toward the storm, what is the difference in frequency between the emitted beam and the beam reflected back from the storm? Be careful to carry plenty of significant figures! (Hint The storm reflects the same frequency that it receives.)

Short Answer

Expert verified
The frequency difference \( \Delta f \approx 26.8 \; \text{kHz} \).

Step by step solution

01

Understand the Problem

The problem involves a Doppler shift due to a moving thunderstorm. We need to find the frequency difference between the emitted radar beam and the reflected beam.
02

Identify the Given Parameters

- Speed of the storm, \( v_s = 20.1 \; \text{m/s} \)- Frequency of emitted radar, \( f_0 = 200.0 \; \text{MHz} \)- Speed of light, \( c = 3 imes 10^8 \; \text{m/s} \)
03

Apply the Doppler Effect Formula

The Doppler effect formula for a wave reflecting off a moving object (approaching source) is given by:\[ f' = f_0 \left( \frac{c + v_s}{c - v_s} \right) \]Where \( f' \) is the frequency received by the storm.
04

Calculate the Frequency Received by the Thunderstorm

Substitute the known values into the formula:\[ f' = 200.0 \times 10^6 \left( \frac{3 imes 10^8 + 20.1}{3 imes 10^8} \right) \] Calculating this value gives us the frequency received by the storm.
05

Calculate the Frequency of the Radar Returning to the Station

Use the same formula again to find the frequency as the storm sends back the radar:\[ f'' = f' \left( \frac{c}{c - v_s} \right) \]
06

Determine the Frequency Difference

The frequency difference \( \Delta f \) is the difference between the emitted frequency \( f_0 \) and the returned frequency \( f'' \):\[ \Delta f = f'' - f_0 \]
07

Compute \( f'' \) and \( \Delta f \)

Calculate \( f'' \) using the values:\[ f'' = 200.0 \times 10^6 \left( \frac{3 imes 10^8 + 20.1}{3 imes 10^8} \right) \left( \frac{3 imes 10^8}{3 imes 10^8 - 20.1} \right) \]Then find the difference using:\[ \Delta f = f'' - 200.0 \times 10^6 \]

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Key Concepts

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

Doppler radar
Doppler Radar is an advanced radar technology used to measure the speed of objects. It works by analyzing changes in the frequency of radar waves that bounce off a moving object. When a weather station sends out radar beams, those beams hit weather patterns like thunderstorms and return with different frequencies. This change is because the storm is moving.

The Doppler Radar is commonly used to track weather phenomena by calculating frequency shifts when radio waves return after bouncing off an object. This effect allows meteorologists to predict the behavior of storms. Doppler radar thus provides real-time data about storms, including their location and velocity. It is a critical tool in weather monitoring because it helps predict severe weather conditions, like thunderstorms and hurricanes.

In essence, Doppler Radar serves as an extra pair of eyes for meteorologists, offering valuable insights that are impossible to see with traditional observational methods.
frequency shift
A frequency shift refers to the change in the frequency of a wave as it is reflected off a moving target. This concept is central to understanding how Doppler Radar works. When a radar wave is emitted, it travels at a constant speed until it hits an object. If this object is moving, the frequency of the reflected wave changes.

In the example of a moving thunderstorm, as the storm moves closer, the waves become compressed, leading to a higher frequency—a phenomenon known as blue shift. Conversely, if the storm were moving away, the waves would stretch out, causing a lower frequency or red shift. This frequency change helps in detecting the speed of the storm.
  • Higher frequency occurs when the object is moving towards the source.
  • Lower frequency occurs when it is moving away from the source.

Understanding frequency shift is crucial for calculating how much a storm or any other object is moving relative to the observer, using the principles outlined in the Doppler effect formula.
physics problem solving
Physics problem-solving often involves applying known formulas and concepts to derive an answer from given data. In Doppler Radar problems, identifying the variables such as the speed of the storm and radar frequency is crucial. The formula for the Doppler Effect helps in calculating the frequency shift.

Key steps in solving a Doppler Radar problem include:
  • Understanding the problem and identifying the given values, such as the speed of the object and emitted frequency.
  • Using the appropriate formula to calculate the frequency shift. For radar waves, it's important to remember the modifications for moving sources or observers in the formula.
  • Computing the values carefully, considering significant figures to maintain accuracy.

Accurate calculations often depend on careful execution of the mathematical steps. With practice, problem-solving in physics can become more intuitive. It’s a skill developed by applying theoretical concepts to practical scenarios like thunderstorms approaching a weather station.
radar waves
Radar waves are electromagnetic waves used extensively in radar systems for detecting objects. In the context of Doppler Radar, these waves help track the speed and movement of objects, like storms or even vehicles.

Radar waves are emitted from a transmitter and sent outwards in all directions. When these waves encounter an object, they reflect off it and return to the receiver. The speed of light governs these waves; hence any frequency shift caused by the motion of the object can be quantified using physics equations like the Doppler Effect.
  • Radar waves can travel long distances, essential for detecting objects far away.
  • They work at different frequencies but commonly in the MHz range (e.g., 200.0 MHz used in our example).
  • The ability to detect frequency shifts in radar waves is what makes Doppler Radar effective in its applications such as weather monitoring and other fields.
Radar waves are integral to many technologies, and their precise manipulation is key to a wide range of scientific and practical applications.

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Most popular questions from this chapter

Two loudspeakers, \(A\) and \(B(\mathrm{Fig} .16 .40),\) are driven by the same amplifier and emit sinusoidal waves in phase. Speaker \(B\) is 2.00 \(\mathrm{m}\) to the right of speaker \(A\) . Consider point \(Q\) along the exten- sion of the line connecting the speakers, 1.00 \(\mathrm{m}\) to the right of speaker \(B .\) Both speakers emit sound waves that travel directly from the speaker to point \(Q .\) (a) What is the lowest frequency for which constructive interference occurs at point \(Q ?\) (b) What is the lowest frequency for which destructive interference occurs at point \(Q ?\)

A standing wave with a frequency of 1100 \(\mathrm{Hz}\) in a column of methane \(\left(\mathrm{CH}_{4}\right)\) at \(20.0^{\circ} \mathrm{C}\) produces nodes that are 0.200 \(\mathrm{m}\) apart. What is the value of \(\gamma\) for methane? (The molar mass of methane is 16.0 \(\mathrm{g} / \mathrm{mol}\) )

One type of steel has a density of \(7.8 \times 10^{3} \mathrm{kg} / \mathrm{m}^{3}\) and a breaking stress of \(7.0 \times 10^{3} \mathrm{N} / \mathrm{m}^{2}\) . A cylindrical guitar string is to be made of 4.00 \(\mathrm{g}\) of this steel. (a) What are the length and radius of the longest and thinnest string that can be placed under a tension of 900 \(\mathrm{N}\) without breaking? (b) What is the highest fundamental frequency that this string could have?

The sound source of a ship's sonar system operates at a frequency of 22.0 \(\mathrm{kH}_{2}\) . The speed of sound in water (assumed to be at a uniform \(20^{\circ} \mathrm{C} )\) is 1482 \(\mathrm{m} / \mathrm{s}\) (a) What is the wavelength of the waves cmitted by the source? (b) What is the difference in frequency between the directly radiated waves and the waves reflected from a whale traveling directly toward the slip at 4.95 \(\mathrm{m} / \mathrm{s} ?\) The ship is at rest in the water.

Example 16.1 (Section 16.1\()\) showed that for sound waves in air with frequency 1000 \(\mathrm{Hz}\) , a displacement amplitude of\( 1.2 \times 10^{-8} \mathrm{m}\) produces a pressure amplitude of \(3.0 \times 10^{-2} \mathrm{Pa}\) . Water at \(20^{\circ} \mathrm{C}\) has a bulk modulus of \(2.2 \times 10^{9} \mathrm{Pa}\) , and the speed of sound in water at this temperature is 1480 \(\mathrm{m} / \mathrm{s}\) . For \(1000-\mathrm{Hz}\) sound waves in \(20^{\circ} \mathrm{C}\) water, what displacement amplitude is produced if the pressure amplitude is \(3.0 \times 10^{-2}\) Pa? Explain why your answer is much less than \(1.2 \times 10^{-8} \mathrm{Pa}\) .

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