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Chapter 37: Problem 68

Measuring Speed by Radar. A baseball coach uses a radar device to measure the speed of an approaching pitched base-ball. This device sends out electromagnetic waves with frequency \(f_{0}\) and then measures the shift in frequency \(\Delta f\) of the waves reflected from the moving baseball. If the fractional frequency shift produced by a baseball is \(\Delta f / f_{0}=2.86 \times 10^{-7}\) , what is the baseball's speed in \(\mathrm{km} / \mathrm{h} ?\) (Hint: Are the waves Doppler- shifted a second time when reflected off the ball?

Short Answer

Expert verified
The baseball's speed is approximately 154.44 km/h.

Step by step solution

01

Understand the Doppler effect

When electromagnetic waves reflect off a moving object, they experience a Doppler shift in frequency. The equation for the Doppler shift of electromagnetic waves is given by:\[ \frac{\Delta f}{f_0} = \frac{2v}{c} \]where \( \Delta f \) is the change in frequency, \( f_0 \) is the original frequency, \( v \) is the speed of the object, and \( c \) is the speed of light.
02

Arrange to solve for speed

We need to rearrange the formula \( \frac{\Delta f}{f_0} = \frac{2v}{c} \) to solve for the speed \( v \): \[ v = \frac{c \cdot \Delta f}{2f_0} \]Substitute the given fractional frequency shift \( \frac{\Delta f}{f_0} = 2.86 \times 10^{-7} \) and the speed of light \( c = 3 \times 10^{8} \) m/s into this equation.
03

Plug values into the equation

Substitute the known values into the speed equation:\[ v = \frac{3 \times 10^{8} \times 2.86 \times 10^{-7}}{2} \]Compute the value to find \( v \).
04

Calculate the speed in m/s and km/h

Upon calculating, we find:\[ v = 42.9 \text{ m/s} \]To convert this speed to km/h, remember that 1 m/s is equivalent to 3.6 km/h:\[ v = 42.9 \times 3.6 \text{ km/h} \]Calculate for \( v \) in km/h.
05

Final calculation and answer

After calculating the above values, you get:\[ v = 154.44 \text{ km/h} \]This is the speed of the baseball.

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Key Concepts

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

Electromagnetic Waves
Electromagnetic waves are an essential part of our daily lives, even though we can鈥檛 see them with the naked eye. They include a vast range of wave types from radio waves to gamma waves. These waves are synchronized oscillations of electric and magnetic fields and can travel through a vacuum at the speed of light, which is approximately \(3 \times 10^8\) meters per second.
One fascinating property of electromagnetic waves is that their speed remains constant in a vacuum. However, when these waves interact with objects, such as a baseball, their frequency can shift due to motion.
This shift, known as the Doppler Effect, is crucial in applications like radar devices which measure speeds by analyzing how waves reflect off moving objects.
Frequency Shift
The frequency shift, or the change in frequency, is a central concept when discussing the Doppler Effect. This occurs when there is a relative motion between the source of electromagnetic wave and an observer.
For example, think about how a car horn sounds as it approaches you and then moves away. The same principle applies to electromagnetic waves used in radar technology.
When a radar device sends out waves towards a moving object, like a baseball, these waves bounce back with a different frequency. The Doppler shift formula \(\Delta f / f_0 = 2v/c\) shows how we relate this frequency change \(\Delta f\) to the object's velocity \(v\) and the wave's original frequency \(f_0\). By measuring \(\Delta f\), we can deduce how fast the object is moving.
Speed Calculation
The calculation of speed from the frequency shift is a direct application of the Doppler Effect.
For instance, in the given exercise, we are provided with a fractional frequency shift value of \(2.86 \times 10^{-7}\). Given this value and knowing the speed of light \(c\) is \(3 \times 10^8\) m/s, we can rearrange the Doppler shift formula to solve for the object's speed \(v\).
The rearranged formula \(v = \frac{c \cdot \Delta f}{2f_0}\) allows us to plug in the known values to find \(v = 42.9\) m/s. Converting to km/h by multiplying \(42.9\) by \(3.6\) yields \(v = 154.44\) km/h. This result means the baseball is moving at a speed of 154.44 kilometers per hour.
  • Remember, understanding how to manipulate these types of formulas is essential for solving physics problems involving speed and waves.

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Most popular questions from this chapter

One of the wavelengths of light emitted by hydrogen atoms under normal laboratory conditions is \(\lambda=656.3 \mathrm{nm},\) in the red portion of the electromagnetic spectrum. In the light emitted from a distant galaxy this same spectral line is observed to be Doppler- shifted to \(\lambda=953.4 \mathrm{nm},\) in the infrared portion of the spectrum. How fast are the emitting atoms moving relative to the earth? Are they approaching the earth or receding from it?

The French physicist Armand Fizeau was the first to measure the speed of light accurately. He also found experimentally that the speed, relative to the lab frame, of light traveling in a tank of water that is itself moving at a speed \(V\) relative to the lab frame is $$v=\frac{c}{n}+k V$$ where \(n=1.333\) is the index of refraction of water. Fizeau called \(k\) the draging coefficient and obtained an experimental value of \(k=0.44 .\) What value of \(k\) do you calculate from relativistic transformations?

A source of electromagnetic radiation is moving in a radial direction relative to you. The frequency you measure is 1.25 times the frequency measured in the rest frame of the source. What is the speed of the source relative to you? Is the source moving toward you or away from you?

An observer in frame \(S^{\prime}\) is moving to the right \((+x\) -direction at speed \(u=0.600 c\) away from a stationary observer in frame \(S .\) The observer in \(S^{\prime}\) measures the speed \(v^{\prime}\) of a particle moving to the right away from her. What speed \(v\) does the observer in \(S\) measure for the particle if \(v^{\prime}=0.400 c\) (b) \(v^{\prime}=0.900 c ;(\mathrm{c}) v^{\prime}=0.990 c ?\)

Space Travel? Travel to the stars requires hundreds or thousands of years, even at the speed of light. Some people have suggested that we can get around this difficulty by accelerating the rocket (and its astronauts) to very high speeds so that they will age less due to time dilation. The fly in this ointment is that it takes a great deal of energy to do this. Suppose you want to go to the immense red giant Betelgeuse, which is about 500 light-years away. (A light- year is the distance that light travels in a year.) You plan to travel at constant speed in a 1000 -kg rocket ship (a little over a ton), which, in reality, is far too small for this purpose. In each case that follows, calculate the time for the trip, as measured by people on earth and by astronauts in the rocket ship, the energy needed in ioules, and the energy needed as a percentage of U.S. yearly use (which is 1.0 \(\times 10^{20} \mathrm{J} ) .\) For comparison, arrange your results in a table showing \(v_{\text { rocket }}, t_{\text { earth }}, t_{\text { rocket }}, E(\) in \(\mathrm{J}),\) and \(E\) (as \(\%\) of U.S. use). The rocket ship's speed is (a) \(0.50 \mathrm{c} ;\) (b) 0.99 \(\mathrm{c}\) ; (c) 0.9999 \(\mathrm{c} .\) On the basis of your results, does it seem likely that any government will invest in such high-speed space travel any time soon?
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