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Chapter 37: Problem 25

(a) How fast must you be spproaching a red traffic light \((\lambda=675 \mathrm{nm})\) for it to appear yellow \((\lambda=575 \mathrm{nm}) ?\) Express your answer in terms of the speed of light. (b) If you used this as a reason not to get a ticket for running a red light, how much of a fine would you get for speeding? Assume that the fine is \(\$ 1.00\) for each kilometer per hour that your speed exceds the posted limit of 90 \(\mathrm{km} / \mathrm{h}\).

Short Answer

Expert verified
(a) 14.815% of the speed of light. (b) \$160,002,000 fine.

Step by step solution

01

Understand the Doppler Effect

The Doppler effect explains the change in wavelength observed when an object is moving relative to an observer. For this problem, we will use the formula \( \lambda' = \lambda (1 \pm v/c) \) where \( \lambda' \) is the observed wavelength, \( \lambda \) is the original wavelength, \( v \) is the velocity of the object, and \( c \) is the speed of light.
02

Set up the equation

For the traffic light to shift from red \((\lambda = 675 \text{ nm})\) to yellow \((\lambda' = 575 \text{ nm})\), we assume the light source is moving towards the observer. Set up the formula as \( 575 = 675 (1 - v/c) \).
03

Solve for velocity (\(v\))

Rearrange the equation \( 575 = 675 (1 - v/c) \) to solve for \( v \): 1. Divide both sides by 675: \( \frac{575}{675} = 1 - \frac{v}{c} \).2. Simplify: \( 0.85185 = 1 - \frac{v}{c} \).3. Solve for \( v/c \): \( \frac{v}{c} = 0.14815 \).4. Thus, \( v = 0.14815c \), meaning the velocity \( v \) is approximately 14.815% of the speed of light.
04

Convert to km/h and calculate the fine

Convert the speed from the fraction of the speed of light to kilometers per hour:1. Speed of light \( c \approx 3 \times 10^8 \text{ m/s} \).2. Calculate \( v = 0.14815 \times 3 \times 10^8 \approx 4.4445 \times 10^7 \text{ m/s} \).3. Convert \( v \) to km/h: \( 4.4445 \times 10^7 \times \frac{1}{1000} \times 3600 \approx 1.60002 \times 10^8 \text{ km/h} \).4. Subtract the speed limit: \(1.60002 \times 10^8 - 90 \approx 1.60002 \times 10^8 \text{ km/h}\).5. Calculate the fine: Fine = \( \\(1 \times 1.60002 \times 10^8 \approx \\)160002000 \), which rounds to \(\$160,002,000\).

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Key Concepts

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

Wavelength Shift
The phenomenon of wavelength shift is a key aspect of the Doppler Effect. It occurs when there is a motion between a wave source and an observer, resulting in an apparent change in the wavelength of the waves. This change depends on the direction of motion:
  • If the source is approaching the observer, the wavelengths compress or get shorter, leading to a blue shift.
  • If the source moves away from the observer, the wavelengths stretch out, resulting in a red shift.
In the context of the traffic light problem, the wavelength shift causes the red light to appear yellow if you are moving fast enough towards it. The relationship between the observed wavelength \(\lambda'\), the original wavelength \(\lambda\), the velocity of the moving observer \(v\), and the speed of light \(c\) is given by the formula:
\[ \lambda' = \lambda (1 \pm \frac{v}{c}) \] In this formula, the positive sign is used when the source is moving away, and the negative sign is used when it is moving towards the observer. Solving the equation \(575 = 675 (1 - \frac{v}{c})\) helps determine the speed at which the observer must approach a red light for it to appear yellow.
Speed of Light
The speed of light is a fundamental constant in physics, signifying the ultimate speed limit in the universe. It is defined as approximately \(3 \times 10^8 \, \mathrm{m/s}\).
  • This constant represents the speed at which all massless particles and associated electromagnetic waves travel in a vacuum.
  • The speed of light is critical not just in physics but also in practical applications like telecommunications and astronomy.
In our problem, knowing the speed of light allows us to calculate how fast a person must travel to make a dramatic color shift in traffic lights due to the Doppler effect. Here, the velocity \(v\) was computed as a fraction of the speed of light, showing the significance of this constant in determining extreme speeds required to observe such phenomena.
Converting percentages of the speed of light into km/h showcases just how fast these values are in real-world terms.
Traffic Light Color Change
The traffic light color change due to the Doppler effect is a fascinating illustration of how relative motion can influence our perception. Normally, a red traffic light has a wavelength of about 675 nm.
  • When approaching the light at a high speed, as calculated in the exercise, the wavelength as perceived by the observer shifts towards the shorter end of the spectrum, turning into yellow, which is around 575 nm.
  • This demonstrates an everyday application of the Doppler effect, beyond its common scientific or astronomical contexts.
In practical terms, such a speed to cause a color shift is not achievable through conventional vehicles. Nevertheless, this theoretical exercise helps in understanding high-energy physics concepts in a more tangible way. It's fascinating to connect such effects to everyday occurrences like a traffic light, reinforcing the real-world implications of scientific principles.

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Most popular questions from this chapter

A \(60.0-\mathrm{kg}\) person is standing at rest on level ground. How fast would she have to run to (a) double her total energy and (b) increase her total energy by a factor of 10\(?\)

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?

Frame \(S^{\prime}\) has an \(x\) -component of velocity \(u\) relative to frame S, and at \(t=t^{\prime}=0\) the two frames coincide (see Fig. 37.3\()\) . A light pulse with a spherical wave front is emitted at the origin of \(S^{\prime}\) at time \(t^{\prime}=0 .\) Its distance \(x^{\prime}\) from the origin after a time \(t^{\prime}\) is given by \(x^{\prime 2}=c^{2} t^{\prime 2} .\) Use the Lorentz coordinate ransformation to S, and at \(t=t^{\prime}=0\) the two frames coincide (see Fig. 37.3\()\) . A light pulse with a spherical wave front is emitted at the origin of \(S^{\prime}\) at time \(t^{\prime}=0 .\) Its distance \(x^{\prime}\) from the origin after a time \(t^{\prime}\) is given by \(x^{\prime 2}=c^{2} t^{\prime 2} .\) Use the Lorentz coordinate ransformation to be spherical in both frames.

Compute the kinctic energy of a proton (mass \(1.67 \times$$10^{-27} \mathrm{kg}\) using both the nonrelativistic and relativistic expressions, and compute the ratio of the two results (relativistic divided by nonrelativistic) for speeds of (a) \(8.00 \times 10^{7} \mathrm{m} / \mathrm{s}\) and (b) \(2.85 \times 10^{8} \mathrm{m} / \mathrm{s}\) .

A photon with energy \(E\) is emitted by an atom with mass \(m\) which recoils in the opposite direction. (a) Assuming that the motion of the atom can be treated nonrelativistically, compute the recoil speed of the atom. (b) From the result of part (a), show that the recoil speed is much less than \(c\) whenever \(E\) is much less than the rest energy \(m c^{2}\) of the atom.
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