Doppler and electromagnetic waves

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SUMMARY

The discussion focuses on calculating the speed of a speeder using the Doppler effect in the context of electromagnetic waves. A police radar gun emits waves at a frequency of 6.0 x 109 Hz, and the reflected frequency measured is 316 Hz lower. The formula used is fo = fs * (1 ± vrel/c), leading to the conclusion that the speeder's speed with respect to the ground is 498 m/s.

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  • Understanding of the Doppler effect in electromagnetic waves
  • Familiarity with frequency and speed calculations
  • Knowledge of basic physics concepts related to motion
  • Ability to manipulate equations involving constants like the speed of light
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  • Study the principles of the Doppler effect in various wave types
  • Learn about the speed of light and its significance in electromagnetic wave calculations
  • Explore practical applications of radar technology in law enforcement
  • Investigate the impact of relative motion on wave frequency shifts
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Physics students, law enforcement professionals using radar technology, and anyone interested in the practical applications of the Doppler effect in real-world scenarios.

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A speeder is pulling directly away and increasing his distance from a police car that is moving at 23 m/s with respect to the ground. The radar gun in the police car emits an electromagnetic wave with a frequency of 6.0 109 Hz. The wave reflects from the speeder's car and returns to the police car, where its frequency is measured to be 316 Hz less than the emitted frequency. Find the speeder's speed with respect to the ground.

fo = fs * (1 +- vrel/c)

I am stressing out and can't get this at all
 
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.Answer: The speed of the speeder with respect to the ground is vrel = (fo - fs)c/fo = (6.0x109-316x106)c/6.0x109 = 4.98 x 10^2 m/s
 

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