How would you solve this physics problem on relativity and the doppler effect?

In summary, the conversation discusses how police radar uses microwaves of a known frequency to detect the speed of a car by measuring the reflected waves and the beat frequency. Part A of the homework asks to show that the reflected wave has a frequency of f=fsource[(c+v)/(c-v)] when reflected from a mirror approaching at speed v. Part B asks to derive the beat frequency when v is much less than c using the approximation f+fsource=2fsource. This can be achieved by simplifying the expression [sqrt(1+v/c)]/[sqrt(1-v/c)] using binomial expansions. The resulting equation for beat frequency is fbeat=2v/(lambda).
  • #1
amr55533
15
0

Homework Statement

Police radar detects the speed of a car as follows: Microwaves of a precisely known frequency are broadcast toward the car. The moving car reflects the waves with a doppler shift. The reflected waves are received and combined with an attenuated version of the tansmitted wave. Beats occur between the two microwave signals. The beat frequency is measured.

a) For an electromagnetic wave reflected back to its source from a mirror approaching at speed v, show that the reflected wave has a frequency: f=fsource[(c+v)/(c-v)]

b) When v is much less than C, the beat frequency is much smaller than the transmitted frequency. In this case use the approximation f+fsource=2fsource and show that the beat frequency can be written as:

fbeat=2v/(lambda)

I figured out part A, but I am unsure how to derive part B.

Thanks!

Homework Equations

f=fsource[(c+v)/(c-v)]

f=fsource=2fsource

f*(lambda)=C

The Attempt at a Solution

For part A, I simply applied fobs=fsource[sqrt(1+v/c)]/[sqrt(1-v/c)] twice, where the mirror would be the first observer. I am just unsure on part B.
 
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  • #2
I don't see how the approximation suggested helps (or that it is valid), but you should be able to simplify [sqrt(1+v/c)]/[sqrt(1-v/c)] based on v << c. Use the binomial expansions.
 
  • #3
What is the equation for beat frequency?
 

1. What is the Doppler effect and how does it relate to relativity?

The Doppler effect is the change in frequency and wavelength of a wave, such as sound or light, as the source or observer moves relative to each other. In relativity, the Doppler effect is used to explain the shift in frequency and wavelength of light as an observer moves at high speeds, approaching the speed of light.

2. How does time dilation affect the Doppler effect in relativity?

Time dilation, a consequence of Einstein's theory of relativity, states that time moves slower for objects in motion compared to stationary objects. This means that the frequency and wavelength of light will appear to change differently for a moving observer compared to a stationary observer due to the difference in passage of time.

3. What is the formula for calculating the Doppler effect in relativity?

The formula for calculating the Doppler effect in relativity is given by f'/f = √[(c±v)/(c∓v)], where f' is the observed frequency, f is the emitted frequency, c is the speed of light, and v is the relative velocity between the source and observer.

4. How does the Doppler effect in relativity impact our understanding of the universe?

The Doppler effect in relativity allows us to measure the relative velocities of objects in the universe, such as stars and galaxies, and determine their distance from us. It also helps us understand the expansion of the universe and the effects of gravity on light.

5. Can the Doppler effect be used to explain redshift and blueshift in the light spectrum?

Yes, the Doppler effect can explain redshift and blueshift in the light spectrum. Redshift occurs when an object is moving away from an observer, causing the light waves to stretch and appear more red. Blueshift occurs when an object is moving towards an observer, causing the light waves to compress and appear more blue. This is a result of the Doppler effect in relativity.

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