# Tag Info

## New answers tagged doppler-effect

0

OK, I get your answer, but as this is a homework question and you have really not shown any attempt to get the solution I cannot show you it. The doppler equation you have is fine: I would write it as $\omega = \omega_{\rm star} \gamma (1-v)$ where we are using units with $c=1$. You then need to use a standard relationship between acceleration in the two ...

14

Yes, gravitational waves will undergo the same red-shift as any wave that propagates at $c$. There were probably very violent gravitational waves in the very early universe. If those waves hadn't been red-shifted, they'd be ripping us apart right now. If so, could observations of them be used like red-shifted electromagnetic waves from distant sources ...

3

If the car with the siren runs you over you won't be hearing anything after it "passes". By definition the line of travel of the sound source must it intersect the observer. But if you write down the expression for frequency vs time as a function of distance of closest approach you will see that in the limit where that distance becomes zero, the step ...

2

How long do you have to listen to a sound to know its frequency? An entire period? Only half a period? This answer can become somewhat complicated, and I doubt that Fourier transforms were discussed in the class this quiz is from. So let's just say you don't have to wait for an entire period in order to know the frequency. Actually, let's just say you know ...

0

Imagine an observer watching a moving rocket carrying on it both, a light source and a clock. If on the rocket the clock is synchronized with the frequency of the light being emitted So we suppose not just any clock being carried on the rocket, but a good clock which is characterized (properly) by some particular constant frequency $f_r$; as is the ...

2

Again, imagine those on the rocket are measuring the distance of one wavelength comparing it with some high precision ruler, then will not the observer watch the ruler contract and then necessarily the wavelength of the light along with it, a contradiction to the original conclusion? No. This is an example of the kind of confusion that can result if one ...

0

If we assume that an observer in the water and one in the air are not moving with respect to each other or the water boundary, then they will both measure the same frequency. When the wave hits the water boundary, the waves can't "pile up" there. So the same number that arrive each second must leave each second. I believe therefore that you could do this ...

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