How to link frequency and speed? Doppler effect

In summary, the person was trying to solve a problem that involved finding the source-speed from the Doppler formula, and then reaching to the result-formula. However, they got stuck when they found a different formula for vs, and realized that the statement was bad formulated.
  • #1
greg_rack
Gold Member
363
79
Homework Statement
At what speed should a plane move to make the roar of its turbines double in frequency for the observers who are in front of it in its path?
Does the result depend on the frequency considered?

RESULT: [vs = vw/2]
Relevant Equations
DOPPLER effect for a moving source heading to a steady receiver: ƒr=ƒs*1/[1-(νs/νw)]

CAPTION: "r = receiver", "s = source", "w = wave"
I had many attempts on trying to solve this one, but I got always stuck in the problem-solving part: how do I manage to find the source-speed from the Doppler formula, in an analytical way, and then reach to the result-formula?
Anyway, I'm pretty sure the only formula needed to solve this problem is the one i attached, which links the received-frequency to the source one, and the source/wave speed.
 
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  • #2
Show your work to see where you get stuck...
 
  • #3
BvU said:
Show your work to see where you get stuck...

IMG_4201.JPG


I hope to have attached the image in the right format and stuff, anyway, that's all I've written and where I exactly got stuck.
Starting from the basic Doppler formula, I just "flipped it around" a bit hoping to reach the resolutive one, but I lost my hopes when I found a totally different formula for vs, compared to the result one.
 
  • #4
I think that this is tripping you up:
1580668931426.png

By doubling the source frequency on the right hand side you're not adjusting the received frequency via the speed of the source.

Instead, divide both sides of the original equation by fs so that you have a ratio of the two frequencies:
$$\frac{f_r}{f_s} = \frac{1}{1 - \frac{v_s}{v_w}}$$
 
  • #5
gneill said:
I think that this is tripping you up:
View attachment 256505
By doubling the source frequency on the right hand side you're not adjusting the received frequency via the speed of the source.

Instead, divide both sides of the original equation by fs so that you have a ratio of the two frequencies:
$$\frac{f_r}{f_s} = \frac{1}{1 - \frac{v_s}{v_w}}$$

Yeah, you're definitely right, but I found out that the statement is bad formulated, since to solve the problem you need to assume that fR=2fS instead of what I've thought, and with this assumption I managed to solve it.
 

1. How does the Doppler effect relate to frequency and speed?

The Doppler effect is a phenomenon that describes the change in frequency of a wave when the source of the wave is moving relative to the observer. This change in frequency is directly proportional to the speed of the source and the speed of the wave.

2. What is the formula for calculating the Doppler effect?

The formula for calculating the Doppler effect is: f' = f(v +/- vs) / (v +/- vo), where f' is the observed frequency, f is the original frequency, v is the speed of the wave, vs is the speed of the source, and vo is the speed of the observer.

3. How does the direction of the source or observer affect the Doppler effect?

The direction of the source or observer affects the Doppler effect because the sign of the velocity term in the formula will change depending on whether the source/observer is moving towards or away from the observer/source. This results in a different observed frequency.

4. Can the Doppler effect be applied to all types of waves?

Yes, the Doppler effect can be applied to all types of waves, including sound waves, light waves, and even water waves. Any wave that has a frequency and is affected by the relative motion of the source and observer can experience the Doppler effect.

5. How is the Doppler effect used in real-life applications?

The Doppler effect has many practical applications in various fields such as astronomy, meteorology, and traffic control. It is used to measure the speed and direction of moving objects, detect the presence of objects, and even study the expansion of the universe.

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