Doppler effect derivation for moving observer and stationary source

In summary, the equation highlighted is a relationship between the number of waves fronts received per unit time and the time.
  • #1
ChiralSuperfields
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Homework Statement
Please see below
Relevant Equations
Please see below
For this,
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1685399385710.png

Does someone please know where they got that ##f'## is number of waves fronts received per unit time from? Also could we write the equation highlighted as ##f' = \frac{n\lambda}{t}## where ##n## is the number of wavefronts in a time ##t##?

I derived that from ##\frac{vt}{\lambda} = n## and ##v = f\lambda##

Many thanks!
 
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  • #2
You have two velocities to consider:
- Propagation of the waves.
- Observer O.

While the observer approaches the non-moving source of sound waves, both velocities have opposite directions, producing a net velocity (which can be converted to a frequency).

While the observer moves away from the source, both velocities have the same direction, producing a net velocity (which can also be converted to another frequency).

Approaching frequency > Distancing frequency
 
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  • #3
Lnewqban said:
You have two velocities to consider:
- Propagation of the waves.
- Observer O.

While the observer approaches the non-moving source of sound waves, both velocities have opposite directions, producing a net velocity (which can be converted to a frequency).

While the observer moves away from the source, both velocities have the same direction, producing a net velocity (which can also be converted to another frequency).

Approaching frequency > Distancing frequency
Thank you for your reply @Lnewqban!

Good idea to think about it as a resultant velocity (I guess relative to the air)? The textbook says assumes that the body air is the reference frame.

Many thanks!
 
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  • #4
ChiralSuperfields said:
Does someone please know where they got that f′ is number of waves fronts received per unit time from?
It's definitional. The frequency observed by a receiver is the rate at which whole cycles are received.
ChiralSuperfields said:
could we write the equation highlighted as ##f' = \frac{n\lambda}{t}## where ##n## is the number of wavefronts in a time ##t##?

I derived that from ##\frac{vt}{\lambda} = n## and ##v = f\lambda##
No, you can’t write it like that for the excellent reason that it is dimensionally inconsistent. The LHS has dimension ##T^{-1}##, while the RHS has dimension ##LT^{-1}##. You could not have derived it correctly from those other two equations because they are consistent.
 
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  • #6
haruspex said:
It's definitional. The frequency observed by a receiver is the rate at which whole cycles are received.

No, you can’t write it like that for the excellent reason that it is dimensionally inconsistent. The LHS has dimension ##T^{-1}##, while the RHS has dimension ##LT^{-1}##. You could not have derived it correctly from those other two equations because they are consistent.
Thank you for your replies @haruspex and @Lnewqban !

Yeah, I can't find where they the textbook where they define frequency as number of wavelengths per unit time ##f' = \frac{n}{t}## where ##n## is the number of wave fronts in a time internal ##t##. I can only find definition ##f = \frac{1}{T}## where the wave speed v is eliminated from ##v = f\lambda## and ##v = \frac{\lambda}{T}## to get the result.

However, using our definition, ##f' = \frac{n}{T}## and comparing with the equation highlighted I get:
##n = \frac{vt}{\lambda} + \frac{v_Ot}{\lambda}##.

I think ##n = \frac{vt}{\lambda} + \frac{v_Ot}{\lambda}## could be rewritten more succinctly as,

##n = \frac{n_1\lambda}{\lambda} + \frac{n_2\lambda}{\lambda}##
##n = n_1 + n_2##

Where, ##n_1## is a positive integer multiple of wavelengths to pass an stationary observer fixed to the frame of the medium and ##n_2## is a positive integer multiple of wavelengths to pass a the observer while they are moving towards the source.

I also tried rewriting it another way,

##n = \frac{d}{\lambda} + \frac{d_O}{\lambda}##

which is also dimensionally consistent and I'm still thinking about

Many thanks!
 
  • #7
ChiralSuperfields said:
##n = n_1 + n_2##

Where, ##n_1## is a positive integer multiple of wavelengths to pass an stationary observer fixed to the frame of the medium and ##n_2## is a positive integer multiple of wavelengths to pass a the observer while they are moving towards the source.
Yes, if you mean "… ##n_2## is the positive integer multiple of wavelengths the observer would pass while moving towards the source if the waves were stationary"
 
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  • #8
haruspex said:
Yes, if you mean "… ##n_2## is the positive integer multiple of wavelengths the observer would pass while moving towards the source if the waves were stationary"
Thank you for catching my mistake there @haruspex!
 
Last edited:

1. How does the Doppler effect work for a moving observer and stationary source?

The Doppler effect is a phenomenon that occurs when there is relative motion between a source of waves (such as sound or light) and an observer. When the source is moving towards the observer, the waves are compressed, resulting in a higher frequency and a shorter wavelength. Conversely, when the source is moving away from the observer, the waves are stretched, resulting in a lower frequency and a longer wavelength.

2. What is the formula for calculating the Doppler effect for a moving observer and stationary source?

The formula for calculating the Doppler effect for a moving observer and stationary source is: f' = f(v + vo) / (v + vs), where f' is the observed frequency, f is the emitted frequency, v is the speed of the waves, vo is the speed of the observer, and vs is the speed of the source.

3. How does the speed of the observer and source affect the Doppler effect?

The speed of the observer and source both play a role in the Doppler effect. The faster the observer is moving towards the source, the higher the observed frequency will be. Similarly, the faster the source is moving away from the observer, the lower the observed frequency will be. The speed of the waves also affects the Doppler effect, as a higher wave speed will result in a larger change in frequency.

4. Can the Doppler effect be observed in both sound and light waves?

Yes, the Doppler effect can be observed in both sound and light waves. In sound waves, the effect is more noticeable, as the speed of sound is relatively slow compared to the speed of light. However, the Doppler effect can also be observed in light waves, such as when the source is a star moving towards or away from Earth, resulting in a shift in the observed wavelength of light.

5. Are there any real-life applications of the Doppler effect for a moving observer and stationary source?

Yes, the Doppler effect has many practical applications in various fields. In astronomy, it is used to determine the speed and direction of stars and galaxies. In medicine, it is used in ultrasound technology, where the Doppler effect is used to measure the velocity and direction of blood flow in the body. It is also used in police radar guns to measure the speed of moving vehicles. Additionally, the Doppler effect is used in weather forecasting to track the movement and intensity of storms.

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