Doppler Effect (Proving equal wavelengths )

Click For Summary
SUMMARY

The discussion centers on proving that the wavelength perceived by an observer is equal to the wavelength of the emitted wave in the context of the Doppler Effect. The emitted wavelength is defined as v / f, while the perceived wavelength is expressed as (u + v) / f, where u is the observer's velocity and v is the wave's velocity. The equality is established through algebraic manipulation, specifically by transforming the perceived wavelength expression and canceling terms to show that both wavelengths are indeed equal.

PREREQUISITES
  • Understanding of the Doppler Effect
  • Familiarity with wave properties (wavelength, frequency, velocity)
  • Basic algebraic manipulation skills
  • Knowledge of relative motion concepts
NEXT STEPS
  • Study the mathematical derivation of the Doppler Effect equations
  • Explore the implications of the Doppler Effect in different mediums (air, water, etc.)
  • Learn about applications of the Doppler Effect in astronomy and radar technology
  • Investigate the effects of relativistic speeds on the Doppler Effect
USEFUL FOR

Students of physics, educators teaching wave mechanics, and anyone interested in the mathematical principles of the Doppler Effect.

Peter G.
Messages
439
Reaction score
0
Hi,

I am doing doppler effect and this question wants me to prove that the wavelength perceived by the observer and that of the wave emitted is the same.

I tried doing it but couldn't, so I checked out what they had as an answer:

They said that the wavelength of the emitted wave is: v / f, which I agree with

They said the perceived wavelength is equal to: u + v / f (1 + u/v), which I also agree with.

Now, how they proved that those are equal I don't understand... I tried but I just can't see it. Can anyone help me with the steps involved?

Thanks,
Peter
 
Physics news on Phys.org
Peter G. said:
Hi,

I am doing doppler effect and this question wants me to prove that the wavelength perceived by the observer and that of the wave emitted is the same.

I tried doing it but couldn't, so I checked out what they had as an answer:

They said that the wavelength of the emitted wave is: v / f, which I agree with

They said the perceived wavelength is equal to: u + v / f (1 + u/v), which I also agree with.

Now, how they proved that those are equal I don't understand... I tried but I just can't see it. Can anyone help me with the steps involved?

Thanks,
Peter

it is just transformation of the expression, using common denominator.

(1 + u/v) = (v/v + u/v) = (v + u)/v = (u + v)/v [I reversed the v+u so it was identical to the numerator]

Put that in the expression and the (u + v) terms cancel, and the v ends up in the numerator to give v/f
 

Similar threads

  • · Replies 8 ·
Replies
8
Views
1K
  • · Replies 6 ·
Replies
6
Views
1K
Replies
8
Views
1K
  • · Replies 1 ·
Replies
1
Views
5K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 4 ·
Replies
4
Views
4K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K