Why Does T0 Appear in the Doppler Effect Equation?

Click For Summary

Homework Help Overview

The discussion revolves around the Doppler effect and the equations related to wave frequency and period as a source moves relative to an observer. Participants are exploring why a specific term, T0, appears in the context of the equations governing the Doppler effect.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants are attempting to understand the relationship between the period of wave emission and the effect of the source's motion on that period. Questions are raised about the derivation of the equations and the significance of the term T0 in the context of the Doppler effect.

Discussion Status

Some participants are providing insights into how the period changes when the source moves, while others express confusion regarding specific lines in the explanations. There is an ongoing exploration of the concepts without a clear consensus on the interpretation of T0.

Contextual Notes

Participants are working within the constraints of homework guidelines, which may limit the depth of discussion and the types of solutions that can be shared. The focus remains on understanding the underlying principles rather than arriving at a definitive answer.

logearav
Messages
329
Reaction score
0

Homework Statement



Revered members,
I have attached the image of Doppler effect explanation.

Homework Equations



I have the following doubts
1)t1 = L/v
2) t2 = T0 +( L + vsT0)/v
But why T0 comes here, instead of ( L + vsT0)/v

The Attempt at a Solution


 

Attachments

  • doppler effect.jpg
    doppler effect.jpg
    43.5 KB · Views: 557
Physics news on Phys.org
A stationary source emits waves of frequency F and period 1/F. The period is the time it takes successive wave fronts to reach you. How does the period change when the source moves away from you? First you need to know how far the source moves in the time 1/F, that is simply 1/F times the source velocity v_s. Now how much is the period lengthened? By the extra time it takes wave to go the extra distance v_s/F which is delta_t =v_s/Fv where v is the wave velocity. So the new period between wave fronts of a moving source is

1/F' = 1/F + v_s/Fv = 1/F(1+ v_s/v)

Makes more sense to me thinking of the period then the frequency?

If you are troubled by what is written derive the formula yourself.
 
A stationary source emits waves of frequency F and period 1/F. The period is the time it takes successive wave fronts to reach you. How does the period change when the source moves away from you? First you need to know how far the source moves in the time 1/F, that is simply 1/F times the source velocity v_s.
Spinnor said:
Now how much is the period lengthened? By the extra time it takes wave to go the extra distance v_s/F which is delta_t =v_s/Fv where v is the wave velocity.

Thanks for the reply Spinnor. I don't understand the line given in quote.
 
Lets say you are in a car and beep your horn once a second. Everyone who is at rest relative to the car will also hear the horn beep once a second. Now suppose you drive away from me at some large velocity, say 1/6 the speed of sound all the while beeping your horn once a second. With this information you should be able to compute the amount of time I hear between beeps. Yes? It will be longer then one second because the beep has longer to travel. You only need three numbers, the period of the beeps, the velocity of the car, and the velocity of sound in air.
 

Similar threads

  • · Replies 4 ·
Replies
4
Views
4K
  • · Replies 1 ·
Replies
1
Views
5K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 16 ·
Replies
16
Views
2K