# Doppler Effects

1. Aug 10, 2007

### ttp88

a Concept Question.
Is there a Doppler Shift if the source and observer move in the same direction with the same velocity? Explain .

Can someone drop their comments and ideas ..? please and thanks you.

2. Aug 10, 2007

### Dick

What do YOU think? Forum rules require your input.

3. Aug 10, 2007

### ttp88

the question is just liddat.. there is no any values... it is a conceptial question..the question given is just mentioned both of them moving in the same directoin and velocity...
does Doppler Effect occurs ?

4. Aug 10, 2007

### Dick

Sure, but what are your conceptual thoughts?

5. Aug 10, 2007

### ttp88

u mean Doppler Effect occurs in the situation ? Can u explain --Why--?
since they have the same values of velocity V, how could the Doppler Effect occur ? the WaveFronts do not overlap, the observer is moving away from the source as well ,as they hv common of velocity.

6. Aug 10, 2007

### Dick

Are you trying to say that the doppler effect involves only the difference in the velocities of the source and observer and since the difference is zero, there is no doppler effect?

7. Aug 10, 2007

### ttp88

i dun know... sorry because my english is poor =s
lets take an example...

Given Values-
Velocity of Sound = 343m/s
Velocity of Source = 10m/s
Velocity of Observer = 10m/s
Frequency = 100Hz

Question-
When the Source is moving towards to the direction of Observer at 10m/s, at the same time, the observer is moving away from the Source.

Problem-
What is the Frequency received by the observer ? Does Doppler Effect occurs? Explain why this happens.

this shld be clearer...

8. Aug 10, 2007

### Dick

I think you know the answer - you are having a hard time expressing it clearly. The difference in velocities of the source and observer is zero. They are not moving towards or away from EACH OTHER. Yes, there is no doppler shift.

9. Aug 10, 2007

### ttp88

No.
Lets see the Formula.. f ' = [(Velocity of Sound + Velocity of Observer ) / ( Velocity of Sound - Velocity of Source)] * Freq of source

that is -
f ' = [(343+10) / (343-10)] * 100
= (353/333)*100
=106 Hz

therefor,
f' > f
there is doppler shift...
BUT, Why ?
what u say is correct. There is no Doppler Shift.
but after proved by the equation.... there is...
WHy???

10. Aug 10, 2007

### Dick

Sorry, you are right. I was thinking of waves like light that don't travel in a medium. Sound is different. I'm trying to think of a way to explain this case clearly...

11. Aug 10, 2007

### ttp88

sorry , u r Right .
f ' = [(343+10) / (343-10)] * 100 --> Wrong
f ' = [(343+10) / (343+10)] * 100 this is the correct one...
finally .... i solved it .. =D
thanx for helping me ya !

12. Aug 10, 2007

### Dick

Thanks for clearing that up for me. You had me confused.

13. Aug 10, 2007

### ttp88

sry -_- i just remember it.. they are moving in the same direction...
If Source is moving Left.... and away from observer... it shld be +10
and
the observer is "chasing" after the source... so.. it shld be +10 too..
sorry ya..