# Doppler effect in different scenarios

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1. May 2, 2016

### takando12

1. The problem statement, all variables and given/known data
A driver of a car travelling towards a wall with a speed of 5m/s, sounds a horn of frequency 256 Hz. If speed of sound= 330m/s, find
a) No. of beats /sec if the observer is between the car and the wall.
b) No. of beats/sec if the car is between the wall the observer.
c) Frequency of reflected sound as heard by driver.

2. Relevant equations
Velocity of observer =Vo
Velocity of source =Vs
Speed of sound= c
1) If source approaches observer
f' = (c/c-Vs)f
2) Source moves away from observer
f' = (c/c+Vs)/f
3) Source and observer move towards each other
f' = (c+Vo/c-Vs)f

3. The attempt at a solution
I have attempted all sections and have got the right answers. But I am still not convinced with which formulas I'm using.I have a little trouble identifying the source and it's details.

a) The observer hears the car on the left and the reflection from the wall on the right. I understand why we use 1) for the incident frequency but for the reflection, the source becomes the wall and it is at rest. So why do we use 1) again and find the frequency heard? Shouldn't it be just the reflected frequency,as it is, without any change?
b) We use 2) for the incident sound and 1) for the reflected sound. Applying the same argument as before, why do we use 1) for reflection?
c) So now the source and the observer are the same. And I get the right answer if I use 3). Again we seem to be considering the source as the wall and that the wall is moving towards us, when it is not.

I am missing some key points of importance here, and that's why it's a messed up. Please help me set my understanding right.

2. May 2, 2016

### andrewkirk

For (a) the reflected frequency is the frequency 'heard' by the wall, which is the number given in formula (1) with car as source and wall as observer (260Hz). The wall then emits as an echo the 260Hz frequency it 'hears' and, as the observer and the wall are both stationary, that's the freq the observer hears.

For c, you have a stationary source emitting 260Hz and the driver as observer moving towards it at speed 5m/s. You apply formula (3) with $V_O=5,V_S=0,f=260$ and you'll get the right answer.

That's the same as what you get if you replace the source and its frequency by a virtual car that is the mirror image of the real car in the wall, with the virtual car moving towards the real one at a speed of 5m/s (rel to the ground) and emitting a 256Hz horn sound.

So it all works out.