Question

[Doppler effect]

Say you have a source and an observer. They are both moving away from each other which the source having a greater velocity than the observer.

when you use the regular Doppler equation:

f'=f(343 +- Vobs)/(343 +- Vsource)

at what time is the f' being related to?

what i mean is that the apparent frequency f' really depends on time.

so when you do this calculation with this formula at what time does this formula take place?

Answer 1

The doppler effect is calculated with dependency on the medium in which the sound travels. Time is not a factor. Both the source and the observer move relative to the medium. The default medium is the air so at about 20 degrees Celsius, sound travels at 343m/s.

Answer 2

General doppler shift equation
\[\large f'={{(v+v_{observer})} \over {(v - v_{source})}}\]

Answer 3

if both the source and the observer are moving away from each other, then the equation you wrote is incorrect. in the denominator, it's supposed to be
v+ v_source because if you think about it, the farther away the source moves, the greater the frequency decreases. To decrease the frequency, the denominator has to increase.

Answer 4

The equation is time independent.. the frequency remains the same.. as long as ur velocities remain the same!

Answer 5

@ mashy

I would think if the source and observer were traveling in the same direction and if the source were traveling with a greater speed than the observer then time would matter.

say for example:

the source traveling at 100m/s and an observer at 10m/s

at t=2 source is 200m and observer is 20m difference is 180m between them.

at t=5 seconds source is 500m and observer is 50m difference is 450m between them.

i would think that the longer the distance between them the further the source is away from the observer then the lower the apparent frequency is that the observer hears.

is it time independent because the speed of sound is so fast? 343m/s

i would think that if the distance between the source and observer is more than 343m than time would matter.

please help me understand this concept. thanks.

Answer 6

no not really.. althought whatever u are sayig is true, that the distance of seperation between them keeps changing.. the relative speeds remain the same right?

Answer 7

in your example you have a source and observer traveling in the same direction with source moving at 100m/s and observer moving at 10m/s
at t=2, the source is 200 m away but the sound has traveled 343 m so the sound is actually 123 m away.

Answer 8

besides, the only things that move that fast are possibly race cars so while it is hypothetically possible, it's not realistic

Answer 9

@mashy yes the distance keeps changing yet the speeds stay the same. i would think eventually the distance would become so great that the formula breaks down.

Answer 10

@Fellowroot
Consider each wavefront to be independent from the rest of the sound wave. This implies that - once emitted, the wave front has nothing to do with the speed of the source.

Hence, the equation will remain same as long as v(source) < v(sound)
(I cant really talk about the other case ;)

Answer 11

@LastDayWork isn't the other case when the speeds become Mach since you break the sound barrier? like Mach1, Mach2 etc?

Answer 12

Yea..but I don't have much insights for that field. XD

Answer 13

@Fellowroot
no it won't break.. cause sound is travelling faster than the source or the observer.. so regardless of how far u go.. the formula will still hold :P

But the formula only holds as long as both the source and the observer are moving along the same straight line!