So do you agree with my post 62 or not? Assuming R=0?vela said:Actually, it does. It's hidden in the last term.
Let's assume the position of the mass is given by ##x(t)##, the source emits a signal ##y(t) = Y \cos(2\pi f t)##, and the observer is located at ##x=R##.
At time ##t=0##, crest 0 is emitted at ##x=0##, and it subsequently propagates toward the observer with speed ##c##. We can describe its position at ##x_0(t) = ct## for ##t\ge 0##.
At time ##t=T=1/f##, crest 1 is emitted from the position ##x(T)##. Its position is subsequently given by ##x_1(t) = x(T) + c(t-T)## for ##t\ge T##.
At time ##t=2T##, crest 2 is emitted, and its position is subsequently given by ##x_2(t) = x(2T) + c(t-2T)## for ##t \ge 2T##.
In general, for ##t \ge nT##, the position of the crest ##n## is given by ##x_n(t) = x(nT) + c(t-nT)##.
The distance ##\lambda_k## between crest ##k-1## and crest ##k## is
$$\lambda_k = x_{k-1}(t) - x_{k}(t) = cT - [x(kT) - x((k-1)T)] = \lambda - \int_{(k-1)T}^{kT} v(t')\,dt',$$ where ##\lambda=cT## is the un-Doppler-shifted wavelength and ##v(t)## is the velocity of the mass. This is essentially the same expression you derived in an earlier post.
So when does crest ##n## arrive at the observer? Just set ##x_n(t_n) = R## and solve for ##t_n##. When we do this, we get
$$t_n = nT + \frac Rc - \frac{x(nT)}{c}.$$ Since ##t=nT## is the time the crest was emitted, this is @Delta²'s expression for the retarded time.
One way of looking at the problem is that the movement of the source distorts the wave. The changing position of the source as crests are emitted causes them to be squeezed together or stretched apart. This modulated wave then propagates to the observer. The other way is to see the changes in the wave as a result of different propagation times. Compared to the time a crest emitted at ##x=0## would take to reach the observer, a crest is emitted from ##x>0## arrives a little early, and one from ##x<0## arrives a little late.