What is the difference between observer moving and source moving in affecting the sound signal?

Question

Why can't we solve all the doppler effect problems for sound in frame of source instead of observer moving using relative velocity.

I want to know why this method does not work?

Answer 1

Let a source $S$ of sound emit waves at frequency $\nu_0$ and a detector $D$ approach it with a uniform velocity $\vec{v}$. If $D$ were still, it would see the compressions of the wave pass by with a speed $c_s$. However, because it is approaching the source with a velocity $v$, it sees compressions passing by with speed $c_s + v$. The wavelength of the wave is seen to be unchanged. Therefore, the apparent frequency is \begin{equation}\tag{1} \nu = \frac{c_s + v}{\lambda} = \frac{c_s}{\lambda} \left(1 + \frac{v}{c_s}\right). \end{equation} Now, $c_s/\lambda = \nu_0$ so that \begin{equation}\tag{2} \nu = \nu_0\left(1 + \frac{v}{c_s}\right). \end{equation}

Now consider the situation when the detector is stationary but the source approaches it with a uniform velocity $\vec{v}$. If $S$ were still the detector will notice a compression every $\lambda$ m apart. However, because $S$ is approaching $D$ with a speed $v$ the distance is reduced by $v/\nu_0$. The apparent wavelength is \begin{equation}\tag{3} \lambda = \lambda_0 - \frac{v}{\nu_0} = \frac{c_s}{\nu_0} - \frac{v}{\nu_0}. \end{equation} Since $\lambda = c_s/\nu$, $\nu$ being the apparent frequency, we have \begin{equation}\tag{4} \nu = \frac{\nu_0}{1 - v/c_s}. \end{equation}

The formulae for the cases (i) source is stationary and the detector is in motion and (ii) the other way round, are not symmetric. There are three velocities involved in the equations, all of them measured with respect to the medium through which sound waves travel. In the case of equation (1) and (2), $c_s + v$ is the relative velocity of the sound waves with respect to $D$. It is $c_s -v$ in case of (3) and (4). If we call $c_{sD}$ the relative velocity in either case, equations (2) and (4) becomes \begin{eqnarray} \nu &=& \nu_0\frac{c_{sD}}{c_s} \\ \nu &=& \nu_0\frac{c_s}{c_{sD}} \end{eqnarray} In either cases, the numerator is the speed of sound with respect to the moving body and the denominator is the speed of sound with respect to the medium.

Answer 2

All that matters is who is moving relative to the air (which incidentally is also usually who's moving relative to the ground). That will always be what "breaks the symmetry" of the two reference frames.