Doppler Effect (Acoustics)

First pick the correct case. If the source moves towards a stationary observer, $f_{B} =\frac{ f_{Q}\cdot{}c}{(c - v_{Q})}$; if it recedes, the denominator is c + v_Q. If the observer moves towards a stationary source, $f_{B} =\frac{ f_{Q}\cdot{}(c + v_{B})}{c}$. Example: an ambulance at 700 Hz approaching at $\frac{30 m}{s}$ gives $f_{B} =\frac{ 700 \times{} 343}{313} \approx{} 767 Hz$. All speeds must be in $\frac{m}{s}$; the speed of sound is $\frac{343 m}{s}$ at 20 °C. Always check the result against the basic rule: approach raises the pitch, receding lowers it.

While the vehicle approaches, it compresses the sound waves ahead of it: the crests arrive at shorter intervals and the frequency is raised. At the moment of passing you briefly hear the true emitted frequency. Afterwards the receding source stretches the waves behind it and the pitch audibly drops. At 700 Hz and $\frac{30 m}{s}$ the tone jumps from about 767 Hz to 644 Hz, a clearly noticeable interval of almost three semitones. Note: during the approach the pitch stays constantly high, it does not glide; the characteristic jump happens only at the moment of passing, when the radial velocity reverses.

No, for sound the two cases are physically different because the medium, the air, forms a preferred reference frame. A moving source changes the wavelength in the medium and appears in the denominator: $f_{B} =\frac{ f_{Q}\cdot{}c}{(c \mp{} v_{Q})}$. A moving observer merely sweeps the unchanged waves faster or slower and appears in the numerator: $f_{B} =\frac{ f_{Q}\cdot{}(c \pm{} v_{B})}{c}$. Numerical example with v = 34.3 m/s (10 % of c) and approach: moving source gives the factor $\frac{1}{0.9} \approx{} 1.111$, moving observer 1.1. Only for speeds small compared with c do both approximations merge; for light the distinction does not exist at all.

A radar gun emits an electromagnetic wave of known frequency. The moving car reflects it and acts twice: once as a moving receiver, once as a moving transmitter. The returning wave is therefore shifted by $\Delta{}f \approx{}\frac{ 2\cdot{}f\cdot{}v}{c}$, twice as much as in the single Doppler effect. The measured frequency shift directly yields the speed: $v =\frac{ \Delta{}f\cdot{}c}{(2f)}$. Since light rather than sound is used, the relativistic formula applies, but for $v << c$ it reduces exactly to this simple approximation. Doppler sonography of blood flow and weather radar use the same principle.

For v_Q → c the formula grows towards infinity and for $v_{Q} > c$ it loses validity: ahead of the source the wave crests can no longer separate. The source overtakes its own wavefronts, which pile up into a cone, the Mach cone. Its half-angle follows from $\sin{} \theta{} =\frac{ c}{v}$. For a supersonic aircraft you hear this cone as a sonic boom when it sweeps over you, continuously along the flight path, not just when breaking the sound barrier. The ratio $\frac{v}{c}$ is the Mach number: Mach 2 means twice the speed of sound and a cone angle of 30°.