Doppler Effect for a Moving Source and Stationary Observer

G14-1. Doppler Effect for a Moving Source and Stationary Observer

This is intended to supplement and amplify the derivation of the Doppler effect equation for a moving source and a stationary observer.

1. First consider a stationary source. We use the textbook notation that u = velocity of the source and v = velocity of the wave. The source is the black dot in the diagram below. The circles represent wave crests emitted one period T apart in time. The perpendicular distance between crests is the wavelength and is given by λ= vT = v/f, where f is the frequency of the source. Keep in mind that the waves are emitted in all directions and are therefore spherical.

2. Now let the source move to the right at a velocity of u = v/2 as shown below. The positions of the source are number-coded and color-coded to the wave crests that were emitted when the source was in the corresponding position. Each wave crest remains centered on the position where the source was when that wave crest was emitted.

We wish to determine the frequencies that observers at P and Q would perceive when the wave reached them. Refer to the diagram below. First note that the source moves a distance uT while the wave moves a distance λ = vT. The wavelength in front of the source is reduced by an amount uT and is λ_P = vT - uT = (v - u)/f. Now λ_P also equals v/f_P, where f_P is the frequency that the observer at P would perceive. Note that the velocity of the wave is unaffected by the motion of the source. The wavelength and frequency are the characteristics that change.

Combining relationships, v/f_P = (v - u)/f. Solving for f_P gives f_P = (1 - u/v)^-1f. Note that f_P is what the textbook calls f '. We choose to distinguish it from f_Q, the frequency perceived by observer Q. We leave it to you to show that f_Q = (1 + u/v)^-1f.

3. If the source moves at the speed of the wave, the wave crests cannot outrun the source in the direction of motion, and we get the result below. The overlap of wave crests ahead of the source produce a region of especially high pressure called a shock front. The observer at P would not perceive a particular frequency but would instead be subjected to a loud boom when the front reached her. The observer at Q would perceive a frequency f/2 determined by the previous equation.

4. If the source moves faster than the wave, then a conical shock front is produced. In the diagram below, the source velocity is twice that of the wave. There is a simple relationship that relates the vertex angle of the cone to v and u. In some cases, the shock fronts can be made visible. In those cases, the speed of the source can be determined easily by knowing the speed of waves in the medium and the vertex angle of the cone.

The geometry of the situation is shown below. We reproduce just the first wave crest for clarity. In time t, the source moves distance d_source = ut, while the wave moves distance d_wave = vt. d_source is the hypotenuse of a right triangle of which d_wave is the side opposite θ. (Note that θ is half of the vertex angle.) Therefore, as shown in the text box, we see that the sine of θ is the ratio of the wave velocity to the source velocity. Inverting this relationship, u/v = 1/sinθ. The ratio u/v is termed the Mach number. In the case shown, θ is 30° and u/v = 2. Therefore, the source is traveling at Mach 2. Here are some good gif animations of the Doppler effect.

A common misunderstanding about shock waves and sonic booms

Many people think that a sonic boom is produced at the instant that a jet exceeds the speed of sound. Actually, the sonic boom is produced continuously. In the diagram below, if the conical shock wave were being produced by a jet overhead, then the boom would be heard when the front passed over the observer on the ground. The front travels along the ground and reaches different locations at different times.