17.4 Doppler effect and sonic booms  (Page 3/9)

Calculate doppler shift: a train horn

Suppose a train that has a 150-Hz horn is moving at 35.0 m/s in still air on a day when the speed of sound is 340 m/s.

(a) What frequencies are observed by a stationary person at the side of the tracks as the train approaches and after it passes?

(b) What frequency is observed by the train’s engineer traveling on the train?

Strategy

To find the observed frequency in (a), $f_{\text{obs}}=f_{\mathrm{s}}\left(\frac{v_{\mathrm{w}}}{v_{\mathrm{w}}\pm v_{\mathrm{s}}}\right),$ must be used because the source is moving. The minus sign is used for the approaching train, and the plus sign for the receding train. In (b), there are two Doppler shifts—one for a moving source and the other for a moving observer.

Solution for (a)

(1) Enter known values into $f_{\text{obs}}=f_{\mathrm{s}}\left(\frac{v_{\mathrm{w}}}{v_{\mathrm{w}}–v_{\mathrm{s}}}\right).$

(2) Calculate the frequency observed by a stationary person as the train approaches.

(3) Use the same equation with the plus sign to find the frequency heard by a stationary person as the train recedes.

(4) Calculate the second frequency.

Discussion on (a)

The numbers calculated are valid when the train is far enough away that the motion is nearly along the line joining train and observer. In both cases, the shift is significant and easily noticed. Note that the shift is 17.0 Hz for motion toward and 14.0 Hz for motion away. The shifts are not symmetric.

Solution for (b)

(1) Identify knowns:

• It seems reasonable that the engineer would receive the same frequency as emitted by the horn, because the relative velocity between them is zero.
• Relative to the medium (air), the speeds are $v_{\mathrm{s}}=v_{\text{obs}}=\mathrm{35.0\; m/s.}$
• The first Doppler shift is for the moving observer; the second is for the moving source.

(2) Use the following equation:

The quantity in the square brackets is the Doppler-shifted frequency due to a moving observer. The factor on the right is the effect of the moving source.

(3) Because the train engineer is moving in the direction toward the horn, we must use the plus sign for $v_{\mathrm{obs}};$ however, because the horn is also moving in the direction away from the engineer, we also use the plus sign for $v_{\mathrm{s}}$ . But the train is carrying both the engineer and the horn at the same velocity, so $v_{\mathrm{s}}=v_{\mathrm{obs}}$ . As a result, everything but $f_{\mathrm{s}}$ cancels, yielding

Discussion for (b)

We may expect that there is no change in frequency when source and observer move together because it fits your experience. For example, there is no Doppler shift in the frequency of conversations between driver and passenger on a motorcycle. People talking when a wind moves the air between them also observe no Doppler shift in their conversation. The crucial point is that source and observer are not moving relative to each other.

Sonic booms to bow wakes

What happens to the sound produced by a moving source, such as a jet airplane, that approaches or even exceeds the speed of sound? The answer to this question applies not only to sound but to all other waves as well.