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Solution for (a)

(1) Enter known values into f obs = f s v w v w v s . size 12{f rSub { size 8{"obs"} } =f rSub { size 8{s} } left ( { {v rSub { size 8{w} } } over {v rSub { size 8{w} } +- v rSub { size 8{s} } } } right )} {}

f obs = f s v w v w v s = 150 Hz 340 m/s 340 m/s – 35.0 m/s size 12{f rSub { size 8{"obs"} } =f rSub { size 8{s} } left ( { {v rSub { size 8{w} } } over {v rSub { size 8{w} } - v rSub { size 8{s} } } } right )= left ("150"" Hz" right ) left ( { {"340"" m/s"} over {"340 m/s-35" "." "0 m/s"} } right )} {}

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

f obs = ( 150 Hz ) ( 1.11 ) = 167 Hz size 12{ {}= \( "150" ital "Hz" \) \( 1 "." "11" \) ="167" ital "Hz"} {}

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

f obs = f s v w v w + v s = 150 Hz 340 m/s 340 m/s + 35.0 m/s size 12{f rSub { size 8{"obs"} } =f rSub { size 8{s} } left ( { {v rSub { size 8{w} } } over {v rSub { size 8{w} } +v rSub { size 8{s} } } } right )= left ("150"" Hz" right ) left ( { {"340"" m/s"} over {"340 m/s-35" "." "0 m/s"} } right )} {}

(4) Calculate the second frequency.

f obs = ( 150 Hz ) ( 0.907 ) = 136 Hz size 12{ {}= \( "150" ital "Hz" \) \( 0 "." "97" \) ="136" ital "Hz"} {}

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 s = v obs = 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:

f obs = [ f s v w ± v obs v w ] v w v w ± v s . size 12{f rSub { size 8{"obs"} } = left [f rSub { size 8{s} } left ( { {v rSub { size 8{w} } +- v rSub { size 8{"obs"} } } over {v rSub { size 8{w} } } } right ) right ] rSub { size 8{s} } left ( { {v rSub { size 8{w} } } over {v rSub { size 8{w} } +v rSub { size 8{s} } } } right )} {}

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 obs ; however, because the horn is also moving in the direction away from the engineer, we also use the plus sign for v s . But the train is carrying both the engineer and the horn at the same velocity, so v s = v obs . As a result, everything but f s cancels, yielding

f obs = f s . size 12{f rSub { size 8{s} } } {}

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.

Suppose a jet airplane is coming nearly straight at you, emitting a sound of frequency f s . The greater the plane’s speed v s , the greater the Doppler shift and the greater the value observed for f obs . Now, as v s approaches the speed of sound, f obs approaches infinity, because the denominator in f obs = f s v w v w ± v s size 12{f rSub { size 8{"obs"} } =f rSub { size 8{s} } left ( { {v rSub { size 8{w} } } over {v rSub { size 8{w} } +- v rSub { size 8{s} } } } right )} {} approaches zero. At the speed of sound, this result means that in front of the source, each successive wave is superimposed on the previous one because the source moves forward at the speed of sound. The observer gets them all at the same instant, and so the frequency is infinite. (Before airplanes exceeded the speed of sound, some people argued it would be impossible because such constructive superposition would produce pressures great enough to destroy the airplane.) If the source exceeds the speed of sound, no sound is received by the observer until the source has passed, so that the sounds from the approaching source are mixed with those from it when receding. This mixing appears messy, but something interesting happens—a sonic boom is created. (See [link] .)

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Source:  OpenStax, College physics. OpenStax CNX. Jul 27, 2015 Download for free at http://legacy.cnx.org/content/col11406/1.9
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