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Video: Relating the Speed of a Sound Wave Source to the Doppler Shift Produced

Ed Burdette

A commuter train approaches a crossing and blows its horn to make a note of frequency 200 Hz. An observer at the crossing measures the frequency of the train’s horn as 208 Hz. The speed of sound in the train’s vicinity is 335 m/s. What is the speed of the train? If the train’s speed remains constant, what frequency does the observer measure as the train recedes from them?

06:24

Video Transcript

A commuter train approaches a crossing and blows its horn to make a note of frequency 200 hertz. An observer at the crossing measures the frequency of the train’s horn as 208 hertz. The speed of sound in the train’s vicinity is 335 meters per second. What is the speed of the train? If the train’s speed remains constant, what frequency does the observer measure as the train recedes from them?

We’re told in this statement that the source frequency of the train’s horn is 200 hertz, which we’ll call 𝑓 sub 𝑠. We’re also told that an observer at rest relative to the train hears the approaching train’s frequency as 208 hertz, which we’ll call 𝑓 sub 𝑜. We’re told that the speed of sound in the train’s vicinity is 335 meters per second, which we’ll call 𝑣 sub 𝑠.

In part one of the problem, we want to know the speed of the train. We’ll call that 𝑣 sub 𝑡. And in part two, we want to know if the train is moving away rather than toward the observer. What frequency does the observer notice? We’ll call this frequency 𝑓 sub 𝑟.

To start, let’s draw a diagram of this scenario. In this scenario, we have a train car moving along the tracks at a speed of 𝑣 sub 𝑡 that we want to solve for. As it approaches an observer standing by the side of the tracks, it blows its horn, which has a frequency 𝑓 sub 𝑠 of 200 hertz. The observer, however, hears a different frequency of 208 hertz.

Then in part two, the train car has passed the observer, still moving at the same speed. When it is past the observer and plays its horn, we want to know what frequency the observer will notice, which we’ve called 𝑓 sub 𝑟. The observed change in frequency of the train’s horn is due to the Doppler effect, which describes the observed frequency of a wave when there is motion relative to that wave front.

The Doppler equation says that the frequency an observer would notice, 𝑓 sub 𝑜, equals the source frequency of the wave, 𝑓 sub 𝑠, multiplied by a combination of speeds. Here, 𝑣 sub 𝑤 is the speed of the wave. 𝑣 sub 𝑜 is the speed of the observer relative to the source. And 𝑣 sub 𝑠 is the speed of the source that emits the wave with frequency 𝑓 sub 𝑠.

The plus and minus signs in the numerator and denominator have to do with whether the source is approaching or moving away from the observer. When the source is approaching, we use the top signs. And when the source is receding, we use those on the bottom.

Let’s apply the Doppler equation to our scenario to solve for 𝑣 sub 𝑡. The observed frequency of the train’s horn, 𝑓 sub zero, as the train approaches the observer is equal to the source frequency times the speed of sound, 𝑣 sub 𝑠, plus the observer’s speed, 𝑣 sub 𝑜, divided by the speed of sound minus the speed of the train, 𝑣 sub 𝑡.

We’ve chosen these signs for the numerator and denominator of the fraction because the train is approaching rather than moving away from the observer. In our example, the observer is stationary. So 𝑣 sub 𝑜 is zero. We want to solve this equation for 𝑣 sub 𝑡, the speed of the approaching train.

We can begin doing that by multiplying both sides of the equation by a quantity 𝑣 sub 𝑠 minus 𝑣 sub 𝑡, which cancels out that term on the right side of the equation. This resulting equation can be rearranged to solve for 𝑣 sub 𝑡. When fully simplified, we find that 𝑣 sub 𝑡, the train’s speed, is equal to 𝑣 sub 𝑠, the speed of sound, times the quantity one minus the source frequency divided by the observer frequency.

Since we’ve been told these three values in the problem statement, we can plug them in at this point. When we calculate this value, we find that 𝑣 sub 𝑡 to three significant figures is 12.9 meters per second. That’s how fast the train is moving along the track.

Now that we know the speed of the train, we can move on to part two: solving for the observed frequency when the train is past the observer and moving away. We’re going to once again apply the Doppler equation to our scenario. But this time, because the source of the sound is moving away from our observer, we’ll use the opposite set of signs than we did before when the train was approaching.

Just like before, our observer is standing still. So 𝑣 sub 𝑜 is zero. So the frequency the observer notices when the train is moving away or receding equals 𝑓 sub 𝑠 times the speed of sound divided by the sum of the speed of sound and the speed of the train 𝑣 sub 𝑡.

We’ve either solved for or been given each one of these values. So we can plug those values in now. When we enter these values on our calculator, we find that 𝑓 sub 𝑟, the frequency observed when the train is moving away from the observer, is 193 hertz. Notice that this frequency is lower than the source frequency, while when the train was approaching the observer, the observed frequency was higher. This is a normal result of the Doppler effect.