Lesson 50: Doppler Effect

You’ve probably noticed that when an ambulance passes you with its siren going you will hear the sound change.

Christian Doppler described the Doppler Effect in 1842.

Bob the Swimming Bug

To see why this change in frequency happens, I want you to imagine a very special bug named “Bob.”

Figure 1

Notice how Bob is in the middle and all the waves are spreading out evenly from him in the water.

After years of training I am able to get Bob to tread water while slowly moving to the right.

Figure 2

Now let’s look at what you would measure about Bob’s frequency if you were standing on the left side compared to the right side. Remember that Bob is actually treading water at 4 Hz.

Imagine that instead of Bob, you now have an ambulance coming towards you with its siren going.

This compression of waves in front of the source of the sound is also why it is difficult to go faster than the speed of sound. As the vehicle gets near the speed of sound, it must push through that wall of compressed air... not an easy thing to do. Click here to watch a quick video of a jet breaking the sound barrier... can you figure out why it looks like clouds form around it?

There is a formula that lets you calculate the apparent frequency of a sound if you know the true frequency of the sound and the difference in velocity between the source and the observer.

f ‘ = apparent frequency (Hz)
f = true frequency (Hz)
v = speed of sound (m/s)
Δv = difference in velocity between source and observer (m/s)

Example 1: A person is running around screaming at a frequency of 8500 Hz after hitting his thumb with a hammer. It is a cold day and the speed of sound is 300m/s.

a) If he is running towards you at 25 m/s (wow, that’s fast!) what frequency will you hear his scream as?

b) If he is running away from you at 20m/s what frequency will you hear?

c) If he is running toward you at 15m/s while you run towards him at 12m/s, what frequency will you hear?

d) If he is running towards you at 17 m/s and you are running away from him at 20m/s, what frequency will you hear?

To figure these out, use the formula you were given above. Be careful with the difference in velocity between the source and the observer. I’m not going to be careful about sig digs in my calculations here because I want you to easily see how the frequency changes. You still need to use sig digs when you do your calculations!

a) The source is approaching the observer at 25m/s, so we will use the positive sign.

b) The source is moving away from the observer at 20 m/s, so we use a negative sign.

c) The source and observer are moving closer together at (15m/s + 12 m/s) 27 m/s, so use the positive sign.

d) The source and observer are moving apart at (20m/s – 17m/s) 3 m/s, so use a negative sign.