# How To Solve Doppler Effect Problems

If you’re having difficulty solving classical Doppler shift problems on your physics homework, you’re not alone! Many textbooks portray the Doppler shift formula in confusing ways that obscure real understanding. It is often stated as either several different formulas (one formula for a moving source and stationary observer, one formula for a stationary observer and moving source, etc) or one formula with a cruel and impossible-to-remember sign convention (for example, “source velocity is positive when the source moves in the direction of the emitted wave”).

I prefer to use what one of my past professors used to call the “Method of Thinking” to solve Doppler shift problems. (This is just one use of a more general Method of Thinking, in which you simply think about a problem to find the answer.) The basic ingredients are one easy-to-remember “incomplete formula,” two simple rules, and an instruction to think:

1. $\displaystyle f_{observed} = f_{emitted} \frac{|v|_{wave} \blacksquare |v|_{observer}}{|v|_{wave} \blacksquare |v|_{source}}$, where the velocities are relative to the medium and $\blacksquare$ stands for either plus or minus. The following two rules determine which sign to choose. (The formula is written in a general way to emphasize that the Doppler shift formula applies to many kinds of waves, but it is good to think of sound as a representative example. In that case, the wave speed is the speed of sound and the medium is air.)
2. If the source and receiver move towards each other (and one is stationary), the observed frequency increases.
3. If the source and receiver move away from each other (and one is stationary), the observed frequency decreases.
4. Think!

# Examples

1. Suppose the observer is stationary with respect to the medium and the source moves towards it. Then $v_{observer} = 0$ and Rule 2 tells us that the frequency must increase. So the correct formula to use is $\displaystyle f_{observed} = f_{emitted} \frac{|v|_{wave}}{|v|_{wave} - |v|_{source}}$. We choose the minus sign because shrinking the denominator of a fraction makes it grow.
2. Suppose the source is stationary with respect to the medium and the source moves away from it. Then $v_{source} = 0$ and Rule 3 tells us that the frequency must decrease. So the correct formula to use is $\displaystyle f_{observed} = f_{emitted} \frac{|v|_{wave} - |v|_{observer}}{|v|_{wave}}$. We choose the minus sign because shrinking the numerator of a fraction makes it decrease.
3. Suppose both the source and observer are moving with respect to the air. This situation involves a little bit more Rule 4 than the first two examples. The trick is to imagine a third entity (call it “The Phantom”) at rest with respect to the air. The Phantom listens to the sound emitted by the source (call this “Stage 1”) and emits an exact copy to the observer (call this “Stage 2”). Thus, The Phantom is both an observer and a receiver. In Stage 1, The Phantom hears a frequency given by $\displaystyle f_{phantom} = f_{emitted} \frac{|v|_{wave}}{|v|_{wave} \blacksquare |v|_{source}}$, where the sign is determined as in the previous examples. In Stage 2, the observer hears a frequency given by $\displaystyle f_{observed} = f_{phantom} \frac{|v|_{wave} \blacksquare |v|_{observer}}{|v|_{wave}}$, where the sign is determined as before. Inserting the equation for the Phantom frequency into the equation for the observed frequency gives the correct formula. Note that the top of the Phantom frequency fraction will cancel with the bottom of the observed frequency fraction.
4. If the source and the observer are at rest relative to the ground, there can still be a Doppler shift if the medium is moving. This is equivalent to the previous case. (This means that on a windy day, your voice could appear to be lower or higher.)

It is not useful to remember the specific equations; instead, try to learn the process that leads to them. This allows you generate formulas as you need them from a few basic concepts, instead of cluttering your head with multiple and/or confusing equations. Note: this is a good example of the power of the physicist’s mode of thinking.

Continue forward to learn where Rules 1, 2, and 3 come from!

# Where Did Rules 2 and 3 Come From?

If you carefully pay attention to this section, you will:

• Learn the derivation for Rules 2 and 3
• Be able to remember Rules 2 and 3 better

It’s always easier to remember a formula or rule if you know why it exists.

First, let’s think about how the observer could measure the frequency of the wave that he observes. The best way is probably to measure the time interval $\Delta T$ between successive waves; then the frequency is given by $f = 1/\Delta T$.

Now, let’s consider the stationary case, where neither the source nor the observer are moving. I have shown this case in the picture below. The source is represented by a guitar, and the observer is represented by an ear. At $t = 0$, the observer has just felt a wave (blue) and another wave is traveling towards it (red). Exactly one period ($T_{source}$) later, the source emits a new wave (yellow) and the red wave arrives at the observer. The observer therefore measures the time interval between waves to be $T_{source}$, so the observed wave has the same frequency as the emitted wave.

Next, let’s think about what happens if the observer starts to move towards the source, as shown in the picture below. Suppose that at $t=0$, the observer has just felt a wave (blue) and that there is another wave (red) exactly one wavelength away. As time progresses, the observer moves to the left and the wave moves to the right until the two meet. This is shown in the second part of the picture (where the previous positions of the observer and the wave are shown in gray). The wave has traveled less than a whole wavelength during this time interval, so $\Delta t$ must be less than $T_{source}$. If the observer measures a smaller period for the wave, then he consequently measures a larger frequency.

Finally, let’s think about what happens if the source starts to move towards the observer. As shown in the picture, the distance between the wavefronts will shrink. Since the wave’s speed remains the same, this means that the time interval that the observer measures between successive waves will also decrease. As before, a smaller time interval means a larger frequency.

These two cases combine to prove Rule 2; it is easy to come up with a similar proof for Rule 3 in which the objects move away from each other. These considerations also form the basis of a proof for the Doppler shift formula, which I will not write out here (you can find it in many textbooks). The main idea of the proof is to quantify exactly how the observed period changes in each case.

# Conclusions

Doppler effect problems are easier to solve if you know beforehand whether the frequency will decrease or increase; then you can simply modify the formula to fit your needs! Don’t forget, this strategy works for other formulas as well.

## 6 comments on “How To Solve Doppler Effect Problems”

1. Leslie says:

I just wanted to show my gratitude – thank you SO much for this excellent explanation. I agree that thinking through problems is much more effective than memorizing equations, especially when one like this changes depending on the situation at hand.

2. Prashant Bikram Shah says:

This is the excellent explanation that I ever received for Doppler effect!!I agree thinking all the problem in a particular way for this effect is quite reliable and best way to learn and solve!!thank you for this explanation !!!!

3. kbtayag says:

what frequency is heard of 440Hz (above middle C). and the observer moves away from the source with a speed of 60 MPH?

• eloch says:

I appreciate your interest in my blog but I will not answer homework questions for you.

4. Tracytron says: