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# Doppler Effect

## Waves emitted from a moving source are perceived are perceived at a higher or lower frequency by a stationary observer.

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Doppler Effect

Students will learn what the Doppler effect is and how to determine frequencies and speeds using the Doppler shift equations.

### Key Equations

Doppler Shifts: \begin{align*} f_o &= f \frac{v+v_o}{v-v_s} && f_o \mathrm{~(observed~frequency)~is~shifted~up~when~source~and~observer~moving~closer}\\ f_o &= f \frac{v-v_o}{v+v_s} && f_o \mathrm{~(observed~frequency)~is~shifted~down~when~source~and~observer~moving~apart, where}\\ & && v \text{ is the speed of sound, } v_s \text{ is the speed of the source, and } v_o \text{ is the speed of the observer} \end{align*}

Guidance
The Doppler Effect occurs when either the source of a wave or the observer of a wave (or both) are moving. When a source of a wave is moving towards you, the apparent frequency of the wave you detect is higher than that emitted. For instance, if a car approaches you while playing a note at 500 Hz, the sound you hear will be slightly higher frequency. The opposite occurs (the frequency observed is lower than emitted) for a receding wave or if the observer moves away from the source. It’s important to note that the speed of the wave does not change –it’s traveling through the same medium so the speed is the same. Due to the relative motion between the source and the observer the frequency changes, but not the speed of the wave. Note that while the effect is similar for light and electromagnetic waves the formulas are not exactly the same as for sound.

#### Example 1

Question: How fast would a student playing an A note (\begin{align*}440\mathrm{Hz}\end{align*}) have to move towards you in order for you to hear a G note (\begin{align*}784\mathrm{Hz}\end{align*})?

Answer: We will use the Doppler shift equation for when the objects are getting closer together and solve for the speed of the student (the source).

\begin{align*} f_o=f(\frac{v+v_o}{v-v_s}) \Rightarrow f_o\times (v-v_s)=f\times (v+v_o) \Rightarrow vf_o-v_sf_o=f\times (v+v_o) \Rightarrow v_s=-(\frac{f\times (v+v_o)-vf_o}{f_o}) \end{align*}

Now we plug in the known values to solve for the velocity of the student.

\begin{align*} v_s=-(\frac{f\times (v+v_o)-vf_o}{f_o})=-(\frac{440\mathrm{Hz}\times (343\mathrm{m/s}+0\mathrm{m/s})-343\mathrm{m/s}\times 784\mathrm{Hz}}{784\mathrm{Hz}})=151\mathrm{m/s} \end{align*}

### Time for Practice

1. What changes about a wave in the Doppler effect: Frequency? Wavelength? Speed? Is it correct to say that if an ambulance is moving towards you the speed of the sound from its siren is faster than normal? Discuss.
2. A friend plays an A note \begin{align*}(440\;\mathrm{Hz})\end{align*} on her flute while hurtling toward you in her imaginary space craft at a speed of \begin{align*}40\;\mathrm{m/s}\end{align*}. What frequency do you hear just before she rams into you? Assume the speed of sound to be 343 m/s.
3. How fast would a student playing an A note \begin{align*}(440\;\mathrm{Hz})\end{align*} have to move towards you in order for you to hear a \begin{align*}G\end{align*} note \begin{align*}(784\;\mathrm{Hz})\end{align*}? Assume the speed of sound to be 343 m/s

Answers

1. .
2. \begin{align*}498 \;\mathrm{Hz}\end{align*}
3. \begin{align*}150 \;\mathrm{m/s}\end{align*}

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