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11.4: Doppler Effect

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Objective

The student will:

• Understand the Doppler effect.

Vocabulary

• Doppler effect:

Introduction

Have you ever heard a car honking its horn or ambulance blaring its siren as it zooms past you at high speed?  It has a distinctive sound, like “eerrrrrrrroooooooom”.  It is not just that the sound gets louder, but it changes pitch, as well.  In other words, it changes in both amplitude and frequency.  The sound is illustrated in the video below:

The car horn distinctly sounds like it has a higher pitch as it is getting closer, and a lower pitch as it goes away.  Why does this happen?

Moving Source, Stationary Observer

Sound is a vibration that travels through the air.  Once the sound leaves the car, it travels at its normal wave speed through the air, regardless of how the car is moving. So if the car is moving quickly, it is catching up to the sound waves that it is sending forwards.  This means that the wavelength is effectively shorter by exactly how far the car moves over one period When the car passes the observer and is moving away, then the car is moving away from the sound waves it sends out, so the wavelength is longer. See Figure below.

When the wavelength is shorter, as seen to the right in the Figure above, the waves arrive together more frequently.  The higher frequency means a higher note or pitch,  When the waves are farther apart, as in the left of the Figure above, the waves arrive less frequently.  Frequency decreases, meaning a lower note or pitch.  This matches what we know from the wave equation:

$v = \lambda f ~~ \rightarrow ~~ \lambda = \frac{v}{f}$

Double the frequency means half the wavelength, and vice versa. The pitch sounds high as the car approaches, and low after it goes past.

Using this, we can figure out exactly what the change is.  When the car is going away, the wavelength is longer by exactly how far the car moves during the time of one period, $T = \frac{1}{f}$.   As before, we will call the wave speed $v$  The car is the source of the wave, so its velocity is labeled $v_s$ for velocity of the source.  Then the effectively longer wavelength $\lambda '$ (called “lambda prime”) is:

$\lambda ' = \lambda + vT = \left( \frac{v}{f} \right) + \left( \frac{v_s}{f} \right) = \frac{v + v_s}{f}$

This can alternately be expressed by what the effective new frequency $f '$ is compared to the original frequency $f$.

$f ' = \frac{ v }{ \lambda ' } = v \left( \frac{f}{v+v_s} \right) = f \frac{v}{v+v_s}$

The denominator here is larger, so this longer wavelength means a smaller frequency.  It will be heard as a lower pitch or note.

If the car is approaching us, then the new wavelength is shorter by the same amount.  We can use the same equation, but put in a negative value for the velocity of the source $v_s$

• Positive $v_s$  is the source moving away for longer wavelength.
• Negative $v_s$  is the source moving closer for shorter wavelength.

Moving Observer, Stationary Source

A similar effect is true if you are in a car going past a source of sound.  There is a small but important difference, though, because we assume that the air is stationary and the sound moves at a fixed speed through the air.  Therefore, the actual wavelength of the sound in the air is the same.  The difference is that as we move toward the source, we move past more waves per second.  So the wavelength is unchanged, but the effective wave speed is different.

In this case, the car is the receiver of the sound rather than the source, so we will call the car's speed $v_r$ for velocity of the receiver.  If the car is heading toward the source, then the waves are effectively moving faster.

$v ' = v + v_r ~~ \rightarrow ~~ f ' = \frac{v'}{\lambda} = \frac{v + v_r}{\lambda}$

The effect is similar.  Moving toward the source means that there is a higher frequency, so a higher pitch or note.

As before, the opposite is true if the car is moving away from the source.  Because the car is moving away, the waves have to catch up to the car, so their effective speed is slower.  We can represent this by putting in a negative number for $v_r$ in the equation above.

• Positive $v_r$ is the receiver moving toward the source for faster effective wave speed
• Negative $v_r$ is the receiver moving away from the source for slower effective wave speed

Source and Observer in Motion

Now let us consider if we are in a car that is chasing after an ambulance.  In this case, the source of sound is moving away with velocity $v_s$, making a longer wavelength.  The receiver is moving toward the sound with velocity $v_r$, making an effectively higher wave speed.

$f^{\prime} = \frac{v'}{\lambda'} = \frac{v + v_r}{\lambda + v_s T} = \frac{v + v_r}{v T + v_s T} = \left( \frac{1}{T} \right) \left( \frac{v + v_r}{v + v_s} \right) = f \left( \frac{v + v_r}{v + v_s} \right)$

Here both the velocity of the source and the velocity of the receiver are defined as before.

a. The siren of an ambulance produces a sound of frequency 700 Hz. If the ambulance approaches a person standing by the road with a velocity of 31.0 m/s, what is the frequency heard by the person? Take the velocity of sound to be $v = 343 ~ \text{m/s}$.

Answer: The ambulance is the source of the sound moving towards a stationary receiver. The source moving toward is an effectively shorter wavelength, so negative $v_s$. $f^{\prime} = f \frac{v}{v + v_s} = (700 ~ \text{Hz}) \left( \frac{343 ~ \text{m/s}}{343 ~ \text{m/s} + (-31 ~ \text{m/s})} \right) = 770 ~ \text{Hz}$

b. The ambulance passes the person standing.  The siren of an ambulance produces a sound of 700 Hz as it moves away from a stationary receiver with a velocity of 31 m/s. What is the frequency perceived by the observer?

Answer: The source is moving away for an effectively longer wavelength, so $v_s$ is positive.

$f^{\prime} = f \frac{v}{v + v_s} = (700 ~ \text{Hz}) \left ( \frac{343 ~ \text{m/s}}{343 ~ \text{m/s} + 31.0 ~ \text{m/s}} \right ) = 642 ~ \text{Hz}$

c.  The ambulance is now stopped by the side of the road still blaring its siren at 700 Hz, and a driver is approaching the ambulance at a velocity of 31 m/s. What frequency does the car driver hear?

Answer: The receiver is moving toward the source for an effectively higher wave speed, so $v_r$ is positive.

$f^{\prime} = f \frac{v + v_r}{v} = (700 ~ \text{Hz}) \left ( \frac{343 ~ \text{m/s} + 31.0 ~ \text{m/s}}{343 ~ \text{m/s}} \right ) = 763 ~ \text{Hz}$

d. The car has now passed the stopped ambulance, and is moving away from the ambulance at 31 m/s.  What frequency does the driver hear?

Answer: The receiver is moving away from the sound, so effective wave speed is smaller and the velocity of the receiver $v_r$ is negative.

$f^{\prime} = f \frac{v + v_r}{v} = (700 ~ \text{Hz}) \left ( \frac{343 ~ \text{m/s} + (-31.0 ~ \text{m/s})}{343 ~ \text{m/s}} \right ) = 637 ~ \text{Hz}$

Date Created:

Mar 11, 2013

Aug 19, 2014
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