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End of chapter exercises

Textbook Exercise 6.2

Write a definition for each of the following terms.

  1. Doppler effect

  2. Redshift

  3. Ultrasound

  1. The Doppler effect occurs when a source of waves and/or an observer move relative to each other, resulting in the observer measuring a different frequency of the waves than the frequency at which the source is emitting.

  2. Redshift is the shift in the position of spectral lines to longer wavelengths due to the relative motion of a source away from the observer.

  3. Ultrasound or ultra-sonic waves are sound waves with a frequency greater than \(\text{20 000}\) \(\text{Hz}\) (or \(\text{20}\) \(\text{kHz}\)).

The hooter of an approaching taxi has a frequency of \(\text{500}\) \(\text{Hz}\). If the taxi is travelling at \(\text{30}\) \(\text{m·s$^{-1}$}\) and the speed of sound is \(\text{340}\) \(\text{m·s$^{-1}$}\), calculate the frequency of sound that you hear when

  1. the taxi is approaching you.

  2. the taxi passed you and is driving away.

Draw a sketch of the observed frequency as a function of time.

  1. The observer is not moving, so \(v_{L}=0\), and the source is approaching. Therefore:

    \begin{align*} {f}_{L}&=\left(\frac{v}{v-{v}_{S}}\right){f}_{S} \\ &=\left(\frac{340}{340-30}\right)\left(500\right) \\ &= \text{548,4}\text{ Hz} \end{align*}
  2. Now the taxi is moving away from the observer, so:

    \begin{align*} {f}_{L}&=\left(\frac{v}{v+{v}_{S}}\right){f}_{S} \\ &=\left(\frac{340}{340+30}\right)\left(500\right) \\ &= \text{459,5}\text{ Hz} \end{align*}
  1. \(\text{548,4}\) \(\text{Hz}\)

  2. \(\text{459,5}\) \(\text{Hz}\)

A truck approaches you at an unknown speed. The sound of the truck's engine has a frequency of \(\text{210}\) \(\text{Hz}\), however you hear a frequency of \(\text{220}\) \(\text{Hz}\). The speed of sound is \(\text{340}\) \(\text{m·s$^{-1}$}\).

  1. Calculate the speed of the truck.

  2. How will the sound change as the truck passes you? Explain this phenomenon in terms of the wavelength and frequency of the sound.

  1. The observer is stationary, so \(v_{L}=0\). Therefore:

    \begin{align*} \frac{{f}_{L}}{f_{S}}&=\frac{v}{v-{v}_{S}} \\ \frac{v-{v}_{S}}{v} &= \frac{f_{S}}{f_{L}} \\ 1 - \frac{v_{S}}{v} &= \frac{f_{S}}{f_{L}} \\ \frac{v_{S}}{v} &= 1 - \frac{f_{S}}{f_{L}} \\ v_{S} &= 340 \left(1-\frac{210}{220} \right) \\ &= \text{15,5}\text{ m·s$^{-1}$} \end{align*}
  2. As the truck goes by, the frequency goes from higher to lower and the wavelength of the sound waves goes from shorter to longer.

  1. \(\text{15,5}\) \(\text{m·s$^{-1}$}\)

  2. As the truck goes by, the frequency goes from higher to lower and the wavelength of the sound waves goes from shorter to longer.

[Extension question] A police car is driving towards a fleeing suspect at \(\frac{v}{35} \text{m}·{\text{s}}^{-1}\), where v is the speed of sound. The frequency of the police car's siren is \(\text{400}\) \(\text{Hz}\). The suspect is running away at \(\frac{v}{68}\). What frequency does the suspect hear?


The source is moving towards the listener, so we need to put:


and the listener is moving away from the source, so we need to put:


So we use the formula as :


We are given that \(v_{L}= \frac{v}{68} \text{m}·{\text{s}}^{-1}\) and \(v_{S}=\frac{v}{35} \text{m}·{\text{s}}^{-1}\). Substituting into the formula:

\begin{align*} {f}_{L}&=\left(\frac{v-\frac{v}{68} }{v-\frac{v}{35}}\right){f}_{S} \\ &= \left(\frac{\frac{68v-v}{68}}{\frac{35v-v}{35}} \right){f}_{S} \\ &= \left(\frac{\left(\frac{67}{68}\right)v}{\left(\frac{34}{35}\right)v} \right){f}_{S} \\ &= \left(\frac{\left(\frac{67}{68}\right)}{\left(\frac{34}{35}\right)} \right)400 \\ {f}_{L}&= \text{405,7}\text{ Hz} \end{align*}

\(\text{405,7}\) \(\text{Hz}\)

Explain how the Doppler effect is used to determine the direction of flow of blood in veins.

An instrument called a Doppler flow meter can be used to transmit ultrasonic waves into a person's body. The sound waves will be reflected by tissue, bone, blood etc., and measured by the flow meter. If blood flow is being measured in an artery for example, the moving blood cells will reflect the transmitted wave and due to the movement of the cells, the reflected sound waves will be Doppler shifted to higher frequency if the blood is moving towards the flow meter and to lower frequency if the blood is moving away from the flow meter.

An person in a car travelling at \(\text{22,2}\) \(\text{m·s$^{-1}$}\) approaches a source emitting sound waves with a frequency of \(\text{410}\) \(\text{Hz}\). What is the frequency of the sound waves observed by the person? Assme the speed of sound in air is \(\text{340}\) \(\text{m·s$^{-1}$}\).

The Doppler formula is:


The source is stationary so \(v_{S}=0\). Rearranging the formula:

\begin{align*} {f}_{L}&=\left(\frac{v+{v}_{L}}{v}\right){f}_{S} \\ {f}_{L}& = \left(\frac{340+\text{22,2}}{340}\right)\left(410\right)\\ & = \text{436,8} \text{Hz} \end{align*}

\(\text{436,8}\) \(\text{Hz}\)