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Wize University Physics Textbook (Master) > Sound & Hearing

Doppler Effect

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Doppler effect





When there is a relative motion between the sound source and the observer the observer receives a sound wave with a different frequency from the original sound wave. This is called the Doppler Effect.


  • When a police car approaches you, the sound wave fronts are compressed, and you hear a sound with a higher frequency.

  • When the car moves away, the wave fronts are stretched out, creating a sound with a lower frequency.

  • This effect is also observed when the observer moves toward/away from the source.

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The frequency received by the receiver is calculated as:


 fr=v±vrv∓vs fs \boxed { \ f_r=\dfrac{v\pm v_r}{v\mp v_s}\ f_s \ }

  • vv is the speed of the sound in the medium
  • vrv_r is the receiver velocity
  • vsv_s is the source velocity
  • frf_r is the frequency perceived by the receiver
  • fsf_s is the frequency generated by the source


Wize Tip
How to use the above equation:
  • vrv_r: use the minus (plus) sign when moving away (towards)
  • vsv_s: use the minus (plus) sign when moving towards (away)




Example 1:

The person is stationary and the police car is moving.

















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Example 2:

The person is moving and the police car is stationary.



















Example 3:

Both the person and the police car are moving.


















Example: Harry Potter and the Moving Wall


Harry Potter lets out a horrified scream of frequency f0=500f_0 = 500 Hz towards a monstrous wall that is approaching at a staggering 2020 m/s. At what frequency does Harry Potter hear his own scream bounce off the wall? (Suppose the speed of sound is 340340 m/s)



Step 1:

From the point of view of the wall, the source is immobile (so vs=0v_s=0 and fs=f0=500f_s=f_0=500), and the observer (the wall) is moving at vr=20v_r=20. Therefore, the frequency perceived by the wall is:

 fr=v±vrv∓vs fs\ f_r=\dfrac{v\pm v_r}{v\mp \cancel{v_s}}\ f_s (use ++ in the numerator since the receiver is
moving towards)

fwall=(340+20340)(500)=529.4f_{wall}=\bigg(\dfrac{340+20}{340}\bigg)(500)= 529.4 (Hz)

Step 2:

Then, treat the wall as source (which is bouncing the sound back towards Harry Potter with a frequencyfs=529.4f_s=529.4 and moving with velocity vs=20v_s=20) and Harry Potter as the observer when receiving his scream (he is at rest so vr=0v_r=0). The frequency he hears is:

 fr=v±vrv∓vs fs\ f_r=\dfrac{v\pm \cancel{v_r}}{v\mp v_s}\ f_s (use −- in the denominator since the source is
moving towards)

fHP=(340340−20)(529.4)=562.5f_{HP}=\Big(\dfrac{340}{340-20}\Big)(529.4)=562.5 (Hz)


Practice: A Bat Chasing a Rodent


A bat is chasing a rodent which is running away at 2020 m/s. The bat is flying at 1010 m/s and is emitting an ultrasound wave with a frequency of 2121 kHz toward the rodent. The ultrasound wave reflects back from the rodent's body and is received by the bat again. What is the frequency heard by the bat? (The speed of ultrasound waves is 15401540 m/s)