The Doppler Effect
Moving Source
•Sound Spreads out in all directions from a moving
source
•When the source moves, the origin of the successive
circles also moves.
The Doppler Effect
•A simple way to think about the Doppler Effect is:
When the moving object is moving towards you,
emitting a sound wave, the actual frequency is higher
than the frequency you hear
When the moving object is moving away from you,
the actual frequency is less than the frequency you
hear
The Doppler Equation
•This is the only equation for the Doppler Effect,
where f’ represents the frequency received by the
observer
•f0 represents the original frequency of the source.
Doppler Equation Explanation of Velocity
•For the ease of understanding, Vo = 0 (Because
observer is stationary)
•However, when the source is moving towards you, for
the denominator we use V-Vs, and vice versa of the
train is moving away from you.
Comparing Frequencies
•When an object is moving towards you, V-Vs will be
smaller, so f’ will be larger. (Inverse relationship)
•When an object is moving away from you, V+Vs will
be larger and f’ will be smaller.
Question 1
•As the truck emits an 880 Hz siren. As he approaches
the observer, the observer perceives a 950 Hz siren.
What frequency does the observer hear when the
truck is moving away from the observer? Assume that
the velocity of the truck is constant and the velocity
of sound in air is 340m/s.
•810 Hz
•820 Hz
•830 Hz
•840 Hz
Determining the Unknowns
•Known: f’ when moving towards observer, Fo (does
not change), V and Vo (Vo = 0 since the observer is
stationary).
•Unknown: Vs (Speed of the source, which in this case
is the truck, f’ when moving away observer (This is
the answer to the question).
Finding Vs
•We simply plug in the numbers to find Vs
•950 = 880 (340 / (340-Vs)) (Minus sign because the
source is moving towards the observer)
•Therefore Vs = 25m/s
Finding f’ when moving away observer
•We substitute the numbers in, except we use V+Vs
for the denominator now instead of V-Vs, since the
truck is moving away.
•f’ = 880(340/(340+25) = 820 Hz
•2-Step that is a simple application of the Doppler
Equation
Question 2
•What best describes the effect of the Doppler
Shift on the Appearance of stars moving towards
Earth?
•Brighter?
•Bluer?
•Smaller?
•Larger?
Applying the Doppler Equation
First we know that the star is moving towards earth,
therefore the frequency is higher than the actual
frequency.
Next we need to determine how frequency relates to the
colour spectrum.
Colour Spectrum
•Higher Frequency means
shorter wavelength
(v=f*wavelength)
•Therefore, a star moving
towards Earth appears blue,
while star moving away
from Earth appears red.
Importance of Doppler Effect
•From the Doppler Shift of the light spectrum,
Astronomists can determine the speed of the star
moving towards or away from us,
References
• Physics for Scientists and Engineers – An Interactive Approach

The Doppler effect

  • 1.
  • 2.
    Moving Source •Sound Spreadsout in all directions from a moving source •When the source moves, the origin of the successive circles also moves.
  • 3.
    The Doppler Effect •Asimple way to think about the Doppler Effect is: When the moving object is moving towards you, emitting a sound wave, the actual frequency is higher than the frequency you hear When the moving object is moving away from you, the actual frequency is less than the frequency you hear
  • 4.
    The Doppler Equation •Thisis the only equation for the Doppler Effect, where f’ represents the frequency received by the observer •f0 represents the original frequency of the source.
  • 5.
    Doppler Equation Explanationof Velocity •For the ease of understanding, Vo = 0 (Because observer is stationary) •However, when the source is moving towards you, for the denominator we use V-Vs, and vice versa of the train is moving away from you.
  • 6.
    Comparing Frequencies •When anobject is moving towards you, V-Vs will be smaller, so f’ will be larger. (Inverse relationship) •When an object is moving away from you, V+Vs will be larger and f’ will be smaller.
  • 7.
    Question 1 •As thetruck emits an 880 Hz siren. As he approaches the observer, the observer perceives a 950 Hz siren. What frequency does the observer hear when the truck is moving away from the observer? Assume that the velocity of the truck is constant and the velocity of sound in air is 340m/s. •810 Hz •820 Hz •830 Hz •840 Hz
  • 8.
    Determining the Unknowns •Known:f’ when moving towards observer, Fo (does not change), V and Vo (Vo = 0 since the observer is stationary). •Unknown: Vs (Speed of the source, which in this case is the truck, f’ when moving away observer (This is the answer to the question).
  • 9.
    Finding Vs •We simplyplug in the numbers to find Vs •950 = 880 (340 / (340-Vs)) (Minus sign because the source is moving towards the observer) •Therefore Vs = 25m/s
  • 10.
    Finding f’ whenmoving away observer •We substitute the numbers in, except we use V+Vs for the denominator now instead of V-Vs, since the truck is moving away. •f’ = 880(340/(340+25) = 820 Hz •2-Step that is a simple application of the Doppler Equation
  • 11.
    Question 2 •What bestdescribes the effect of the Doppler Shift on the Appearance of stars moving towards Earth? •Brighter? •Bluer? •Smaller? •Larger?
  • 12.
    Applying the DopplerEquation First we know that the star is moving towards earth, therefore the frequency is higher than the actual frequency. Next we need to determine how frequency relates to the colour spectrum.
  • 13.
    Colour Spectrum •Higher Frequencymeans shorter wavelength (v=f*wavelength) •Therefore, a star moving towards Earth appears blue, while star moving away from Earth appears red.
  • 14.
    Importance of DopplerEffect •From the Doppler Shift of the light spectrum, Astronomists can determine the speed of the star moving towards or away from us,
  • 15.
    References • Physics forScientists and Engineers – An Interactive Approach