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    How is the formula
    v/c=delta f/f

    Actually formed/derived?
    Thanks
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    Consider a wave, velocity c and wavelength \lambda, emitted from a stationary source:

    

c = f_0 \lambda

    

f_0 = \dfrac{c}{\lambda}

    Now, let's say the source begins to move at a velocity v. By extension, the velocity of the wave is now c+v:

    

c + v = f \lambda

    

f = \dfrac{c+v}{\lambda}

    Taking ratios:

    

\dfrac{f}{f_0} = \dfrac{c+v}{\lambda} \times \dfrac{\lambda}{c}

    

\dfrac{f}{f_0} = \dfrac{c+v}{c}

    

\dfrac{f}{f_0} = 1 + \dfrac{v}{c}

    I have a "1+" in my equation because I've used f instead of \Delta f, but you'll get what you have if you do \Delta f = f - f_0.



    NB: remember 'c' is the speed of the wave relative to the source, not the speed of light
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    (Original post by Qwertish)
    Consider a wave, velocity c and wavelength \lambda, emitted from a stationary source:

    

c = f_0 \lambda

    

f_0 = \dfrac{c}{\lambda}

    Now, let's say the source begins to move at a velocity v. By extension, the velocity of the wave is now c+v:

    

c + v = f \lambda
    With the final part, I am a bit confused. So you are talking about RELATIVE movement of the wave right?
    Thanks a lot for helping
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    (Original post by krisshP)
    With the final part, I am a bit confused. So you are talking about RELATIVE movement of the wave right?
    Thanks a lot for helping
    Relative velocity of wave with respect to a static observer.
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    (Original post by agostino981)
    Relative velocity of wave with respect to a static observer.
    Okay. But I am still struggling to unerstand WHY you adding c and v together
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    (Original post by krisshP)
    Okay. But I am still struggling to unerstand WHY you adding c and v together
    Without doing any relativity and suppose that the wave is linear, velocity is linear, i.e. additive.

    First, consider a moving particle with velocity v_p. If you are moving in the opposite direction with velocity v_o, you will find that the particle is moving faster than it actually is, and that the velocity v' you observed is actually related by v'=v_o+v_p (recall that velocity is a vector, which gives the linear property of velocity.)

    Then consider the wave velocity v and the wavelength is \lambda = vT, where T is the period of wave. If the source of the wave is moving with velocity c, then one would find that the wavelength is related by \lambda = (c+v)T. We can thus deduce that the velocity of wave is actually v'=c+v. For better illustration, you can actually draw a diagram with a moving source.
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    (Original post by krisshP)
    Okay. But I am still struggling to unerstand WHY you adding c and v together
    If I stand on the side of the road, and an ambulance passes, sounding it's siren. The ambulance is moving at 10m/s (v=10) and the speed of sound is 340m/s (c=340). From my point of view, the wave is moving at 350m/s.
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    (Original post by Qwertish)
    If I stand on the side of the road, and an ambulance passes, sounding it's siren. The ambulance is moving at 10m/s (v=10) and the speed of sound is 340m/s (c=340). From my point of view, the wave is moving at 350m/s.
    No I don't think so. The speed of sound is relative to the air. You would need to be moving (wrt the air), not the ambulance, to measure a different speed. Either that or the wind was blowing in your direction at 10m/s.
    There's a difference between moving source and moving observer in the case of sound.
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    (Original post by Stonebridge)
    No I don't think so. The speed of sound is relative to the air. You would need to be moving (wrt the air), not the ambulance, to measure a different speed. Either that or the wind was blowing in your direction at 10m/s.
    There's a difference between moving source and moving observer in the case of sound.
    The principal still holds though, the effective speed of the wave from your POV is v+c, otherwise the Doppler effect would be different for sound waves, which isn't the case...
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    (Original post by Qwertish)
    The principal still holds though, the effective speed of the wave from your POV is v+c, otherwise the Doppler effect would be different for sound waves, which isn't the case...
    The "effective speed of the wave" is not very helpful in this context. The wave speed doesn't change with a moving source. That's the fundamental principle.
    You only talk about effective speed (as measured by the observer) when the observer is moving. Not the source. The moving source effect is explained by the compression of the waves thus reducing the wavelength (for an approaching source) and a consequent increase in frequency.
    This needs the velocity to be constant (which it is) as you are saying if wavelength decreases then frequency increases; and using v=fλ with constant v.

    In this case of moving source it's the same with light.

    The difference with light is that the measured speed of light is constant even when the observer is moving. With light you cannot differentiate between the movement of the source and the movement of the observer. With sound you can, with reference to the medium.
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    (Original post by Stonebridge)
    The "effective speed of the wave" is not very helpful in this context. The wave speed doesn't change with a moving source. That's the fundamental principle.
    You only talk about effective speed (as measured by the observer) when the observer is moving. Not the source. The moving source effect is explained by the compression of the waves thus reducing the wavelength (for an approaching source) and a consequent increase in frequency.
    This needs the velocity to be constant (which it is) as you are saying if wavelength decreases then frequency increases; and using v=fλ with constant v.

    In this case of moving source it's the same with light.

    The difference with light is that the measured speed of light is constant even when the observer is moving. With light you cannot differentiate between the movement of the source and the movement of the observer. With sound you can, with reference to the medium.
    The speed of the wave relative to the medium is a constant, but when the source begins to move relative to the observer, the entire wave as a whole also moves faster (or slower, depending on direction) with respect to the observer, so the speed of a particular point on a the wave (say, a peak) will traverse the same displacement with less time.

    You could say the wavelength is squashed, but they're mathematically equivalent statements, otherwise the maths I wrote wouldn't work, but it does... :confused:
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    (Original post by Qwertish)
    The speed of the wave relative to the medium is a constant, but when the source begins to move relative to the observer, the entire wave as a whole also moves faster (or slower, depending on direction) with respect to the observer, so the speed of a particular point on a the wave (say, a peak) will traverse the same displacement with less time.

    You could say the wavelength is squashed, but they're mathematically equivalent statements, otherwise the maths I wrote wouldn't work, but it does... :confused:
    Yes there is confusion here between the speed of the vibrations of the medium and the speed of propagation of the wave.
    It's the speed of propagation of the wave in the medium that doesn't change; and that I'm referring to.
    What do you mean by "the entire wave moves faster". Yes it has to vibrate at a greater frequency (approaching source) so the particles of the medium do indeed vibrate (move) faster. But the wave itself does not propagate through the medium any faster.
    It's the propagation speed that we need for the Doppler Effect.
    Confusion resolved?
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    (Original post by agostino981)
    Without doing any relativity and suppose that the wave is linear, velocity is linear, i.e. additive.

    First, consider a moving particle with velocity v_p. If you are moving in the opposite direction with velocity v_o, you will find that the particle is moving faster than it actually is, and that the velocity v' you observed is actually related by v'=v_o+v_p (recall that velocity is a vector, which gives the linear property of velocity.)

    v'=c+v. For better illustration, you can actually draw a diagram with a moving source.
    This doesn't make sense for me.
    Consider a person who moves to the left at 5m/s and a car which moves to the left at -5m/s. Note the negative sign for opposite direction. The sum equals to 0m/s. Since 0>-5, so the car is observed moving faster. However logically the person wouldn't realistically observe the car at rest. :confused:
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    (Original post by krisshP)
    This doesn't make sense for me.
    Consider a person who moves to the left at 5m/s and a car which moves to the left at -5m/s. Note the negative sign for opposite direction. The sum equals to 0m/s. Since 0>-5, so the car is observed moving faster. However logically the person wouldn't realistically observe the car at rest. :confused:
    If a person moves to the left at 5m/s and the car moves to the left at -5m/s the sum equals 10m/s, not 0...

    This is not the same at the v+c formula. This one is the Galliean Transformation for adding two velocities together, which is:

    

v_{resultant} = v_1 - v_2

    Note the - sign.
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    (Original post by Qwertish)
    If a person moves to the left at 5m/s and the car moves to the left at -5m/s the sum equals 10m/s, not 0...

    This is not the same at the v+c formula. This one is the Galliean Transformation for adding two velocities together, which is:

    

v_{resultant} = v_1 - v_2

    Note the - sign.
    Thanks
 
 
 
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