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    I am studying Edexcel AS Physics. I cannot understand what is the doppler effect. This is in unit 2: topic 3. Any one help me by providing some resources relating to this.
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    (Original post by Rashina)
    I am studying Edexcel AS Physics. I cannot understand what is the doppler effect. This is in unit 2: topic 3. Any one help me by providing some resources relating to this.
    The Doppler effect is the difference in observed frequency heard than the actually frequency itself.

    A good example would be if an ambulance is emitting its siren.
    The way it works is that if a stationary observer is standing and an ambulance is travelling towards the observer while it's siren is playing then the frequency heard by the observer while appear greater than the actual frequency. E.g. If the ambulances siren is emitting a 300Hz signal and is travelling toward the observer then the frequency heard depending on the velocity of the ambulance is travelling at, it could be 330Hz.

    When the ambulance is travelling away from the stationary observer then it will appear to have a lesser frequency.

    Hope this helped.


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    Oh yeah I forgot to mention there is also another area of the Doppler effect which ties into stars and their wavelength which is called redshift and blueshift.


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    Imagine that you've got a speaker in the middle of the room. As you know, sound is a pressure wave and like all waves, it has a wavelength, frequency and velocity.

    So let's do some basic calculations. The speed of sound in air is about 340m/s and let's say that this speaker is playing a note at 1Hz (which would be too low to hear, but we'll use that to simplify calculations). We know that v=f\lambda so \lambda = \frac{v}{f}. The wavelength is therefore \frac{340}{1} = 340 m. You can intuitively understand this. The speaker drum is vibrating at 1Hz so it sends out one pressure pulse per second. This means that between the first pulse and the second pulse, the first pulse has travelled for one second = 340m before the second pulse is made.

    Now, we're going to start moving the speaker. Specifically, we're going to start moving the speaker at a constant speed of 10m/s, directly away from us. The speaker drum goes up, and a pressure wave is made, that travels towards us at the speed of sound, i.e. 340m/s. The frequency of the drum has not changed so one second later, the second pressure wave is made. However, remember that this time, the sound source has been moving away from us. So whilst the first pressure wave has moved 340m away from the original location of the speaker, the speaker is now 10m further away. So the second pressure wave is 340+10m = 350m from the first, so rather than experiencing a sound wave with a wavelength of 340m, you experience a sound wave with a wavelength of 350m. Plugging this into our v=f\lambda formula, we find that the new frequency is 0.97Hz, i.e. slightly lower than the sound you heard when the speaker was at rest. So if you imagine that you could hear it, it would have a lower pitch.

    You can repeat the same experiment, this time imagining that the speaker is moving towards you. This time, the separation between wavefronts will be 340m-10m = 330m, so the frequency you would hear is 1.03Hz.

    That's the basic principle. A source emitting waves at a constant frequency has its frequency shifted up (and its wavelength shifted down) when it is moving towards you, and its frequency shifted down (and its wavelength shifted down) when it is moving away from you. This principle applies to all waves, e.g. electromagnetic radiation, not just sound waves.

    Does that help?
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    Here is a demonstration of the Doppler effect:

    https://www.youtube.com/watch?v=imox...avidrobert2007

    Recorded on a potato!
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    (Original post by RossB1702)
    The Doppler effect is the difference in observed frequency heard than the actually frequency itself.

    A good example would be if an ambulance is emitting its siren.
    The way it works is that if a stationary observer is standing and an ambulance is travelling towards the observer while it's siren is playing then the frequency heard by the observer while appear greater than the actual frequency. E.g. If the ambulances siren is emitting a 300Hz signal and is travelling toward the observer then the frequency heard depending on the velocity of the ambulance is travelling at, it could be 330Hz.

    When the ambulance is travelling away from the stationary observer then it will appear to have a lesser frequency.

    Hope this helped.


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    This was the one I needed
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    (Original post by Plagioclase)
    Imagine that you've got a speaker in the middle of the room. As you know, sound is a pressure wave and like all waves, it has a wavelength, frequency and velocity.

    So let's do some basic calculations. The speed of sound in air is about 340m/s and let's say that this speaker is playing a note at 1Hz (which would be too low to hear, but we'll use that to simplify calculations). We know that v=f\lambda so \lambda = \frac{v}{f}. The wavelength is therefore \frac{340}{1} = 340 m. You can intuitively understand this. The speaker drum is vibrating at 1Hz so it sends out one pressure pulse per second. This means that between the first pulse and the second pulse, the first pulse has travelled for one second = 340m before the second pulse is made.

    Now, we're going to start moving the speaker. Specifically, we're going to start moving the speaker at a constant speed of 10m/s, directly away from us. The speaker drum goes up, and a pressure wave is made, that travels towards us at the speed of sound, i.e. 340m/s. The frequency of the drum has not changed so one second later, the second pressure wave is made. However, remember that this time, the sound source has been moving away from us. So whilst the first pressure wave has moved 340m away from the original location of the speaker, the speaker is now 10m further away. So the second pressure wave is 340+10m = 350m from the first, so rather than experiencing a sound wave with a wavelength of 340m, you experience a sound wave with a wavelength of 350m. Plugging this into our v=f\lambda formula, we find that the new frequency is 0.97Hz, i.e. slightly lower than the sound you heard when the speaker was at rest. So if you imagine that you could hear it, it would have a lower pitch.

    You can repeat the same experiment, this time imagining that the speaker is moving towards you. This time, the separation between wavefronts will be 340m-10m = 330m, so the frequency you would hear is 1.03Hz.

    That's the basic principle. A source emitting waves at a constant frequency has its frequency shifted up (and its wavelength shifted down) when it is moving towards you, and its frequency shifted down (and its wavelength shifted down) when it is moving away from you. This principle applies to all waves, e.g. electromagnetic radiation, not just sound waves.

    Does that help?
    Yes it helps, thank you
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    (Original post by johnTroy1)
    Here is a demonstration of the Doppler effect:

    https://www.youtube.com/watch?v=imox...avidrobert2007

    Recorded on a potato!
    Thank you.
 
 
 
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