Tommy asked in Science & MathematicsPhysics · 10 months ago

Hi guys, i need a bit of help with an physics assigment ( can't solve it and my brother needs it for tomorrow)?

It's assumed that 1 car is moving with v1=72km/h which passes a biker who is moving with 36km/h. If the frequency of the car is 500 Hz, what will be the frequency of the engine that the biker will hear before it's passed, and after it's passed given the speed of sound of 340m/s ? any help would be appreciated because it will literally blow my mind if i keep trying to solve it. Thank you )

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  • 10 months ago
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    f’ = observed frequency

    f = source frequency

    v = speed of sound

    vₛ = source’s speed= 72 km/h = 20m/s

    v₀ = observer’s speed = 32km/h = 10m/s

    If the vehicles are getting closer use:

    f’ = f(v+v₀)/(v-vₛ)

    = 500 * (340 + 10)/(340 – 20) = 547Hz (before car passes biker)

    If the vehicles are moving apart use:

    f’ = f(v-v₀)/(v+vₛ)

    = 500 * (340 - 10)/(340 + 20) = 458Hz (after car passes biker)

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  • 10 months ago

    The topic is known as Doppler Effect.

    There's plenty of video on YouTube with

    simplification.

    Convert kph to m/s

    (72-36)/3.6 = 10 m/s

    500 × [(340+10)/(340)] = 514.7 approaching

    500 × [(340-10)/(340)] = 485.3 passing

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  • 10 months ago

    The car needs to keep 1 meter from the biker. Or 2. I’m not sure

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