# 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 )

Relevance

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

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