A fire engine is rapidly approaching the corner. Explain what happens to the frequency of the sound wave as the fire engine draws closer. What do you hear?
Can some please help
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I Meant to say can someone please help
Reed posted those websites to help you.
What have you learned from them?
Sure, I can help you with that! When a fire engine approaches and passes by you, you will notice a change in the frequency of the sound waves it produces. This phenomenon is due to the Doppler effect.
The Doppler effect is the shift in frequency or wavelength of a wave as observed by an observer moving relative to the source of the wave. In the case of sound waves, it refers to the change in pitch or frequency that we perceive when a sound source is approaching or moving away from us.
When the fire engine approaches you, the sound waves it produces are compressed or "squeezed" because the source of the waves is moving closer to you. As a result, the frequency of the sound waves increases. This means that the pitch or tone of the siren's sound becomes higher than its original frequency.
So, what you hear when a fire engine approaches is a higher-pitched and more intense sound compared to when it is stationary. The sound becomes louder and more shrill until the fire engine passes by.
If you would like to calculate the change in frequency due to the Doppler effect, you can use the following formula:
f' = (v + vo) / (v - vs) * f
Where:
- f' is the observed frequency
- v is the speed of sound in air (approximately 343 m/s)
- vo is the speed of the observer relative to the medium (you)
- vs is the speed of the source relative to the medium (fire engine)
- f is the frequency emitted by the source (fire engine siren)
By plugging in the appropriate values, you can calculate the change in frequency and estimate how much higher the pitch will be when the fire engine approaches.
I hope this explanation helps! Let me know if you have any further questions.