How Does the Doppler Effect Change the Sound Frequency of a Passing Airplane?

AI Thread Summary
An airplane traveling at 154 m/s emits a sound frequency of 5.20 kHz, and the speed of sound is 344 m/s. To calculate the frequency heard by a stationary listener as the plane approaches, the formula used is f' = f * (v / (v - v_source)), resulting in a higher frequency. After the plane passes, the frequency is calculated using f' = f * (v / (v + v_source)), leading to a lower frequency. The discussion highlights the importance of using the correct Doppler effect equations for different scenarios. Understanding these concepts is crucial for solving related physics problems effectively.
PhysicsHatesMe01
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I've got another one.

An airplane traveling at v = 154 m/s emits a sound of frequency 5.20 kHz. At what frequency does a stationary listener hear the sound during each of the following times? (Use 344 m/s as the speed of sound.)

1) as the plan approaches (kHz)
2) after it passes (kHz)

What equation do I use to solve this?

Basically, I'm at a loss for almost all of my homework problems. My prof goes through the chapters during lecture, but he doesn't do very many examples which is why this class is causing me problems.

-- NVM. I figured it out :)
 
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PhysicsHatesMe01 said:
I've got another one.

-- NVM. I figured it out :)
Good job. See! Physics loves you! :biggrin:
 
PhysicsHatesMe01 said:
I've got another one.

An airplane traveling at v = 154 m/s emits a sound of frequency 5.20 kHz. At what frequency does a stationary listener hear the sound during each of the following times? (Use 344 m/s as the speed of sound.)

1) as the plan approaches (kHz)
2) after it passes (kHz)

For approaching, the formula for the Doppler-effect is different than for passing, since for approaching the frequencty the listener hears is getting higher --> mulitply with \frac {v} {v-v_{source}} and when passing the frequence gets lower --> multiply with \frac {v} {v + v_{source}}
 
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