Speed of a jet (challenging doppler effect)

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Homework Help Overview

The problem involves a fighter jet at an air show, where the frequency of sound from the jet engine drops by one octave as it approaches and then passes overhead. The context is related to the Doppler effect and the calculation of the jet's speed based on observed frequency changes.

Discussion Character

  • Exploratory, Assumption checking, Mixed

Approaches and Questions Raised

  • Participants discuss the relationship between the observed frequency and the original frequency, noting the significance of the frequency drop by one octave. There are attempts to set up equations based on the Doppler effect, but some express confusion about the number of variables compared to equations available. Others question the notation used and suggest clarifying the equations.

Discussion Status

Some participants have reached a numerical conclusion regarding the speed of the jet, with one suggesting a Mach number of 0.5. However, there is no explicit consensus on the correctness of this conclusion, as others express uncertainty about the reasoning and calculations involved.

Contextual Notes

There are indications of confusion regarding the notation and the setup of equations, as well as a lack of clarity about the assumptions made in the problem. The discussion reflects varying interpretations of the situation described.

Ethan_Tab
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Homework Statement


At an air show, a fighter jet does manuevers past the crowd. Your increible hearing notes that the frequnecy of the sound coming from the jet engine drops exactly by one octave when it is approaching you to when it is directly above you (not moving relitive to you). How fast is the jet flying?

Homework Equations


Mach #= velocity/velocity sound
ƒobs=ƒo(v+-d/v+-s) where ƒo= frequency original, d= velocity of detector, s= velocity of source and v is the velocity of sound.

The Attempt at a Solution


[/B]
Since a drop by one octave is the same as half the frequency, we can represent the frequency observed when the plane approaches as ƒ and the frequency heard directly above as f/2.

we can then make two equations;
Approaching--- ƒ=ƒo(v/v-s)
Directly above--- ƒ/2=ƒo(v/v) (no s since its not moving relative to you at that exact moment)

However there are still more variables then equation, not sure how to go on from here.
 
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Hello ET,

Scary story! jet plane overhead, not moving relative to you. Hanging from a thread ? With engine running ?

Never mind. Your notation is awful, but in ##f/2 = f_0## I see one variable disappearing, so you are left with one frequency only. Write out the equation and have a eureka moment ! :woot:
 
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Eureka indeed.

I got the answer as being mach 0.5 for the speed of the plane. Would you concur?
 
The picture of a jet plane coming straight at you at half the speed of sound is also rather terrifying! Better step aside real fast !

And your answer is exactly what I got -- so either we're both right or both wrong :wink:
 
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