Doppler effect sensitive parabolic sound collector

AI Thread Summary
Tom uses a parabolic sound collector to record the frequency of a tuning fork dropped into the Grand Canyon, noting a frequency of 1887.0 Hz at t=7.880 seconds. The discussion centers on applying the Doppler effect to determine the natural frequency of the tuning fork, with calculations indicating a discrepancy in the expected results. Participants emphasize the importance of accounting for the fork's speed at the moment of sound emission, rather than its speed when the sound is detected. The need to calculate the distance the fork falls and the time it takes for the sound to travel back to the observer is highlighted as crucial for accurate frequency determination. Understanding the timing and speed of the fork during the sound emission is essential for applying the Doppler formula correctly.
Rasine
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Using a highly sensitive parabolic sound collector, Tom records the frequency of a tuning fork as it drops into the Grand Canyon. He drops the vibrating tuning fork from rest at t=0. He records a frequency of 1887.0 Hz at t=7.880 s. What is the natural frequency of the tuning fork? Use Vsound=343.0 m/s


ok so... i want to use fr=fs(v-us)/(v)

since tom is not moving ur=0 so i did not include that in the equation and i put the negitve becuase the fork is moving away from tom and the waves would be longer than it it was moving towards him

us=v=vo+at so us=77.224 m/s

1887=fs(343-77.224)/343

and i get 1462.05 Hz

what did i do wrong?
 
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Rasine said:
He records a frequency of 1887.0 Hz at t=7.880 s.
That's the time that he records the incoming sound. When did the tuning fork emit that sound?
 
With your equation I get 2435.28Hz.
 
Rasine said:
ok so... i want to use fr=fs(v-us)/(v)
The Doppler equation for a receding sound source should be:
f_{obs} = f_{source}\frac{c}{c + v}

Where c is the speed of sound; v is the speed of the source at time of emission.
 
the sound was emitted from the fork at t=0 so how do i take that into account?
 
if the fork is falling at t=7.880 it has a v=77.224 which the equation calls for that + the speed of sound

dosent that take it into account?
 
i don't know!
 
Rasine said:
the sound was emitted from the fork at t=0 so how do i take that into account?
No, t=0 is the time that the fork was dropped. It continually vibrates as it falls, picking up speed along the way. Note that it takes time for the sound to travel from the fork back to the top of the canyon.

Rasine said:
if the fork is falling at t=7.880 it has a v=77.224 which the equation calls for that + the speed of sound
True, at t=7.88 the fork has a speed of about 77 m/s. But that's not relevant, since you want to know how fast the fork was moving when it emitted the sound that was detected (at the top of the canyon) at t=7.88.

Before you can use the Doppler formula, you need to first figure out what speed the fork had when it emitted the sound that arrived at the detector at t=7.88. Hints for figuring that out: If the fork falls for T seconds it travels a distance D. Sound emitted at that time takes D/c seconds to reach the detector. The total time must equal t=7.88 seconds.
 

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