Using a highly sensitive parabolic sound collector, Frank 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 1747.0 Hz at t=5.110 s. What is the natural frequency of the tuning fork? Use Vsound=343.0 m/s.
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this is a problem in uniform acceleration coupled with the doppler shift
we want to find the velocity of the tuning fork at t=5.11s, assuming it is dropped from rest and neglecting friction, the speed of the fork at t=5.11 s is
v = 9.81m/s/s x 5.11s = 50.13 m/s
the fork is receding from the observer, so we expect that the observed frequency is shorter than the natural frequency
we find the natural frequency from:
f(observed) = [c/(v+c)] f(natural)
where v is the speed of the object and c is the speed of sound
we have
f(natural) = (v+c) f(observed)/c = (50.13m/s + 343m/s) x 1747Hz/343m/s = 2002Hz
of course...if you want to be completely accurate, you will take into account the fact that sound has a finite speed and that at t=5.11s, the fork is approx 125 m below the rim of the canyon, so that it took about 0.36s for the sound to reach you...therefore, the sound that you observe left the fork at some time before t=5.11, meaning the speed of the fork was not quite as large, and the doppler shift will not be quite so much...
so...the more exact analysis...
if the sound from the fork arrives at t=5.11s, we find the distance the fork has dropped from
t1 = time for fork to drop a distance D
t2= time for sound to travel D back to your ears
t1 = Sqrt[2D/g] t2= D/343m/s
therefore, Sqrt[2D/g] + D/343 = 5.11
Sqrt[2D/g] = 5.11 - D/343
square both sides:
2D/g = (5.11 - D/343)^2
this yields a quadratic in D whose physically meaningful solution is D =112.1 m
so the velocity of the fork after falling this distance is v = Sqrt[2 g D] = 46.87m/s
now, use the doppler shift equation to find the natural frequency of the fork:
f(natural) = (343m/s + 46.87m/s)x1747Hz/343m/s = 1986Hz
we want to find the velocity of the tuning fork at t=5.11s, assuming it is dropped from rest and neglecting friction, the speed of the fork at t=5.11 s is
v = 9.81m/s/s x 5.11s = 50.13 m/s
the fork is receding from the observer, so we expect that the observed frequency is shorter than the natural frequency
we find the natural frequency from:
f(observed) = [c/(v+c)] f(natural)
where v is the speed of the object and c is the speed of sound
we have
f(natural) = (v+c) f(observed)/c = (50.13m/s + 343m/s) x 1747Hz/343m/s = 2002Hz
of course...if you want to be completely accurate, you will take into account the fact that sound has a finite speed and that at t=5.11s, the fork is approx 125 m below the rim of the canyon, so that it took about 0.36s for the sound to reach you...therefore, the sound that you observe left the fork at some time before t=5.11, meaning the speed of the fork was not quite as large, and the doppler shift will not be quite so much...
so...the more exact analysis...
if the sound from the fork arrives at t=5.11s, we find the distance the fork has dropped from
t1 = time for fork to drop a distance D
t2= time for sound to travel D back to your ears
t1 = Sqrt[2D/g] t2= D/343m/s
therefore, Sqrt[2D/g] + D/343 = 5.11
Sqrt[2D/g] = 5.11 - D/343
square both sides:
2D/g = (5.11 - D/343)^2
this yields a quadratic in D whose physically meaningful solution is D =112.1 m
so the velocity of the fork after falling this distance is v = Sqrt[2 g D] = 46.87m/s
now, use the doppler shift equation to find the natural frequency of the fork:
f(natural) = (343m/s + 46.87m/s)x1747Hz/343m/s = 1986Hz