A stationary bat sends out an ultrasonic tone at 60,000Hz searching for food. At what frequency does the bat hear the echo from a dragonfly moving away from the bat at 5m/s?
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If the observer is moving away at velocity v and the normal velocity of sound wrt air is c,
then the apparent frequency of the sound wrt to the dragonfly is f(1- v/c) where f is the frequency of the original sound source which is 600 kHz. f1 = f°(1-v/c) v = 5 m/s; c = 330 m/s
Now when emitting the echo, the dragonfly acts as the new sound source and this time it is the source which is moving away from the stationary observer and the frequency heard by the bat is say f2 then
f2 = f1/(1 + v/c) ~ f1*(1-v/c) since v is only 5 m/s and velocity of sound in air is 331 m/s
this eventually leads to f2 = f°(1-v/c)^2 ~ f°(1-2v/c) = 60*(1-10/330) ~ 58.18 kHz
I am no physicist; will someone please check and confirm my solution and help this student?
then the apparent frequency of the sound wrt to the dragonfly is f(1- v/c) where f is the frequency of the original sound source which is 600 kHz. f1 = f°(1-v/c) v = 5 m/s; c = 330 m/s
Now when emitting the echo, the dragonfly acts as the new sound source and this time it is the source which is moving away from the stationary observer and the frequency heard by the bat is say f2 then
f2 = f1/(1 + v/c) ~ f1*(1-v/c) since v is only 5 m/s and velocity of sound in air is 331 m/s
this eventually leads to f2 = f°(1-v/c)^2 ~ f°(1-2v/c) = 60*(1-10/330) ~ 58.18 kHz
I am no physicist; will someone please check and confirm my solution and help this student?
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Double the bat's speed, = 10m/sec.
(343 + 10) = 353.
60,000 x (343/353) = 58,300.28Hz.
I used 343m/sec as speed of sound.
(343 + 10) = 353.
60,000 x (343/353) = 58,300.28Hz.
I used 343m/sec as speed of sound.