Suppose that the ultrasound source placed on the mother's abdomen produces sound at a frequency 2 MHz (a megahertz is 10^610 6
1 answer:
Answer:
the maximum frequency observed is 2.0044 10⁶ Hz
Explanation:
This is a Doppler effect exercise. Where the emitter is still and the receiver is mobile, therefore the expression that describes the process is
f ’=
the + sign is used when the observer approaches the source
typical speeds of a baby's heart stop are around 200 m / min
let's reduce to SI units
v₀ = 200 m / min (1 min / 60 s) = 3.33 m / s
let's calculate
f ’= 2 10⁶ ()
f ’= 2.0044 10⁶ Hz
f ’= 1,9956 10⁶ Hz
therefore the maximum frequency observed is 2.0044 10⁶ Hz
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where
By using the data of the problem and re-arranging (1), we find
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