In each of the four scenarios to the right, a large bat lets out a short burst o
ID: 1390944 • Letter: I
Question
In each of the four scenarios to the right, a large bat lets out a short burst of ultrasonic sound, which the smaller bat hears a moment later. If the large bat flies at 3.090 m/s and the small bat flies at 14.80 m/s, rank the frequency that the smaller bat detects in the four scenarios, from highest to lowest. Assume that the speed of sound is 343.0 m/s.
In scenario C, some of the large bat's signal reflects off of the small bat and returns to the large bat, warning it of the smaller bat's presence. If the initial signal has a frequency of 24.30 kHz, what return frequency will the large bat detect? Calculate the final frequency to four significant figures.
Explanation / Answer
velocity of sound ( velocity of large bat)= Vs = 3.090 m/s
velocity of listner ( velocity of small bat) = Vl =14.80 m/s
V = velocity of sound in air = 343 m/s
f = actual frequency of sound = 24.30 Khz
(1) here large bat is moving towards small bat , small bat moving away from large bat:
so apprent freuqency ( freq. detect by small bat) = f' = ( V-Vl / V-Vs ) * f
f' = ( 343- 14.80/ 343- 3.090)*24.30
f' = 23.46 Khz
(2) here large and small bat moving away from each other:
f' = ( V-Vl / V- (-Vs) )*f
= (343-14.80/ 343+3.090)*24.30
= 23.04 KHz
(3) here large and small bat moving towards each other
f' = (V- (-Vl) / V-Vs ) *f
= (343+14.80/ 343-3.090)*24.30
= 25.57Khz
(4) here large bat moving away from small bat , small bat moving towards large bat
f' = (V- (-Vl) / V-(-Vs) ) *f
= 343+14.80/ 343+3.090) * 24.30
= 25.12 Khz
from above four cases higher frequency will be obtain in third case which is = 25.57 Khz
lowest frequency will be in second case which is = 23.04 Khz
high freq. / low freq = 1.10
ans (B) in case c apprent freq. is 25.57 Khz as calculated above
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