A bat, flying at 5.1 m/s, pursues an insect that is flying at 1.1 m/s in the same direction. The bat emits a 47000-Hz sonar pulse. Take the speed of sound to be 343 m/s. show answer No Attempt At what frequency, in hertz, does the bat hear the pulse reflected back from the insect? f = | sin() cos() tan() cotan() asin() acos() atan() acotan() sinh() cosh() tanh() cotanh() Degrees Radians π ( ) 7 8 9 HOME E ↑^ ^↓ 4 5 6 ← / * 1 2 3 → + – 0 . END √() BACKSPACE DEL CLEAR Grade Summary Deductions 0% Potential 100% Submissions Attempts remaining: 3 (4% per attempt) detailed view Hints: 4% deduction per hint. Hints remaining: 2 Feedback: 5% deduction per feedback.
Question
Answer:
The frequency of the sona-pulse reflected back is [tex]f_b = 48109.22Hz[/tex]
Explanation:
From the question we are told that
The speed of the bat is [tex]v = 5.1 m/s[/tex]
The speed of the insect is [tex]v_i = 1.1 m/s[/tex]
The frequency emitted by the bat is [tex]f = 47000 \ Hz[/tex]
The speed of sound is [tex]v_s = 343 m/s[/tex]
Let look at this question in this manner
At the first instant the that the bat emits the sonar pulse
Let the bat be the source of sound
Let the insect be the observer
This implies that the frequency of sound the the insect would receive is mathematically represented as
[tex]f_a = [\frac{v_s – v_i}{v_s – v}] f[/tex]
Substituting values
[tex]f_a = [\frac{343 – 1.1}{345 -5.1} ] * 47000[/tex]
[tex]f_a = 47556.4 Hz[/tex]
Now at the instant the sonar pules reaches the insect
Let the bat be the observer
Let the insect be the source of the sound
Here the sound wave is reflected back to the bat
This implies that the frequency of sound the the bat would receive is mathematically represented as
[tex]f_b =[ \frac{x}{y} ] * f_a[/tex]
[tex]f_b =[ \frac{v_s + v}{v_s + v_i} ] * f_a[/tex]
[tex]f_b =[ \frac{343 + 5.1}{343 + 1.1} ] * 47556.4[/tex]
[tex]f_b = 48109.22Hz[/tex]