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3 years ago

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After enjoying a tasty meal of the first moth, the bat goes after another moth. Flying with the same speed and emitting the same

frequency, this time the bat detects a reflected frequency of 55.5 kHz. How fast is the second moth moving

1 answer:

Aleksandr-060686 [28]3 years ago

4 0

This question is incomplete, the complete question is;

A bat flies towards a moth at 7.1 m/s while the moth is flying towards the bat at 4.4 m/s. The bat emits a sound wave of 51.7 kHz.

After enjoying a tasty meal of the first moth, the bat goes after another moth. Flying with the same speed and emitting the same frequency, this time the bat detects a reflected frequency of 55.5 kHz. How fast is the second moth moving

Answer:

the second moth is moving at 5.062 m/s

Explanation:

Given the data in the question;

Using doppler's effect

$f_{moth$ = f₀( $v_{s$ ± $v_{observer$ / $v_{s$ ± $v_{source$ )

f₁ = f₀( ( $v_{s$ + v₂) / ( $v_{s$ - v₁ ) )

frequency reflected from the moth,

Now, moth is the source and the bat is the receiver

f₂ = f₁( ( $v_{s$ + v₁ ) / ( $v_{s$ - v₂ ) )

hence, f = f₀[ ( ( $v_{s$ + v₁ ) / ( $v_{s$ - v₂ ) ) ( ( $v_{s$ + u₂ ) / ( $v_{s$ - u₁ ) )

we know that, the velocity of sound $v_{s$ = 343 m/s.

given that v₁ and v₂ { velocity of bat } = 7.1 m/s, f₀ = 51.7 kHz and f = 55.5 kHz.

we substitute

55.5 = 51.7[ ( ( 343 + 7.1 ) / ( 343 - 7.1 ) ) ( ( 343 + u ) / ( 343 - u ) ) ]

55.5 = 51.7[ ( 350.1 / 335.9 ) ( ( 343 + u ) / ( 343 - u ) ) ]

55.5 = 51.7[ 1.04227 ( ( 343 + u ) / ( 343 - u ) ) ]

55.5 = 53.885359 ( ( 343 + u ) / ( 343 - u ) ) ]

55.5 / 53.885359 = ( 343 + u ) / ( 343 - u )

1.02996 = ( 343 + u₂ ) / ( 343 - u )

( 343 + u₂ ) = 1.02996( 343 - u )

343 + u = 353.27628 - 1.02996u

u + 1.02996u = 353.27628 - 343

2.02996u = 10.27628

u = 10.27628 / 2.02996

u = 5.062 m/s

Therefore, the second moth is moving at 5.062 m/s

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