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The change in frequency noticed by observers on sub b is : 2.66 Hz before the submarines pass each other.
Given data:
New speed of Sub b ( Vb ) = 12.4 m/s
F₁ = 1416.44 Hz
Frequency of sonar wave ( Fₐ ) = 1400 Hz
Speed of sound in water ( V ) = 1533 m/s
Speed of submarine A ( Va ) = 8.4 m/s
Determine the change in frequency observed on sub B
The change in frequency can be calculated using the formula below
F₁' = Fₐ ( V + Vb / V - Vₐ )
= 1400 ( 1533 + 12.4 / 1533 - 8.4 )
= 1400 ( 1545.4 / 1524.6 )
= 1419.1 Hz
Final step : determine the change in frequency
ΔF = F₁' - F₁
= 1419.1 - 1416.44
= 2.66 Hz.
Hence we can conclude that The change in frequency noticed by observers on sub b is : 2.66 Hz.
learn more about change in frequency : https://brainly.com/question/254161
The frequency is the number of waves that travel through a particular place in one unit of time. The change in frequency noticed by observers on sub b is 2.66 Hz.
What is the frequency?
The frequency is the number of waves that travel through a particular place in one unit of time. It may also be described as the number of cycles or vibrations experienced by a body in periodic motion in one unit of time.
Given that the new speed of the sub B is 12.4 m//sec, while the frequency of the submarine is 1416.44 Hz. The frequency of the sonar wave of submarine A is 1400 Hz. Also, the speed of sound in water, V is 1533 m/sec, and the speed of submarine A is 8.4 m/s.
Now, the frequency of the sub B after the speed is increased will be equal to,
[tex]F_1' = F_a \times \dfrac{(V+V_b)}{(V-V_a)}[/tex]
[tex]\rm = 1400 \times \dfrac{(1533+12.4)}{(1533-8.4)}\\\\=1400 \times \dfrac{1545.4}{1524.6}\\\\=1419.1\ Hz[/tex]
Further, the change in frequency noticed by observers on sub b,
[tex]\rm \triangle F = F_1'-F_1 = 1419.1-1416.44 = 2.66\ Hz[/tex]
Hence, the change in frequency noticed by observers on sub b is 2.66 Hz.
Learn more about Frequency:
https://brainly.com/question/5102661
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