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kateykob01
10.04.2021 •
Mathematics
An experiment in a wind tunnel generates cyclic waves. When the experiment starts (at time the wind is moving 13 feet per second (the slowest speed recorded). The speed maxes out at 71 feet per second and returns to 13 feet per second after 44 seconds. Let V represent the wind speed (velocity) in feet per second and let t represent the time in seconds. Write a sine equation that describes the wave.
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Ответ:
The sine equation for the wave is;
y = 29·sin[(π/22)·(x - 11)] + 42
Step-by-step explanation:
The given wind parameters in the wind tunnel are;
The slowest speed of the wind in the wind tunnel, v₁ = 13 feet per second
The maximum speed of the wind, v₂ = 71 feet per second
The time it takes the wind to complete one cycle = 44 seconds
Therefore, we have;
The general form of the sine equation is given as follows;
The period, T = 44 seconds = 2·π/B
∴ B = 2·π/44 = π/22
Given that the wave starts at the slowest speed, the horizontal shift
The amplitude, A = (v₂ - v₁)/2
∴ A = (71 ft./s - 13 ft./s)/2 = 29 ft./s
The vertical shift, D = v₁ + a = 13 ft./s + 29 ft./s = 42 ft./s
D = 42 ft./s
When x = 0, sin(B·C) = -1
∴ -B × C = -π/2
∴ C = π/2 ÷ B = π/2 ÷ (π/22) = π/2 × 22/π = 11
Therefore, the sine equation for the wave is;
the sine equation for the wave is;
y = 29·sin[(π/22)·(x - 11)] + 42
Ответ:
29.5735296
Step-by-step explanation:
hope this helped