Question

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.

Answer 1

Answer:

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;

$y = A \cdot sin \left [B(x - C)\right ] + D$

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