A wave with a period of 1.2 seconds and an oscillation amplitude of 2 cm propagates at a speed of 15 m / s

A wave with a period of 1.2 seconds and an oscillation amplitude of 2 cm propagates at a speed of 15 m / s. What is the displacement of a point located at a distance of 45 meters from the wave source at the moment when a time of 4 s has elapsed from the beginning of the source oscillations?

To find the displacement of the indicated point, we will use the plane wave equation: x (t) = A * sin (ω * t – k * x + φ0) = A * sin (2Π / T * t – ω / V * x + φ0) = A * sin (2Π / T * t – 2Π / (T * V) * x + φ0).

Variables: A – vibration amplitude (A = 2 cm = 0.02 m); T – period (T = 1.2 s); t – elapsed time (t = 4 s); V is the propagation speed (V = 15 m / s); x is the distance from the source (x = 45 m); φ0 is the initial phase (φ0 = 0).

Let’s calculate: x (4) = A * sin (2Π / T * t – 2Π / (T * V) * x + φ0) = 0.02 * sin (2 * Π / 1.2 * 4 – 2 * Π / (1.2 * 15) * 45 + 0) = 0.02 * sin (1.66 (6) Π) = 0.02 * sin 300º = 0.01732 m = 17.32 mm.

Answer: The offset of the indicated point is 17.32 mm.