Final answer:
The equation of a harmonic wave with an amplitude of 5 m and a wavelength of 50 m traveling in the x-direction at t=0 is y(x, 0) = 5 m sin((2π/50 m⁻¹)x).
Step-by-step explanation:
The question revolves around writing the equation of a harmonic wave traveling in the x-direction at the specific time t=0. Given the amplitude of 5 m and a wavelength of 50 m, the wave equation can be expressed in the form:
y(x, t) = A sin(kx - ωt).
Since the wavelength (λ) is 50 m, then the wave number (k) is given by k = 2π/λ. Using the value of λ, we find that k = 2π/50 m⁻¹. At t=0, the angular frequency (ω) term drops out, simplifying our equation. Therefore, the equation of the wave at t=0 is:
y(x, 0) = 5 m sin((2π/50 m⁻¹)x).
This equation provides a snapshot of the wave along the x-axis at the initial moment.