A simple-harmonic wave train of amplitude 3 cm and frequency 200 Hz travels in a +ve direction of x-axis with a velocity of 20 m/s?

Let y be the displacement at distance x and time t.

Then y = Asin{(2π/λ)(x - vt)}, where

A = amplitude

λ = wavelength

v = wave velocity

Here A = 0.03m, v = 20 m/s, f = 200 Hz

We have λf = v → λ = v/f = 20/200 = 0.1m

So y = 0.03sin{2π/0.1(x - 20t)} = 0.03sin{20π(x - 20t)}

u (particle) = ∂y/∂t = -0.03 x 400πcos{20π(x - 20t)}

a = ∂²y/∂t² = -0.03 x 400π x 400πsin{20π(x - 20t)}

We want y, u and a when x = 0.5, t = 2

Subsituting:

y = -2 x 10^(-8)m = -2 x 10^(-6)cm

u = 37.7m/s

a = 0.03m/s²

The particle is 2 x 10^(-6)cm below the x-axis moving at 37.7m/s upwards and accelerating at 0.03m/s² upwards.