Find the equation of motion for a particle at x = 8.8 meters along a longitudinal sine wave driven by an oscillator whose equilibrium position is at x = 0, which is its t = 0 position, and whose t = 0 velocity is positive. The wave is traveling along the x axis at 90 m/s with a wavelength of 1.4 m and amplitude of particle motion equal to .22 meters.
The particles at x = 0 move in SHM with modeled by the equation A sin(`omega t), since this function implies that at t = 0 we have x = 0 and motion in the positive x direction.
We begin by finding the frequency of the wave. A 1-second segment of the wave will be 90 m long and will have peaks 1.4 m apart, so there will be 90/ 1.4 = 64.28 peaks/second. The frequency is therefore 64.28 Hz. The displacement of a particle from equilibrium will therefore be ( .22 meters) sin(2`pi * 64.28 (t-timeLag)).
Since the equilibrium position is x = 8.8 meters, and since the displacement is in the x direction, the motion of the particle is characterized by the equation
The time lag between what happens at x = 0 and what happens at position 8.8 is the time required for the wave to propagate through this distance. The required time is
The position of the particle is therefor
which can also be written
The frequency of a wave with wavelength 64.28571 and propagation velocity v is f = v / 64.28571. The equation of simple harmonic motion for the x = 0 particle under the given conditions must be
where A is the amplitude of motion.
The particle at position x lags the x = 0 particle by the time required for the wave to propagate over the distance x. This time is x / v. So the function governing the simple harmonic motion of the particle whose equilibrium position is x will be
Since its equilibrium position is x and it oscillates in the x direction about this position, the particle will at time t be at position