Set 56 Problem number 10

Problem (not required for Physics 122)

A longitudinal wave is one in which the particles move parallel to the direction of motion. In this case, a sine wave will move each particle from its equilibrium position as dictated by a sine function.

If a longitudinal sine wave travels along the x axis at 40 m/s with a wavelength of .4 m and amplitude of particle motion equal to .23 meters, then what is the position vs. time function of a particle whose equilibrium position is at x = 8.2 meters? Assume that at time t = 0 the x = 0 particle is at its equilibrium position and moving in the positive direction.

Solution

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 40 m long and will have peaks .4 m apart, so there will be 40/ .4 = 100 peaks/second. The frequency is therefore 100 Hz. The displacement of a particle from equilibrium will therefore be ( .23 meters) sin(2`pi * 100 (t-timeLag)).

Since the equilibrium position is x = 8.2 meters, and since the displacement is in the x direction, the motion of the particle is characterized by the equation

• position(x)  = 8.2 meters + ( .23 meters) sin(2`pi * 100 (t-timeLag)).

The time lag between what happens at x = 0 and what happens at position 8.2 is the time required for the wave to propagate through this distance. The required time is

• time lag = 8.2 m / 40 m/s = .205 sec.

The position of the particle is therefor

• position(x) = 8.2 meters + ( .23 meters) sin(2`pi * 100 (t- .205)),

which can also be written

• position(x) = 8.2 meters + ( .23 meters) sin( (2`pi * 100) t - (2 `pi / 100) x ),

Generalized Solution

The frequency of a wave with wavelength 100 and propagation velocity v is f = v / 100. The equation of simple harmonic motion for the x = 0 particle under the given conditions must be

• displacement = A sin( (2 `pi f) t),

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

• displacement = A sin( (2 `pi f) (t - x/v) ).

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

• position(x) = x + A sin( (2 `pi f) (t -x/v) ) = x + A sin( (2 `pi f) t - (2 `pi / 'lambda) x ).