Set 56 Problem number 9

Problem (not required for Physics 122)

A transverse sine wave which propagates at 78 m/s is driven by a simple harmonic oscillator whose motion is described by the form y = A sin(`omega t).  Find the equation of motion for a particle at x = 4.6 meters given the amplitude .157 meters.

Solution

A transverse wave is one for which the particles move in a direction perpendicular to the direction of motion. In this case, since the direction of motion is perpendicular to the x direction, particles must move in some combination of the y and z directions (the only directions perpendicular to the x direction).

It would be possible for the particle to move only in the y direction, with a frequency of 84 Hz. Since the wave is a sine wave the particle would undergo simple harmonic motion with amplitude .157 m, so would have an equation of motion of form

• y = .157 m sin( 2`pi * 84 (t - timeLag) )

(note that the 2 `pi * 84 factor achieves a frequency of 84 Hz, since this factor ensures that when t changes by 1, 2`pi * 84 * t changes by through 84 * 2`pi , representing 84 complete cycles, 84 times).

The quantity timeLag is the time lag required for the disturbance to propagate from the x=0 point to the given position; it takes until t = timeLag for the motion of the particle to reach the same phase as the motion of the particle located at x=0. The time lag to a position on the string is the time required for the wave to propagate from x = 0 to that point.

At position x the time lag is given by

• timeLag(x) = dist / vel = x / v = 4.6 m /   ( 78 m/s) = .05897 sec

so that the equation of motion is

• y = .157 m sin(2 `pi * 84 ( t - .05897 sec) ).

This can also be written as

• y = .157 m sin(2 `pi * 84 * t  - 31.1 ).

Note that 2 `pi 84 = `omega, consistent with the circular model of Simple Harmonic Motion.

Generalized Solution

If the position and motion of the x = 0 particle are as described, and if the motion of the particle is all in the y direction, then the angular velocity of the point on the reference circle is 2 `pi f, since every  complete cycle corresponds to 2 `pi radians. (NOTE:  If the reference circle model of SHM is not clear, you should review the first few Physics I problems on Simple Harmonic Motion).  In this case, at x = 0 we have

• y motion at x = 0:  y = A sin ( (2 `pi f) t).

There is a time lag between the x = 0 point and an arbitrary coordinate x, corresponding to the time  required for the disturbance to travel the distance x. When t is equal to this time lag, the point at x will  begin its cycle just as the x = 0 particle did at t = 0. Thus the motion of the particle must be governed   by the equation

• y motion at arbitrary x:  y = A sin( (2 `pi f) (t - timeLag) ).

Note that at t = timeLag we yave y = A sin(0) = 0, just as we have at x = 0 when t = 0.  The motion of the two points on the wave will therefore be the same, except that the second starts at a time timeLag later.

The quantity timeLag is easily found.  It is just the time for the wave to propagate at velocity v from 0 to x, a distance x.   Thus

• timeLag = x / v

and the equation is

• y motion at arbitrary x:  y = A sin( (2 `pi f) (t - x / v) ),

which can also be written as

• y motion at arbitrary x:  y = A sin( (2 `pi f t - 2 `pi f / v * x ).

Since v / f is the wavelength `lambda, f / v is the reciprocal of `lambda, so we have

• y motion at arbitrary x:  y = A sin( (2 `pi f * t - 2 `pi / `lambda * x ).

This is the general equation of motion for a wave propagating in the positive x direction at velocity v = f * `lambda.

University Physics Students Note:

• We can write this equation explicitly showing the dependence of y on x and t:

• y(x, t)  = A sin( (2 `pi f * t - 2 `pi / `lambda * x ).

• In this form the equation satisfies the wave equation `d^2 y / `dt = 1/v^2 `d^2 y / `dx, where the 0 notation here is taken to express the partial derivative (e.g., `dy / `dt is the derivative with respect to t, where x is taken as a constant; `dy / `dx treats t as a constant).

• If the motion is not in the y direction, but is rather a combination of y and z motions, then there is a  some amplitude yAmplitude in the y direction and another amplitude zAmplitude in the z direction.  These amplitudes, being at right angles, add up by way of the Pythagorean theorem to the amplitude of motion. That is, yAmplitude^2 + zAmplitude^2 = amplitude^2. In this case the displacement from equilibrium can be described by y = yAmplitude sin( (2 `pi f) (t - timeLag) ) and z =  zAmplitude sin( (2 `pi f) (t - timeLag) ).