Wave Phase
and the
The reference circle is the primary tool for drawing and analyzing displacement versus position (snapshot) and Displacement versus time (history) graphs of a travelling harmonic wave and determining the phase of any point on any of the graphs.
The reference circle for a travelling sinusoidal wave has one vectors drawn on it, the displacement vector of size D. Normally the reference circle is drawn at x = 0 and t = 0, as shown in Figure 1. The phase constant φ0 is the positive (counterclockwise) angle between the x-axis and the position vector.
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Using the
The equation for a travelling
sinusoidal wave in one dimension is 
for
a wave travelling in the positive direction and 
for a wave travelling in the negative direction. The snapshot, or y-x,
graph is the same no matter which way the wave is travelling
since the equations are identical when t = 0. In Fig. 2(a), the
displacement vector D is in the first quadrant. Since we are dealing
with a sine function we are interested in the y-component of the
displacement vector. The y-component is 0.6D initially. As x
increases, we turn the displacement vector counterclockwise. So the y-component
will reach the maximum value of D. So the snapshot graph in Fig. 2b
starts at 0.6D and goes to a maximum.
The history, or y-t, graph is different for the two directions because of the sign difference for the wt term. Each is still simple to draw though as can be seen in Fig 3. The displacement vector D is still in the first quadrant and its y-component is 0.6D in Fig. 3(a). For a wave travelling to the right, the wt term has a negative sign in front. That means that increasing time turns the vector clockwise. So for the wave travelling to the right, the history graph starts at 0.6D but decreases to zero as shown in Fig. 3(b). For a wave travelling to the left, the wt term has a positive sign in front. That means that increasing time turns the vector counterclockwise. So for the wave travelling to the left, the history graph starts at 0.6D but increases to a maximum as shown in Fig. 3(c).
In summary start with x = 0 and t = 0 reference circle and determine y-component of the displacement. For a snapshot, y-x, graph see if the subsequent behaviour of the displacement vector as it turns counterclockwise is going to zero, or going to a maximum, or going to a minimum. For a history, y-t, graph of a wave travelling to the right see if the subsequent behaviour of the displacement vector as it turns clockwise is going to zero, or going to a maximum, or going to a minimum. For a history, y-t, graph of a wave travelling to the left see if the subsequent behaviour of the displacement vector as it turns counterclockwise is going to zero, or going to a maximum, or going to a minimum.
Using a Snapshot Graph to Determine the Phase Constant or Phase
Conversely a snapshot graphs can be used to find the phase constant f0 at x = 0 and t = 0. First, read the value of the displacement off the graph at x = 0 note whether it is positive or negative. Next see if the graph is going to zero, or a maximum, or a minimum after x = 0. This identifies which quadrant your vector is in. From the maximum value of the vector and from its value at x = 0, you have the hypotenuse and vertical side of a right triangle. It is simple trigonometry to find the angle of the right triangle. It is then usually simple trigonometry to relate this known angle to the phase constant f0. Figure 4 is an example. You are given a y-x graph. The maximum displacement is 10. The value of the displacement at x = 0 is –3. As you move in the positive direction (to the right), the displacement goes to a minimum. Thus we know that the displacement vector is in the third quadrant. Since the height is 3 and the hypotenuse is 10, the angle a for the known right triangle is a = arcsin(3/10) = 17.5°. That is the displacement vector, and thus the phase constant f0, is 197.5° from the positive x-axis.
If the snapshot graph is not shown at t = 0 but at some later time t = ¼T, there is an additional step as shown in Fig. 5. First we know the orientation of the displacement vector D at the current time ¼T. It is the same as Fig. 4 but recall that the phase constant f0 is measured from the x-axis at x = 0 and t = 0 not at t = ¼T. Now ¼T is equivalent to one-quarter of the way around the reference circle or 90°. For a wave travelling to the right, as time goes by the vector rotates clockwise. So to find the displacement vector’s orientation at t = 0 we must rotate 90° counterclockwise. Thus the phase constant would have been f0 = 197.5° + 90° = 287.5°. Similarly for a wave travelling to the left we would need to rotate 90° clockwise and the phase constant would have been f0 = 197.5° – 90° = 107.5°.
The phase q = kx ± wt + f0 = kx + f0, since t = 0, of any point at any other position x > 0 can also be found using the reference circle. Note that the phase q is measured counterclockwise from the positive x-axis to the displacement vector. Figure 6 shows a d-x graph and we are interested in knowing the phase at the point shown by the dashed vertical line. Since the displacement is at a maximum at the dashed line, the displacement vector must be lying along the positive y-axis. The phase q, from the x-axis to the position vector, is thus 90°.
Using a History Graph to Determine the Phase Constant or Phase
Similarly a history graph can be used to find the phase constant f0 at x = 0 and t = 0. However you need to know if the wave is travelling to the right or to the left since the displacement vector respectively rotates either clockwise or counterclockwise as time goes by. First, read the value of the displacement off the graph at x = 0 and t = 0 and note whether it is positive or negative. Next see if the graph is going to zero, or a maximum, or a minimum after x = 0 and t = 0. This identifies which quadrant your vector is in. From the maximum value of the vector and from its value at x = 0, you have the hypotenuse and vertical side of a right triangle. It is simple trigonometry to find the angle of the right triangle. It is then usually simple trigonometry to relate this known angle to the phase constant f0. Figure 7 is an example for a wave travelling to the right, v > 0. You are given a y-t graph. The maximum displacement is 10. The value of the displacement at x = 0 and t = 0 is –3. As time goes by (to the right on the graph), the displacement goes to a minimum. Since the displacement vector must rotate clockwise, we know that the displacement vector is in the fourth quadrant. Since the height is 3 and the hypotenuse is 10, the angle a for the known right triangle is a = arcsin(3/10) = 17.5°. That is the displacement vector, and thus the phase constant f0, is 342.5° from the positive x-axis.
The phase q = kx ± wt + f0 = ± wt + f0, since x = 0, of any point at any other time t > 0 can also be found using the reference circle. Note that the phase q is measured counterclockwise from the positive x-axis to the displacement vector. Figure 8 shows a d-x graph for a wave travelling to the right and we are interested in knowing the phase at the point shown by the dashed vertical line. Since the displacement is zero at the dashed line and increasing with time, the displacement vector must be lying along the negative x-axis because the vector rotates clockwise for a wave travelling to the right. The phase q, from the x-axis to the position vector, is thus 180°. If this had been a wave travelling to the left, the vector would have to lie along the positive x-axis since the vector would be rotating counterclockwise with time and the phase would be 0°.
The following link is to an interactive webpage where you can vary the phase constant and see the motion diagrams that this produces. Note that it will only work with MS Internet Explorer with the appropriate Office Web Components plugin.