17.7.2 A Problem-Solving Strategy for Interference Problems

The information in this section is the basis of a strategy for solving interference problems. This strategy applies equally well to interference in one dimension if you use $Δ x$ instead of $Δ r$ .

PROBLEM-SOLVING STRATEGY 17.1
Interference of two waves

MODEL Model the waves as linear, circular, or spherical.

VISUALIZE Draw a picture showing the sources of the waves and the point where the waves interfere. Give relevant dimensions. Identify the distances $r_{1}$ and $r_{2}$ from the sources to the point. Note any phase difference $Δ ϕ_{0}$ between the two sources.

SOLVE The interference depends on the path-length difference $Δ r = r_{2} - r_{1}$ and the source phase difference $Δ ϕ_{0}$ .

\begin{array}{l} Constructive: Δ ϕ = 2 π \frac{Δ r}{λ} + Δ ϕ_{0} = m \cdot 2 π \\ m = 0, 1, 2, \dots \\ Destructive: Δ ϕ = 2 π \frac{Δ r}{λ} + Δ ϕ_{0} = (m + \frac{1}{2}) \cdot 2 π \end{array}

For identical sources $(Δ ϕ_{0} = 0)$ , the interference is maximum constructive if $Δ r = m λ$ , maximum destructive if $Δ r = (m + \frac{1}{2}) λ$ .

REVIEW Check that your result has correct units and significant figures, is reasonable, and answers the question.

Workbook Exercise 18

EXAMPLE 17.10 Two-dimensional interference between two loudspeakers

Two loudspeakers in a plane are 2.0 m apart and in phase with each other. Both emit 700 Hz sound waves into a room where the speed of sound is $341 m / s$ . A listener stands 5.0 m in front of the loudspeakers and 2.0 m to one side of the center. Is the interference at this point maximum constructive, maximum destructive, or in between? How will the situation differ if the loudspeakers are out of phase?

MODEL The two speakers are sources of in-phase, spherical waves. The overlap of these waves causes interference.

VISUALIZE FIGURE 17.28 shows the loudspeakers and defines the distances $r_{1}$ and $r_{2}$ to the point of observation. The figure includes dimensions and notes that $Δ ϕ_{0} = 0 rad$ .

The figure illustrates a pictorial representation of the interference between two loudspeakers emitting sound waves with a frequency of 700 Hertz.

FIGURE 17.28 Full Alternative Text

SOLVE It’s not $r_{1}$ and $r_{2}$ that matter, but the difference $Δ r$ between them. From the geometry of the figure we can calculate that

\begin{array}{l} r_{1} = \sqrt{{(5.0 m)}^{2} + {(1.0 m)}^{2}} = 5.10 m \\ r_{2} = \sqrt{{(5.0 m)}^{2} + {(3.0 m)}^{2}} = 5.83 m \end{array}

Thus the path-length difference is $Δ r = r_{2} - r_{1} = 0.73 m$ . The wavelength of the sound waves is

λ = \frac{v}{f} = \frac{341 m / s}{700 Hz} = 0.487 m

In terms of wavelengths, the path-length difference is $Δ r / λ = 1.50$ , or

Δ r = \frac{3}{2} λ

Because the sources are in phase $(Δ ϕ_{0} = 0)$ , this is the condition for destructive interference. If the sources were out of phase $(Δ ϕ_{0} = π rad)$ , then the phase difference of the waves at the listener would be

Δ ϕ = 2 π \frac{Δ r}{λ} + Δ ϕ_{0} = 2 π (\frac{3}{2}) + π rad = 4 π rad

This is an integer multiple of $2 π rad,$ so in this case the interference would be constructive.

REVIEW Both the path-length difference and any inherent phase difference of the sources must be considered when evaluating interference.