17.6 The Mathematics of Interference

Let’s look more closely at the superposition of two waves. As two waves of equal amplitude and frequency travel together along the x-axis, the net displacement of the medium is

\begin{array}{l} D = D_{1} + D_{2} & = a sin (k x_{1} - ω t + ϕ_{10}) + a sin (k x_{2} - ω t + ϕ_{20}) \\ = a sin ϕ_{1} + a sin ϕ_{2} \end{array}

(17.24)

where the phases $ϕ_{1}$ and $ϕ_{2}$ were defined in Equations 17.20.

A useful trigonometric identity is

sin α + sin β = 2 cos [\frac{1}{2} (α - β)] sin [\frac{1}{2} (α + β)]

(17.25)

This identity is certainly not obvious, although it is easily proven by working backward from the right side. We can use this identity to write the net displacement of Equation 17.24 as

D = [2 a cos (\frac{Δ ϕ}{2})] sin (k x_{avg} - ω t + {(ϕ_{0})}_{avg})

(17.26)

where $Δ ϕ = ϕ_{2} - ϕ_{1}$ is the phase difference between the two waves, exactly as in Equation 17.21. $x_{avg} = (x_{1} + x_{2}) / 2$ is the average distance to the two sources and ${(ϕ_{0})}_{avg} = (ϕ_{10} + ϕ_{20}) / 2$ is the average phase constant of the sources.

The sine term shows that the superposition of the two waves is still a traveling wave. An observer would see a sinusoidal wave moving along the x-axis with the same wavelength and frequency as the original waves.

But how big is this wave compared to the two original waves? They each had amplitude a, but the amplitude of their superposition is

A = | 2 a cos (\frac{Δ ϕ}{2}) |

(17.27)

where we have used an absolute value sign because amplitudes must be positive. Depending upon the phase difference of the two waves, the amplitude of their superposition can be anywhere from zero (perfect destructive interference) to 2a (maximum constructive interference).

The amplitude has its maximum value $A = 2 a$ if $cos (Δ ϕ / 2) = \pm 1$ . This occurs when

Δ ϕ = m \cdot 2 π (maximum amplitude A = 2 a)

(17.28)

where m is an integer. Similarly, the amplitude is zero if $cos (Δ ϕ / 2) = 0$ , which occurs when

Δ ϕ = (m + \frac{1}{2}) \cdot 2 π (minimum amplitude A = 0)

(17.29)

Equations 17.28 and 17.29 are identical to the conditions of Equations 17.22 and 17.23 for constructive and destructive interference. We initially found these conditions by considering the alignment of the crests and troughs. Now we have confirmed them with an algebraic addition of the waves.

It is entirely possible, of course, that the two waves are neither exactly in phase nor exactly out of phase. Equation 17.27 allows us to calculate the amplitude of the superposition for any value of the phase difference. As an example, FIGURE 17.22 shows the calculated interference of two waves that differ in phase by $40 °$ , by $90 °$ , and by $160 °$ .

The figure consists of three parts that illustrate the interference of two waves for three different values of phase difference.

FIGURE 17.22 Full Alternative Text

EXAMPLE 17.8 More interference of sound waves

Two loudspeakers emit 500 Hz sound waves with an amplitude of 0.10 mm. Speaker 2 is 1.00 m behind speaker 1, and the phase difference between the speakers is $90 °$ . What is the amplitude of the sound wave at a point 2.00 m in front of speaker 1?

MODEL The amplitude is determined by the interference of the two waves. Assume that the speed of sound has a room-temperature $(20 ° C)$ value of $343 m / s$ .

SOLVE The amplitude of the sound wave is

A = | 2 a cos (Δ ϕ / 2) |

where $a = 0.10 mm$ and the phase difference between the waves is

Δ ϕ = ϕ_{2} - ϕ_{1} = 2 π \frac{Δ x}{λ} + Δ ϕ_{0}

The sound’s wavelength is

λ = \frac{v}{f} = \frac{343 m / s}{500 Hz} = 0.686 m

Distances $x_{1} = 2.00 m$ and $x_{2} = 3.00 m$ are measured from the speakers, so the path-length difference is $Δ x = 1.00 m$ . We’re given that the inherent phase difference between the speakers is $Δ ϕ_{0} = π / 2 rad$ . Thus the phase difference at the observation point is

Δ ϕ = 2 π \frac{Δ x}{λ} + Δ ϕ_{0} = 2 π \frac{1.00 m}{0.686 m} + \frac{π}{2} rad = 10.73 rad

and the amplitude of the wave at this point is

A = | 2 a cos (\frac{Δ ϕ}{2}) | = | (0.200 mm) cos (\frac{10.73}{2}) | = 0.121 mm

REVIEW The interference is constructive because $A > a$ , but less than maximum constructive interference.