At a school function, two loudspeakers on the stage play the same steady tone. Walk slowly along the front row. The sound is loud at one seat, almost silent two seats later, and loud again beyond. Each speaker alone sounds the same at every seat. The quiet seats appear only when both speakers play.
At one point, two waves are two sinusoids
Fix one point, say one seat. Each wave makes the air at that point oscillate with the frequency of its source. So at that point the two waves are two sinusoids in time, \(y_1 = a_1\sin\omega t\) and \(y_2 = a_2\sin(\omega t + \phi)\). The two waves add at every instant, which is the principle of superposition. Both have the same \(\omega\), because both speakers play the same tone. So the sum at the point is one sinusoid, and its amplitude comes from the phasor triangle.
Move the slider through its range and read the amplitude. At \(0°\) the two arrows line up and \(A = 2a\). At \(60°\), \(A = 1.73a\); at \(90°\), \(A = 1.41a\); at \(120°\), \(A = a\). At \(180°\) the arrows point opposite ways and the solid curve is flat, \(A = 0\). Beyond \(180°\) the resultant grows again and returns to \(2a\) at \(360°\).
When two waves meet in phase, \(\phi = 0, 2\pi, 4\pi, \ldots\), their phasors lie along one line and the amplitudes add, \(A = a_1 + a_2\). This is constructive interference. When they meet in opposite phase, \(\phi = \pi, 3\pi, 5\pi, \ldots\), the phasors point opposite ways and \(A = |a_1 - a_2|\). This is destructive interference.
\[ A^2 = a^2 + a^2 + 2a^2\cos\phi \]
Why. the phasor rule with two equal amplitudes, \(a_1 = a_2 = a\).
\[ A^2 = 2a^2(1 + \cos\phi) \]
Why. take \(2a^2\) out as a common factor.
\[ 1 + \cos\phi = 2\cos^2\dfrac{\phi}{2} \]
Why. the double-angle formula \(\cos\phi = 2\cos^2(\phi/2) - 1\).
\[ A = 2a\left|\cos\dfrac{\phi}{2}\right| \]
Read it. \(A^2 = 4a^2\cos^2(\phi/2)\); take the positive square root, since an amplitude is never negative.
The same result comes straight from the picture. Two equal phasors and their resultant make an isosceles triangle. The resultant bisects the angle \(\phi\) between the two phasors. Each phasor therefore contributes \(a\cos(\phi/2)\) along the resultant, and the two together give \(2a\cos(\phi/2)\).
Two waves of equal intensity \(I_0\) meet at a point with a phase difference of \(90°\). What is the resultant intensity?
From amplitude to intensity
The energy of an oscillation grows as the square of its amplitude. A spring stretched twice as far stores four times the energy. So the intensity of a wave, the energy it carries across unit area per second, is proportional to the square of its amplitude, \(I = kA^2\). Multiply the phasor rule by \(k\). Then \(ka_1^2 = I_1\), \(ka_2^2 = I_2\) and \(ka_1a_2 = \sqrt{I_1I_2}\), which gives the intensity formula at the top of this lesson.
Look at the dashed line in Fig 3.1. The term \(2\sqrt{I_1I_2}\cos\phi\) is positive at some places and negative at others. Across a pattern its average is zero, so the average intensity is just \(I_1 + I_2\). Interference does not destroy energy. It moves energy away from the dark places and puts it into the bright ones.
Two coherent waves have amplitudes in the ratio \(3 : 4\). What is the ratio of the maximum to the minimum intensity in their interference pattern?
At a dark place in an interference pattern of two equal waves, the intensity is zero. What has happened to the energy of the two waves there?
Where the phase difference comes from
A wave changes its phase by \(2\pi\) over each wavelength of path. Suppose two sources are in phase and the two paths to a point differ by \(\Delta x\). Then the waves arrive with the phase difference \(\phi = (2\pi/\lambda)\Delta x\). A path difference of \(n\lambda\) gives \(\phi = 2n\pi\), a loud or bright place. A path difference of \((n + \tfrac12)\lambda\) gives \(\phi = (2n + 1)\pi\), a quiet or dark place.
The formulas of this lesson need coherent sources: the phase difference must stay fixed in time. Two ordinary bulbs change their phase at random millions of times a second. Then \(\cos\phi\) averages to zero at every point, the intensities simply add, and no pattern appears. Add amplitudes, never intensities, for coherent waves: two equal waves in phase give \(4I_0\), not \(2I_0\).
This is the worked example of Superposition and Reflection, lesson 2, done with the tool. Two waves travel along one string: \(y_1 = 4\sin(2\pi x - 10\pi t)\) and \(y_2 = 4\sin(2\pi x - 10\pi t + \pi/3)\), with \(y\) in mm, \(x\) in m and \(t\) in s. Find the amplitude of the resultant wave, its phase ahead of \(y_1\), and its intensity compared with that of one wave.
The two waves share \(k\), \(\omega\) and direction, and their amplitudes are equal, \(a = 4\,\text{mm}\). At every point they are two sinusoids with the phase difference \(\phi = \pi/3\).
Equal phasors: \(A = 2a\cos(\phi/2)\). The resultant bisects the angle, so its phase is \(\phi/2\) ahead of \(y_1\). The intensity ratio is \(I/I_0 = 4\cos^2(\phi/2)\).
\(A = 2 \times 4 \times 0.866 = 6.93\,\text{mm}\). The phase lead is \(\pi/6\). \(I/I_0 = 4 \times 0.75 = 3\).
The string carries \(y = 6.93\sin(2\pi x - 10\pi t + \pi/6)\) mm, with three times the intensity of one wave alone.
Two coherent sources oscillate in phase and emit light of wavelength \(600\,\text{nm}\). A point P is \(5.0\,\mu\text{m}\) from one source and \(6.2\,\mu\text{m}\) from the other. What is the intensity at P?
Two loudspeakers play in phase at \(500\,\text{Hz}\). The speed of sound is \(340\,\text{m/s}\). A seat is \(1.02\,\text{m}\) farther from one speaker than from the other. How does the seat sound?
Two waves of equal amplitude meet at a point. For which phase differences is the resultant amplitude zero?
Superposition and Reflection uses the phasor rule for constructive and destructive interference on a string. Interference and Reflection of Sound finds the loud and quiet places near two speakers from the path difference. Interference and Young's Double Slit builds the intensity of two light waves and the fringe pattern from it. Thin Films, Newton's Rings and Other Interference explains the colours of a soap film and the anti-reflection coating the same way. Superposition, Damped and Forced Oscillations uses it for two SHMs that reinforce or cancel.
- Adding the intensities of two coherent waves.Add the amplitudes as phasors first, then square. Two equal waves in phase give \(4I_0\), not \(2I_0\).
- Thinking a large path difference always means a quiet place.Only the part of a wavelength left over matters. \(\Delta x = 2\lambda\) is as loud as \(\Delta x = 0\).
- Expecting zero intensity at a dark fringe when the waves are unequal.The minimum is \((\sqrt{I_1} - \sqrt{I_2})^2\), which is zero only when \(I_1 = I_2\).