Interference
The combining of waves according to the principle of superposition is called interference. The two pulses of the last lesson interfered, first constructively and then destructively. Two long trains of waves interfere in the same way, but at every point and at every instant.
A single instant of interference is easy to make. A pattern that stays is harder. For a stable interference pattern, the sources of the waves must be coherent sources. They must emit waves of the same frequency, with a constant phase difference between them. If the phase difference drifted at random, the places of reinforcement and cancellation would wander along the string. Averaged over time, they would wash out, and no steady pattern would be seen.
Two sources are coherent when they emit waves of the same frequency with a phase difference that stays constant in time.
Phase difference and path difference
The kind of interference at a point depends on the phase difference \(\Delta\varphi\) between the waves that arrive there. The phase of a harmonic wave \(A\sin(kx - \omega t)\) is the angle \(kx - \omega t\), where \(k = 2\pi/\lambda\) is the angular wave number and \(\omega\) the angular frequency. Often the phase difference comes from a path difference \(\Delta x\): one wave has travelled further than the other to reach the point.
The two are linked through the wavelength \(\lambda\). The phase \(kx\) grows by \(2\pi\) over every wavelength of travel. So a path difference \(\Delta x\) gives a phase difference
\[ \Delta\varphi = k\,\Delta x = \dfrac{2\pi}{\lambda}\,\Delta x. \]
A path difference of one whole wavelength gives \(2\pi\), which is the same as no difference at all.
Which pairs of sources are coherent?
The two types of interference
The phase difference decides the type of interference. There are two extreme cases, and every other phase difference lies between them.
When waves arrive in phase, with a phase difference of \(0, 2\pi, 4\pi, \dots\), their amplitudes add. The result is a larger amplitude.
When waves arrive out of phase, with a phase difference of \(\pi, 3\pi, 5\pi, \dots\), their amplitudes subtract. The result is a smaller amplitude, or complete cancellation.
With \(\Delta\varphi = (2\pi/\lambda)\Delta x\), the same two conditions can be written for the path difference. A phase difference of \(2\pi n\) needs \(\Delta x = n\lambda\), a whole number of wavelengths. A phase difference of \((2n + 1)\pi\) needs \(\Delta x = (n + \tfrac12)\lambda\), an odd number of half wavelengths.
A wave pulse \(y_1 = f(x - vt)\) overlaps with another pulse \(y_2 = -f(x + vt)\). At the moment they are perfectly centred on each other, what is the displacement of the string?
Two coherent waves of wavelength \(0.4\,\text{m}\) start in phase and reach a point along paths that differ by \(0.6\,\text{m}\). What kind of interference happens there?
Two harmonic waves travelling the same way
Now take two harmonic waves of the same amplitude \(A\), the same \(k\) and the same \(\omega\), travelling the same way along one string. The second leads the first by a constant phase difference \(\Delta\varphi\):
\[ y_1 = A\sin(kx - \omega t), \qquad y_2 = A\sin(kx - \omega t + \Delta\varphi). \]
The sum can be found exactly, one step at a time.
\[ y = A\sin(kx - \omega t) + A\sin(kx - \omega t + \Delta\varphi) \]
Why. the principle of superposition: the string moves with the sum of the two displacements.
\[ \sin C + \sin D = 2\sin\dfrac{C + D}{2}\cos\dfrac{C - D}{2} \]
Why. the identity for the sum of two sines, used with \(C = kx - \omega t\) and \(D = kx - \omega t + \Delta\varphi\).
\[ \dfrac{C + D}{2} = kx - \omega t + \dfrac{\Delta\varphi}{2}, \qquad \dfrac{C - D}{2} = -\dfrac{\Delta\varphi}{2} \]
Why. add and subtract the two angles, then halve.
\[ y = 2A\cos(\Delta\varphi/2)\,\sin(kx - \omega t + \Delta\varphi/2) \]
Read it. put step 3 into step 2, and use \(\cos(-\theta) = \cos\theta\), since cosine is an even function.
Look at the result. The factor \(\sin(kx - \omega t + \Delta\varphi/2)\) is a harmonic wave with the same \(k\) and \(\omega\). So the resultant travels the same way at the same speed, with the same wavelength and frequency. Its phase lies halfway between the two. The factor \(2A\cos(\Delta\varphi/2)\) contains neither \(x\) nor \(t\). It is a constant, so it is the amplitude of the resultant wave.
The chapter Superposition, Damped and Forced Oscillations found the same amplitude for two SHMs, with a phasor triangle. The agreement is no accident. At any fixed point of the string, each wave makes the particle perform an SHM, and the two SHMs differ in phase by \(\Delta\varphi\).
Try the slider in the figure above. In phase, \(\Delta\varphi = 0\), the resultant has amplitude \(2.00\,\text{cm}\), twice one wave. At \(90°\) it is \(2\cos 45° = 1.41\,\text{cm}\). At \(120°\) it is \(2\cos 60° = 1.00\,\text{cm}\), the same as one wave alone. At \(180°\) the string stays flat, although both waves keep moving. At \(360°\) the amplitude is \(2.00\,\text{cm}\) again, because a whole cycle of phase changes nothing.
Intensity grows as the square of the amplitude
The chapter Propagating Waves on a String showed that the power carried by a harmonic wave is \(P = \tfrac12\mu v\omega^2A^2\). With the medium and the frequency fixed, the power grows as the square of the amplitude. The same is true of the intensity, the power carried per unit area. So a resultant of amplitude \(2A\cos(\Delta\varphi/2)\) has an intensity \(4\cos^2(\Delta\varphi/2)\) times that of one wave, \(I_0\).
Amplitudes add; intensities do not. Two equal waves in phase give twice the amplitude and so four times the intensity, \(4I_0\), not \(2I_0\). In opposite phase they give zero, not \(2I_0\). Add the displacements first, then square.
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.
Same \(k = 2\pi\,\text{rad/m}\), same \(\omega = 10\pi\,\text{rad/s}\), same direction and equal amplitudes \(A = 4\,\text{mm}\). The phase difference is \(\Delta\varphi = \pi/3\).
\(A_R = 2A\cos(\Delta\varphi/2)\), the phase of the resultant is \(\Delta\varphi/2\) ahead of \(y_1\), and \(I/I_0 = 4\cos^2(\Delta\varphi/2)\).
\(A_R = 2 \times 4 \times 0.866 = 6.93\,\text{mm}\). The phase lead is \(\pi/6\). \(I/I_0 = 4 \times 0.75 = 3\).
The string carries \(y = 6.93\sin(2\pi x - 10\pi t + \pi/6)\,\text{mm}\), with three times the intensity of one wave alone.
Two harmonic waves of amplitude \(2\,\text{mm}\) each travel the same way along a string. They have the same frequency and wavelength, and a phase difference of \(60°\). What is the amplitude of the resultant wave, in mm?
Two equal coherent waves, each of intensity \(I_0\), travel the same way along a string. One lags the other by a path difference of a quarter wavelength. What is the resultant intensity?
In 1801 the English physician Thomas Young stated the principle of interference in a lecture to the Royal Society in London. He had watched water waves reinforce and cancel, and he argued that light did the same. He later showed the same with light from two coherent sources. Bright bands appear where the waves arrive in phase, and dark bands where they arrive half a wavelength apart. Light added to light could give darkness.
- Adding the intensities of two coherent waves.Add the displacements or amplitudes first, then square. In phase, two waves give \(4I_0\).
- Using \(2A\cos\Delta\varphi\) instead of \(2A\cos(\Delta\varphi/2)\).The sum-to-product identity halves the angle. At \(\Delta\varphi = \pi\) the half angle is \(\pi/2\), and the amplitude is zero.
- Expecting a steady pattern from two sources of slightly different frequency.Coherence needs one frequency and a constant phase difference; otherwise the reinforcement drifts.