You stand on a crowded platform at a big railway junction. An announcement booms from the speakers, a tea-seller calls out, and a train sounds its horn. All three sounds are waves in the same air, and they cross one another on the way to your ears. Yet you can still pick out each sound on its own.
The golden rule of wave interactions
The rule of this lesson is the golden rule of wave interactions. Almost everything that waves do together rests on it. It holds in a linear medium, a medium whose waves obey a linear wave equation. In practice this needs one condition: the amplitude must not be so large that the wave changes the properties of the medium itself.
Why must the medium be linear? The wave equation of a string, \(\partial^2 y/\partial x^2 = (1/v^2)\,\partial^2 y/\partial t^2\), from the chapter Introduction to Waves, has \(y\) only to the first power. So if \(y_1\) and \(y_2\) each satisfy it, their sum satisfies it too. Pull a string so far that its tension changes, and the equation gains higher powers of \(y\). Then the sum of two waves is no longer a solution, and the rule of this lesson fails.
Let two or more waves overlap in a medium. The resultant displacement of any particle at any instant is then the vector sum of the separate displacements. Each separate displacement is the one that a single wave alone would give the particle there. For displacements along one line, as on a string, the vector sum is the algebraic sum, signs included.
This is the principle of superposition. Written for two or more waves on a string, it says:
\[ y_{net}(x, t) = y_1(x, t) + y_2(x, t) + \dots \]
Here \(y_1\) is the displacement that the first wave alone would give at the position \(x\) and the time \(t\). Likewise \(y_2\) is that of the second wave alone.
The principle looks simple, but it is a powerful tool for understanding how waves behave. Take two pulses travelling in opposite directions along one rope. When the pulses meet and overlap, their displacements add at every point, as the principle says. Before you press play, predict the result.
Two equal upright pulses, each \(0.8\,\text{cm}\) high, run towards each other along one string. At the instant their centres meet, how high is the string there?
In the figure above, the string is \(10\,\text{m}\) long. The two pulses start \(6\,\text{m}\) apart and move at \(2\,\text{m/s}\), so their centres meet at \(x = 5\,\text{m}\) after \(1.5\,\text{s}\). The dashed curves are \(y_1\) and \(y_2\) on their own, and the solid curve is the string, \(y_1 + y_2\). As the pulses pass through each other, their heights add. For a moment the string forms one combined pulse, \(1.6\,\text{cm}\) high. This is constructive interference: the net displacement is greater than the displacement of either pulse alone.
When one pulse is inverted
Now invert one of the pulses, so that it points down while the other points up. When such pulses meet, their displacements partly or completely cancel. The net displacement is then smaller than that of either pulse alone. This is destructive interference, and Fig 1.1 shows it.
Press play in Fig 1.1 and watch the readout. Under the peak of \(y_1\) the string is at \(0.80\,\text{cm}\) while the pulses are apart. At \(1.5\,\text{s}\) the readout falls to zero, and the whole string is flat for an instant. Then the two pulses separate, each with its own shape again.
Where is the energy at the flat instant? The string is not stretched, so it stores no potential energy. But it is still moving. In the overlap both pulses push each piece of string the same way, so the pieces move at their fastest. For that instant all the energy is kinetic, and this motion builds the two pulses again.
An upright pulse \(3\,\text{mm}\) high and an inverted pulse \(5\,\text{mm}\) deep travel towards each other on a string. At the instant their peaks are at the same point, what is the displacement of the string there, in mm?
The same rule in a table
In both figures, the two waves travel in the same medium and overlap each other. At every point, the resultant displacement is the sum of the displacements that each wave alone would cause there, \(y_{net}(x, t) = y_1(x, t) + y_2(x, t)\). The same principle can be read from numbers.
Fig 1.2 freezes two waves at one instant. The first is \(y_1 = 3\sin(2\pi x/8)\) and the second is \(y_2 = 2\cos(2\pi x/8)\), with \(y\) in millimetres and \(x\) in centimetres. Five positions are marked, \(2\,\text{cm}\) apart.
The table below lists the same values. The second column is the displacement \(y_1\) caused by the first wave. The third column is the displacement \(y_2\) caused by the second wave. These values change with time as the waves move. At any one instant, the fourth column is the resultant, the algebraic sum of the two before it. The positions run from \(x_1 = 0\) to \(x_5 = 8\,\text{cm}\). Every displacement is in millimetres.
\[ \begin{array}{c|c|c|c} x & y_1 & y_2 & y_1 + y_2 \\ \hline x_1 & 0 & 2 & 2 \\ x_2 & 3 & 0 & 3 \\ x_3 & 0 & -2 & -2 \\ x_4 & -3 & 0 & -3 \\ x_5 & 0 & 2 & 2 \end{array} \]
Read the row of \(x_3\). The first wave gives nothing there, and the second wave gives \(-2\,\text{mm}\), so the medium is at \(-2\,\text{mm}\). Every row demonstrates the principle numerically.
At one point of a string, the first wave alone would give a displacement of \(4\,\text{mm}\), and the second wave alone \(-1.5\,\text{mm}\). What is the displacement of the string there, in mm?
Two laser beams that ignore each other
The figures, the table and the equations all say one thing. The disturbance caused by one wave is not affected by the presence of the other waves. The principle holds for all waves, light included. Let a red laser beam travel along the \(x\)-axis and a blue laser beam along the \(y\)-axis, in the same medium. The two beams of light, which are electromagnetic waves, meet at the origin. Neither affects the other's travel: the red beam goes on exactly as it would without the blue beam.
A red laser beam and a blue laser beam cross at right angles in a room. What does a screen in the path of the red beam, beyond the crossing, show?
True or false: when two waves collide, they bounce off each other like billiard balls and change direction.
So waves can pass through each other without being destroyed or permanently altered. During the overlap, the displacement of the medium is their combined effect. After they pass, each goes on with its original shape and properties. This principle is what lets us hear many sounds at once. It also lets us see many sources of light without their light scrambling together.
A tall, narrow pulse moves right along a string, and a short, wide pulse moves left. They cross. After they have separated completely, what is on the string?
Music is playing in a room, and a friend speaks to you across it. Why can you still hear your friend's words clearly?
The music and the voice are two sound waves in the same air. By the principle of superposition, the air at the ear moves with the sum of the two displacements. Neither wave changes the other, so the voice arrives with its own shape. The ear and brain can then pick it out from the sum.
A plucked string does not vibrate as one simple curve. In a memoir of 1753, Daniel Bernoulli argued that its motion is a sum of simple sine-shaped vibrations. Each vibration goes on as if the others were absent. Euler doubted that such a sum could describe every shape of a string. Yet Bernoulli's idea was the principle of superposition. It later became the method of Fourier.
In any Indian city, the air in your room carries the waves of many FM stations, mobile towers and Wi-Fi routers at once. They cross one another all the time. A radio set tuned to one station still receives that station's wave unharmed. Each wave passes through all the others unchanged, so the set can pick out the one it wants.
Now return to the railway platform. The announcement, the tea-seller's call and the horn are three waves in one air. Where they overlap, the air follows the sum of the three displacements. None of them changes the others, so each wave reaches your ears with its own shape. That is why each sound can still be picked out.
- Adding the sizes of two displacements and forgetting their signs.Up is positive and down is negative. An upright \(3\,\text{mm}\) and an inverted \(5\,\text{mm}\) give \(-2\,\text{mm}\), not \(8\,\text{mm}\).
- Thinking that two waves bounce off each other.Waves pass through each other. The sum exists only during the overlap, and each wave leaves unchanged.
- Thinking that a flat string at the meeting of two opposite pulses means the waves have gone.The string is flat but moving. The energy is all kinetic for that instant, and both pulses reappear.