A student on a noisy Delhi Metro coach puts on noise-cancelling earphones, and the steady hum of the motors almost disappears. The earphones do not block the hum with thick padding. A tiny microphone listens to the hum, and the speaker adds a second sound of its own. Adding a sound has made the ear quieter.
Two causes, one displacement
Suppose a particle is pushed by two separate causes. Each cause alone would make the particle perform an SHM along the same line: \(x_1(t)\) from the first and \(x_2(t)\) from the second. With both acting together, the displacement of the particle at every instant is the sum, \(x = x_1 + x_2\). This is the principle of superposition.
Let a particle be subject to two or more simple harmonic motions along the same line at the same time. Its displacement at every instant is the algebraic sum of the displacements that each motion alone would give.
The rule holds whenever the restoring forces are proportional to the displacement, as they are in every SHM. A small patch of air at your ear is one example. The hum of the motors pushes it to and fro with one SHM, and the earphone's speaker pushes it with another. The eardrum follows the sum of the two.
Why does the sum work? The equation of an SHM, \(d^2x/dt^2 = -\omega^2 x\), is linear: \(x\) and its derivatives appear only to the first power. So if \(x_1\) and \(x_2\) each satisfy it, their sum satisfies it too. A large stretch of a real spring or a large swing of a pendulum makes the equation non-linear, and then superposition is only approximate.
Adding two SHMs of the same frequency
Take two SHMs of the same angular frequency \(\omega\) along the same line: \(x_1 = A_1\sin\omega t\) and \(x_2 = A_2\sin(\omega t + \Delta\varphi)\). The first is taken to start at phase zero, which only sets the zero of time. The sum \(x = x_1 + x_2\) looks like a mixture. In fact it is a single, clean SHM, and the algebra shows why.
\[ x = A_1\sin\omega t + A_2(\sin\omega t\cos\Delta\varphi + \cos\omega t\sin\Delta\varphi) \]
Why. expand \(\sin(\omega t + \Delta\varphi)\) by the formula for the sine of a sum.
\[ x = (A_1 + A_2\cos\Delta\varphi)\sin\omega t + (A_2\sin\Delta\varphi)\cos\omega t \]
Why. collect the terms in \(\sin\omega t\) and in \(\cos\omega t\). The two brackets are constants.
\[ A_R\cos\delta = A_1 + A_2\cos\Delta\varphi, \qquad A_R\sin\delta = A_2\sin\Delta\varphi \]
Why. name the two constants as the sides of a right triangle with hypotenuse \(A_R\) and angle \(\delta\).
\[ x = A_R(\cos\delta\sin\omega t + \sin\delta\cos\omega t) = A_R\sin(\omega t + \delta) \]
Why. substitute step 3 into step 2 and use the sine of a sum backwards.
\[ A_R^2 = A_1^2 + A_2^2 + 2A_1A_2\cos\Delta\varphi, \qquad \tan\delta = \dfrac{A_2\sin\Delta\varphi}{A_1 + A_2\cos\Delta\varphi} \]
Read it. square and add the two equations of step 3, using \(\sin^2 + \cos^2 = 1\); divide one by the other for \(\tan\delta\).
So the sum is an SHM of the same angular frequency \(\omega\). Its amplitude is \(A_R\), the resultant amplitude, and its phase is \(\delta\) ahead of the first motion. Nothing in the result depends on \(t\): the sum keeps its shape for ever. This is true only because both motions share one \(\omega\).
The same result as a triangle of phasors
The algebra has a picture. Draw each SHM as its phasor: an arrow of length \(A_1\) at angle \(\omega t\), and an arrow of length \(A_2\) at angle \(\omega t + \Delta\varphi\). Place the second arrow with its tail at the head of the first. The arrow from the centre to the far head is the vector sum, of length \(A_R\).
The height of a sum of arrows is the sum of their heights. So the height of the resultant's tip is \(x_1 + x_2\), the displacement we want. Both phasors turn at the same \(\omega\), so the triangle they make turns as one rigid piece. The resultant keeps its length \(A_R\) and turns at \(\omega\) too. Its shadow is therefore an SHM of amplitude \(A_R\).
Set \(\Delta\varphi = 60°\) in the figure above. The resultant is \(4.36\,\text{cm}\) long, and it stays \(23.4°\) ahead of the \(3\,\text{cm}\) phasor as they turn. On the graph the solid curve is a clean sine curve of height \(4.36\,\text{cm}\), with the same period of \(4\,\text{s}\). The law of cosines on the triangle gives the same \(A_R\) as the algebra, because the angle inside the triangle at the joint is \(180° - \Delta\varphi\).
A particle is subject to \(x_1 = 3\sin\omega t\) and \(x_2 = 5\sin(\omega t + \pi/3)\), both in centimetres, along the same line. Find the amplitude of the resultant motion and its phase ahead of \(x_1\).
Same \(\omega\), so the sum is an SHM of that \(\omega\). Here \(A_1 = 3\,\text{cm}\), \(A_2 = 5\,\text{cm}\) and \(\Delta\varphi = \pi/3 = 60°\).
\(A_R^2 = A_1^2 + A_2^2 + 2A_1A_2\cos\Delta\varphi\) and \(\tan\delta = A_2\sin\Delta\varphi/(A_1 + A_2\cos\Delta\varphi)\).
\(A_R^2 = 9 + 25 + 2 \times 3 \times 5 \times 0.5 = 49\), so \(A_R = 7\,\text{cm}\). Then \(\tan\delta = (5 \times 0.866)/(3 + 2.5) = 4.33/5.5 \approx 0.787\), so \(\delta \approx 38.2°\).
The particle moves as \(x = 7\sin(\omega t + 38.2°)\) centimetres. The phase \(38.2°\) lies between \(0\) and \(60°\), nearer the larger phasor.
Two SHMs of the same frequency along one line have amplitudes \(5\,\text{cm}\) and \(12\,\text{cm}\) and a phase difference of \(\pi/2\). What is the amplitude of the resultant motion, in cm?
A particle moves as \(x = 6\sin\omega t + 8\cos\omega t\), in centimetres. What is the amplitude of its motion?
The largest and the smallest sum
The term \(2A_1A_2\cos\Delta\varphi\) decides everything. At \(\Delta\varphi = 0\), \(\cos\Delta\varphi = 1\), and \(A_R^2 = (A_1 + A_2)^2\). The amplitudes simply add: \(A_R = A_1 + A_2\). The phasors point the same way, and the two pushes always help each other. This is constructive superposition.
At \(\Delta\varphi = \pi\), \(\cos\Delta\varphi = -1\), and \(A_R = |A_1 - A_2|\). The phasors point opposite ways, and one motion always undoes part of the other. This is destructive superposition. If the two amplitudes are equal as well, \(A_R = 0\): the particle does not move at all. Every other phase difference gives an amplitude between these two limits, so \(|A_1 - A_2| \le A_R \le A_1 + A_2\).
Step through the chips of Fig 2.1. In phase, \(3\) and \(2\,\text{cm}\) give a tall curve of \(5\,\text{cm}\). In opposite phase the same pair gives only \(1\,\text{cm}\). At \(90°\) apart the sum is \(\sqrt{13} \approx 3.61\,\text{cm}\), and its peaks fall between those of the two parts. Two SHMs of \(3\,\text{cm}\) at \(120°\) apart give exactly \(3\,\text{cm}\). The same two in opposite phase give a flat line.
Two equal amplitudes
When \(A_1 = A_2 = A\), the phasor triangle has two equal sides, and the result takes a simple form.
\[ A_R^2 = A^2 + A^2 + 2A^2\cos\Delta\varphi \]
Why. put \(A_1 = A_2 = A\) in the formula for the resultant amplitude.
\[ A_R^2 = 2A^2(1 + \cos\Delta\varphi) \]
Why. take \(2A^2\) out as a common factor.
?
Your turn. use the half-angle identity. Which line is it?
\[ A_R = 2A\cos(\Delta\varphi/2), \qquad \delta = \Delta\varphi/2 \]
Read it. take the square root for \(0 \le \Delta\varphi \le \pi\); by symmetry the resultant bisects the angle between the two equal phasors.
Two SHMs along one line have the same frequency and the same amplitude \(A\). Their resultant also has amplitude \(A\). What is the phase difference between them?
Two SHMs of the same frequency, each of amplitude \(2\,\text{cm}\), act along one line with a phase difference of \(60°\). What is the resultant amplitude, in cm?
A particle is subject to three SHMs along one line, each of amplitude \(A\) and the same frequency, with phases \(0\), \(2\pi/3\) and \(4\pi/3\). What is the amplitude of its motion?
If the two SHMs have different frequencies, the phasors turn at different rates. The angle between them grows, so the resultant changes its length as it turns, and the sum is not an SHM. For two nearly equal frequencies the resultant swells and shrinks slowly. For sound, that slow swell is heard as beats, the subject of the chapter Beats and the Doppler Effect in the Waves unit.
Alternating currents in a circuit are SHMs in time, all of one frequency but with different phases. Adding them with trigonometry was slow and full of errors. At the International Electrical Congress in Chicago in 1893, the engineer Steinmetz showed a better way. He treated each current as a rotating arrow and added the arrows like vectors. Electrical engineers have used these phasors ever since.
Electricity reaches many Indian buildings as a three-phase supply at \(50\,\text{Hz}\). The three currents have the same frequency and phases \(120°\) apart. If the three loads are equal, the three currents return through one shared neutral wire. There they add like three equal phasors \(120°\) apart, and the sum is zero. So a well-balanced building sends almost no current through its neutral wire.
Now the earphones on the Metro make sense. The hum pushes the air at the ear with one SHM. The speaker adds a second SHM of the same frequency and equal amplitude, in opposite phase. The resultant amplitude is \(|A - A| = 0\), and the air at the eardrum hardly moves. Adding a second sound has cancelled the first.
- Adding the amplitudes of two SHMs as plain numbers.\(A_R = A_1 + A_2\) only in phase. In general use \(A_R^2 = A_1^2 + A_2^2 + 2A_1A_2\cos\Delta\varphi\).
- Using the angle inside the phasor triangle as \(\Delta\varphi\) in the law of cosines.The angle between the phasors is \(\Delta\varphi\); the angle inside the triangle at the joint is \(180° - \Delta\varphi\). The formula already allows for this.
- Treating \(a\sin\omega t + b\cos\omega t\) as two unrelated motions.The cosine term is an SHM leading by \(\pi/2\). The sum is one SHM of amplitude \(\sqrt{a^2 + b^2}\).