Two wires meet at a junction in an AC circuit. The current in one wire is \(i_1 = 3\sin\omega t\) amperes. The current in the other wire is \(i_2 = 4\sin(\omega t + \pi/2)\) amperes. By Kirchhoff's junction rule, the current that leaves the junction is \(i_1 + i_2\) at every instant. Yet the peak of the current that leaves is \(5\,\text{A}\), not \(7\,\text{A}\). No charge is lost. The two currents simply do not reach their peaks at the same instant.
Add the heights, instant by instant
Take two sinusoids of the same angular frequency, \(y_1 = A_1\sin\omega t\) and \(y_2 = A_2\sin(\omega t + \phi)\). Their sum at each instant is \(y = y_1 + y_2\). Draw the phasor of \(y_1\) from O. Then draw the phasor of \(y_2\) from the tip of the first, turned \(\phi\) ahead of it. The height of the far tip is the height of the first tip plus the rise along the second arrow. So the height of the far tip is \(y_1 + y_2\), the sum we want.
The arrow from O to the far tip is called the resultant. Both phasors turn at the same rate \(\omega\), so the angle \(\phi\) between them never changes. The triangle of the two phasors and the resultant therefore turns as one rigid piece. Its shape does not change, so the resultant has a fixed length \(A\). It also turns at the rate \(\omega\). The height of its tip is therefore \(A\sin(\omega t + \delta)\), one sinusoid of the same \(\omega\).
Two sinusoids of the same angular frequency add to a sinusoid of the same angular frequency. The phasor of the sum is the vector sum of the two phasors, drawn tip to tail. The amplitude of the sum is the length of the resultant, and its phase is the angle of the resultant.
The triangle, worked out
Freeze the turning diagram at the instant when the first phasor lies along the horizontal. The shape of the triangle is the same at every instant, so nothing is lost by this choice. The derivation below reads the two formulas off the triangle of Fig 2.2, using components.
\[ \vec{A}_1 = (A_1,\ 0) \]
Why. the first phasor lies along the horizontal at the frozen instant.
\[ \vec{A}_2 = (A_2\cos\phi,\ A_2\sin\phi) \]
Why. the second phasor has length \(A_2\) and makes the angle \(\phi\) with the horizontal.
\[ \vec{A} = (A_1 + A_2\cos\phi,\ A_2\sin\phi) \]
Why. adding two arrows tip to tail adds their components.
\[ A^2 = (A_1 + A_2\cos\phi)^2 + (A_2\sin\phi)^2 \]
Why. the length of an arrow from its two perpendicular components, by Pythagoras.
\[ A^2 = A_1^2 + A_2^2 + 2A_1A_2\cos\phi \]
Why. expand the square and use \(\cos^2\phi + \sin^2\phi = 1\).
\[ \tan\delta = \dfrac{A_2\sin\phi}{A_1 + A_2\cos\phi} \]
Read it. the angle of the resultant is fixed by its upright part divided by its horizontal part, from step 3.
Three special cases follow at once. In phase, \(\phi = 0\), the arrows lie along one line and \(A = A_1 + A_2\). In opposite phase, \(\phi = \pi\), they point opposite ways and \(A = |A_1 - A_2|\). A quarter turn apart, \(\phi = \pi/2\), the triangle has a right angle and \(A = \sqrt{A_1^2 + A_2^2}\). Every other \(\phi\) gives an amplitude between \(|A_1 - A_2|\) and \(A_1 + A_2\).
The rule works only for one angular frequency. With two different values of \(\omega\) the arrows turn at different rates. The triangle then changes shape, and the sum is not a sinusoid of fixed amplitude. Write both quantities as sines before reading \(\phi\): a cosine is a sine a quarter turn ahead. The \(\phi\) of the formula is the angle between the two arrows drawn from one point. The angle inside the triangle, between the arrows drawn tip to tail, is \(180° - \phi\).
Two currents \(i_1 = 3\sin\omega t\) and \(i_2 = 4\sin(\omega t + \pi/2)\) amperes meet at a junction. What is the peak of the current that leaves the junction, in amperes?
What is the amplitude of \(4\sin\omega t + 4\sin(\omega t + 120°)\)?
Two sinusoids \(3\sin\omega t\) and \(5\sin(\omega t + \phi)\) are added, for some phase difference \(\phi\) between \(0\) and \(\pi\). Which statements must be true?
A sine plus a cosine
A cosine is a sine whose phasor is a quarter turn ahead. So the sum \(a\sin\omega t + b\cos\omega t\) has two phasors at right angles: \(a\) along the reference and \(b\) straight up. The triangle is right-angled, so the resultant has length \(R = \sqrt{a^2 + b^2}\) and angle \(\tan\delta = b/a\). For example, \(3\sin\omega t + 4\cos\omega t = 5\sin(\omega t + 53.1°)\). The largest value of the sum is \(5\), not \(3 + 4 = 7\).
This form appears whenever an oscillation starts at an awkward moment. A particle in SHM that starts at \(x_0\) with velocity \(v_0\) moves as \(x = x_0\cos\omega t + (v_0/\omega)\sin\omega t\). The phasor rule turns this into one sine, whose amplitude is the largest displacement.
\(3\sin\omega t + 4\cos\omega t\) is written as \(5\sin(\omega t + \delta)\). What is \(\delta\)?
Can \(3\sin\omega t + 4\sin 2\omega t\) be written as one sinusoid by the phasor rule?
This is the worked example of Superposition, Damped and Forced Oscillations, lesson 2, done with the tool. 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 \(\phi = \pi/3 = 60°\).
\(A^2 = A_1^2 + A_2^2 + 2A_1A_2\cos\phi\) and \(\tan\delta = A_2\sin\phi/(A_1 + A_2\cos\phi)\).
\(A^2 = 9 + 25 + 2 \times 3 \times 5 \times 0.5 = 49\), so \(A = 7\,\text{cm}\). 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 longer phasor.
In the junction of the opening problem, the currents \(3\sin\omega t\) and \(4\sin(\omega t + \pi/2)\) amperes meet. Kirchhoff's rule holds at every instant. Why is the peak of the current that leaves only \(5\,\text{A}\)?
Kirchhoff's rule adds the two currents at each instant. The two currents are a quarter cycle apart, so when one is at its peak the other is zero. Their peaks never come together, so the sum never reaches \(3 + 4\). The sum is the height of the resultant phasor, which has length \(\sqrt{3^2 + 4^2} = 5\,\text{A}\) because the two phasors are at right angles.
Superposition, Damped and Forced Oscillations adds two SHMs of one frequency with exactly this triangle. It also uses the triangle to compare two SHMs by their phase difference. Kinematics of SHM writes a sine plus a cosine as one sine to find an amplitude. Superposition and Reflection adds two waves on one string. Interference and Young's Double Slit finds the intensity of two light waves the same way. Alternating Current and the LCR Circuit adds currents at a junction and voltages round a loop as phasors.
- Adding the amplitudes as numbers, \(A = A_1 + A_2\).That is right only in phase. In general, add the phasors: \(A^2 = A_1^2 + A_2^2 + 2A_1A_2\cos\phi\).
- Using the angle inside the triangle in the formula.\(\phi\) is the angle between the two phasors drawn from one point. The inside angle of the tip-to-tail triangle is \(180° - \phi\).
- Adding a sine and a cosine as if both were sines at the same phase.Write the cosine as a sine a quarter turn ahead first: \(\cos\omega t = \sin(\omega t + \pi/2)\).