16-6 Phasors

16-6 Phasors#

Prompts

  • What is a phasor? How does a rotating vector represent a sinusoidal wave? What does its projection on the vertical axis give?

  • The trig method in section 16-5 works only for waves with identical amplitudes. Why? How do phasors generalize to waves with different amplitudes?

  • Sketch a phasor diagram for two waves with amplitudes \(y_{m1}\) and \(y_{m2}\) and phase difference \(\phi\). How do you find the amplitude and phase of the resultant wave?

  • Two waves have \(y_{m1} = 4\;\text{mm}\), \(y_{m2} = 3\;\text{mm}\), and \(\phi = \pi/3\). Use phasor addition (by components) to find the resultant amplitude and phase constant.

Lecture Notes#

Overview#

  • A phasor is a rotating vector that represents a sinusoidal wave. Its magnitude = amplitude; its projection on the vertical axis = displacement.

  • Phasor addition generalizes the trig method of section 16-5: it works for waves with different amplitudes as well as different phases.

  • To find the resultant of two waves: add their phasors as vectors. The magnitude of the sum = resultant amplitude; the angle = resultant phase constant.

What is a phasor?#

A phasor is a vector of magnitude \(y_m\) (the wave amplitude) that rotates about the origin at angular speed \(\omega\) (the wave’s angular frequency).

  • The projection of the phasor on the vertical axis equals the displacement \(y\) of a point as the wave passes.

  • As the phasor rotates, the projection varies sinusoidally: \(y = y_m \sin(\theta)\) where \(\theta = kx - \omega t + \phi\) advances with time.

Connection to circular motion

This is the same idea as SHM being the projection of uniform circular motion (section 15-4). The phasor is the rotating “radius”; its \(y\)-component is the displacement.

Phasor diagram for two waves#

For two waves with the same \(k\) and \(\omega\) but possibly different amplitudes and phases:

(104)#\[ y_1 = y_{m1} \sin(kx - \omega t), \qquad y_2 = y_{m2} \sin(kx - \omega t + \phi) \]

  • Draw phasor 1: magnitude \(y_{m1}\), phase 0 (e.g., along horizontal when \(kx - \omega t = 0\)).

  • Draw phasor 2: magnitude \(y_{m2}\), at angle \(\phi\) from phasor 1.

  • \(\phi > 0\): phasor 2 lags phasor 1. \(\phi < 0\): phasor 2 leads phasor 1.

  • Both rotate at \(\omega\); the angle between them stays \(\phi\).

Finding the resultant wave#

The resultant is \(y'(x,t) = y'_m \sin(kx - \omega t + \beta)\). To find \(y'_m\) and \(\beta\):

Vector addition: Add the two phasors (place tail of 2 at head of 1). The vector sum has:

  • Magnitude = \(y'_m\) (amplitude of resultant)

  • Angle from phasor 1 = \(\beta\) (phase constant of resultant)

By components (with phasor 1 along the horizontal axis):

  • Horizontal: \(y'_{mh} = y_{m1} \cos 0 + y_{m2} \cos\phi\)

  • Vertical: \(y'_{mv} = y_{m1} \sin 0 + y_{m2} \sin\phi\)

  • \(y'_m = \sqrt{y'_{mh}^2 + y'_{mv}^2}\), \(\quad \beta = \arctan(y'_{mv}/y'_{mh})\)

Important

Phasors work for any amplitudes. The trig identity \(\sin a + \sin b = 2\sin\frac{a+b}{2}\cos\frac{a-b}{2}\) in section 16-5 required \(y_{m1} = y_{m2}\). Phasor addition has no such restriction.

Summary#

  • Phasor: rotating vector; magnitude = amplitude, projection = displacement.

  • Phasor diagram: each wave → phasor; phase difference = angle between phasors.

  • Resultant: vector sum of phasors; magnitude = \(y'_m\), angle = \(\beta\).

  • Advantage: combines waves with different amplitudes (and phases).