A 2 sin(ωt+α 2 ). When we come to complex numbers we will see how this sort of
thing can be done using objects calledphasors(371
➤
), but really such methods
are simply shorthand for the ideas covered in this chapter. In particular we can
add two waves that are 90°out of phase using the results of Section 6.2.9, since
A 1 sin(ωt+ 90 °)+A 2 sin(ωt) is equivalent to acos(ωt)+bsin(ωt). For additions
such asA 1 sin(ωt+α)+A 2 sin(ωt),whereαis other than 90°, phasor methods are
equivalent to constructing a parallelogram with sidesA 1 andA 2 and included angleα
and taking the combined amplitude as the length of the diagonal,r, and the combined
phase to be the angle,θ, made by the diagonal with the sideA 2 (see Figure 6.15).A 1 sin(ωt+α)+A 2 sin(ωt)=rsin(ωt+θ)R
a qA 1A 2Figure 6.15Use the above methods to find the sine waves representing:(i) 4 sin(ωt)+3cos(ωt) (ii) 6 sin(ωt)+4sin(ωt+ 45 °)Answers to reinforcement exercises
6.3.1 Radian measure and the circle
A. (i)
π
5(ii)101 π
180(iii)2 π
3(iv)25 π
18(v)17 π
9(vi) −π
4(vii) −11 π
18(viii) −π
12(ix)3 π
20(x)91 π
60B. (i) 120° (ii) 0° (iii) 270° (iv) 60°
(v) 30° (vi) 90° (vii) 40° (viii) 225°(ix) 308° (x) 15°C. (i)
π
3,2 π
3(ii)2 π
3,4 π
3(iii) π,2π (iv)4 π
3,8 π
3(v) 2π,4π (vi)8 π
3,16 π
3(vii)32 π
9,64 π
9(viii) 4π,8πNote: If you are adept with ratios you will have noticed that with the given radius the area
will always have a magnitude double that of the arc length.