Physics 5B Winter 2009 How to add sine functions of different amplitude and phase In these notes I will show you how to add two sinusoidal waves each of different amplitude and phase to get a third sinusoidal wave That is we wish to show that given E1 E10 sin t E2 E20 sin t 1 2 the sum E E1 E2 can be written in the form E E10 sin t E20 sin t E 0 sin t 3 where the amplitude E 0 and phase are determined in terms of E10 E20 and In these notes we shall derive that the amplitude E 0 is given by E 0 q 2 E 2 2E E cos E10 10 20 20 and the phase is determined modulo 2 by1 sin E20 sin E 0 cos E10 E20 cos E 0 By definition the amplitudes E10 and E20 are positive numbers If we divide the last two equations then can be determined modulo from tan E20 sin E10 E20 cos To determine modulo 2 we need to supplement the result for tan with sign sin sign sin sin 6 0 where sign sin literally means the sign i e either 1 or 1 of the quantity sin Finally if sin 0 then cos 1 and can be fixed modulo 2 by 1 cos 1 cos 1 cos 1 and E10 E20 1 cos 1 and E10 E20 Note that in the case of cos 1 and E10 E20 we have E 0 in which case E 0 0 and is no longer meaningful In fact this is the only circumstance in which E 0 can vanish I shall provide two different derivations of the above formulae Finally in an appendix I will provide a mathematically more advanced derivation that makes use of complex numbers If you are unfamiliar with complex numbers you can skip the appendix for now and return to this last derivation later after you take Physics 116A or an equivalent course This last method also provides the real motivation for the method of phasors introduced in Section 2 below 1 The phrase is determined modulo 2 means that is determined up to an additive integer multiple of 2 This is all that is needed since adding a multiple of 2 to the phase angle does not change the value of sin t 1 1 Algebraic method First set t 0 in eq 3 to obtain E20 sin E 0 sin Solving for sin yields sin E20 sin E 0 Next set t 2 in eq 3 Noting that sin 2 cos it follows that E10 E20 cos E 0 cos Solving for cos yields cos E10 E20 cos E 0 Finally using cos2 sin2 1 and inserting the expressions for cos and sin just obtained one finds 2 2 E 0 E10 E20 cos 2 E20 sin2 2 2 E10 2E10 E20 cos E20 cos2 sin2 2 2 E20 2E10 E20 cos E10 By definition E 0 is a non negative number 2 Thus we take the positive square root to obtain q 2 E 2 2E E cos E 0 E10 10 20 20 This completes our derivation 2 Geometric method the method of phasors Consider a fictitious vector in a two dimensional space whose length is E0 which makes an angle with respect to the x axis as shown below y E0 E0 sin x If we project this vector onto the y axis then its projected length is E0 sin as shown above This vector is called a phasor and represents a quantity with an amplitude E0 and an angle The utility of such a representation is that we can perform the sum of eq 3 by considering the phasors corresponding to each sine term in the sum and then adding the phasors vectorially The projection of the vector sum of the two phasors onto the y axis is just the sum of the two sine functions that we wish to compute This vector sum can be carried out geometrically and provides a second method for evaluating E 0 and 2 Warning This is a matter of convention which Giancoli chooses not to follow without warning you 2 To see how this works consider the computation of eq 3 by the method of phasors We represent E1 and E2 cf eqs 1 and 2 as shown in the figure below Then the phasor representation of E is just the vector sum shown above We identify E10 E20 and E 0 as the lengths of the phasors representing E1 E2 and E respectively To evaluate E 0 and we focus on the triangle in the figure above First using the law of cosines 2 2 2 E 0 E10 E20 2E10 E20 cos since is the angle between the phasors representing E1 and E2 Using cos cos we end up with q 2 E 2 2E E cos E 0 E10 10 20 20 as before Next using the law of sines sin sin E20 E 0 Using sin sin we can solve for sin We find that sin E20 sin E 0 which again agrees with our previous result This equation only fixes modulo In order to fix modulo 2 we employ the law of sines again noting that the angle in the triangle between the phasors representing E2 and E is given by sin sin E20 E10 This equation can be rearranged in the following form E10 sin E20 E10 sin cos cos sin sin sin cos tan 3 One can solve this equation easily for tan to obtain tan E20 sin E10 E20 cos in agreement with our previous result Finally we can use our results above for sin and tan to compute cos as follows sin E10 E20 cos E10 E20 cos E20 sin cos tan E 0 E20 sin E 0 which completes the derivation 3 The limit of equal amplitudes As a check consider the case of equal amplitudes E10 E20 E0 Then using the above results p E 0 2E0 1 cos Recalling the trigonometric identity cos2 2 12 1 cos we end up with E 0 2E0 cos 2 Note the absolute value sign since by definition the p amplitude E 0 is defined to be non negative which means we must take the positive square root cos2 2 cos 2 If cos 1 then E 0 0 and the angle is undefined Otherwise we may use the results derived above for sin and cos to obtain sin 2 sin 2 cos 2 sin 2 if cos 2 0 sin sin 2 if cos 2 0 2 cos 2 2 cos 2 cos cos2 2 1 cos cos 2 2 cos 2 cos 2 after using the trigonometric identity sin 2 sin 2 cos 2 Note that cos 2 1 2 1 cos 6 0 for cos 6 1 which must hold if E 0 6 0 The results above for sin and cos imply that 2 if cos 2 0 2 if cos 2 0 In Section 34 4 of Giancoli a convention is chosen in which E 0 can be of either sign so in …