1. Please go to View, then select Slide Show
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2. E1 = (4 N/C) sin (w1t + p/6) E2 = (3 N/C) sin (w2t +p/3).
Adding together waves using the Method of Phasors.
Let’s say that you want to add two waves together. In this example, they will be the electric field part of an EM wave. We’ll say the wave is harmonic and plane polarized. At some point P, we have that the waves are
E1 = (4 N/C) sin (w1t + p/6)
and
E2 = (3 N/C) sin (w2t +p/3).
If the (angular) frequencies of the waves are the same (ie. if w1 = w2), then you can use the
METHOD OF PHASORS.
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3. METHOD OF PHASORS.
Step 1: Can you use the Method of Phasors? If the (angular) frequencies are the same, then YES. (Of course, the waves must overlap in space and time, and be the same kind of wave!)
Step 2: Represent each wave by a VECTOR, whose MAGNITUDE is the AMPLITUDE of the wave. The DIRECTION of the vector is given by the PHASE of the wave. The PHASE is the angle of the VECTOR as measured COUNTER-CLOCWISE with respect to the POSITIVE x-axis. These vectors are called PHASORS.
Step 3: Add the PHASORS together. Remember they are VECTORS!
Step 4: Determine the amplitude Atot of the resultant vector (phasor) you obtain by adding the individual phasors together. Determine the angle of the resultant phasor (measured counter-clockwise from the + x-axis); this will be ftot.
Step 5: The resultant wave from the superposition of the individual waves will be: Atot sin (wt + ftot)
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4. We were given E1 = (4 N/C) sin (w1t + p/6)
and E2 = (3 N/C) sin (w2t + p/3) at the point P.
Since they are the same kind of wave, with the same angular frequency, and we see they overlap in space, we can use the Method of Phasors. We’ve just checked Step 1. Now, we’ll use Step 2 and represent the waves by phasors:
The MAGNITUDE of the phasor is the AMPLITUDE of the wave.
The DIRECTION of the phasor is given by the PHASE of the wave.
(For convenience, the numerical values will be represented symbolically.)
p/3
3 N/C
p/6
4 N/C f2 A2 f1 A1
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5. Atot Atot ftot Atot ≠ A1 + A2 Atot ftot ≠ f1 + f2 ftot
A2y
A2y
Atot
Atot A2 f2 f1 A1
A2x
A2x
ftot
Atot ≠ A1 + A2
A2y
Atot
ftot ≠ f1 + f2 f1 A1
A1y
A1y
Must do VECTOR addition
A1y
ftot
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A1x
A1x
A1x
A2x

6. Atot ftot Atot,x = A1x + A2x = A1 cos f1 + A2 cos f2
Since all the angles are measured counter-clockwise from the +x axis, then can use our standard trig definitions to relate magnitudes and components. Note that with this convention, the SIGN of the component will automatically be taken care of (still should check that it makes sense, though.)
Atot,x = A1x + A2x
= A1 cos f1 + A2 cos f2
Atot,y = A1y + A2y
= A1 sin f1 + A2 sin f2
Atot,x = (4 N/C) cos p/6 + (3N/C) cos p/3 = N/C
Atot,y = (4 N/C) sin p/6 + (3N/C) sin p/3 = N/C
A1x
A2x
A1y
A2y
Atot
ftot
N/C
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7. Atot A2 f2 Df A1 f1 Df = f2 – f1 = p/3 – p/6 = p/6
Sometimes, when adding waves together, you only care about the amplitude. Then, the only thing that matters is the phase difference between one wave and the next. You lay the first phasor along the horizontal, then draw the second phasor with an angle (counterclockwise from the +x axis) that is the difference in the phases of the two waves: Df = f2 – f1. If there are multiple waves, you continue this way, always rotating the next phasor with respect to the previous phasor. A2 f2 f1 A1
Atot Df
Df = f2 – f1 = p/3 – p/6 = p/6
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8. THE END Atot,x = A1 cos 0 + A2 cos Df Atot,y = A1 sin 0 + A2 sin Df
Df = f2 – f1 = p/3 – p/6 = p/6 Df
Atot,x = A1 cos 0 + A2 cos Df
Atot,y = A1 sin 0 + A2 sin Df
Atot,x = (4 N/C) cos 0 + (3N/C) cos p/6 = N/C
Atot,y = (4 N/C) sin 0 + (3N/C) sin p/6 = N/C
N/C
EXACT same amplitude as before!!!
NOTE: Can NOT get correct phase shift with this picture
THE END