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Phasor Addition in Physics: Intensity of Sound Waves from Stereo Speakers, Exams of Quantum Physics

University of Illinois - Urbana-ChampaignQuantum Physics

A solution to a problem about phasor addition in physics, specifically regarding the intensity of sound waves coming from stereo speakers. The problem involves an audiophile testing the balance between the two channels of her stereo system and adjusting the phase difference between them. The solution involves drawing phasor diagrams and applying the procedure for finding the intensity with multiple sources. The steps for finding the intensity when the signals are in phase, out of phase by 180 degrees, and when the desired intensity is 2 times the initial intensity.

Typology: Exams

Pre 2010

Uploaded on 03/10/2009

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Solution
Physics 214 Problem 1 Week 2
Phasor Addition
We solved a problem similar to this last week. The idea here is to solve it using phasors.
An audiophile is seated in front of her stereo system, equidistant from her two high-end speakers.
She is using a test CD that can put a pure harmonic oscillation into either or both channels. The
balance (equality of intensity) between the two channels of her system is not correctly adjusted.
In the channel-separation test, in which the same signal is first put on one channel and then on
the other channel, she first hears intensity I1 from the right speaker, and then she hears intensity
I1/2 from the left speaker.
For each of the following questions, first draw the appropriate phasor diagram. Make sure to
label the phasor lengths.
To find the intensity due to more than one source, our procedure has four steps.
1. Draw a phasor diagram, incorporating whatever information is given on the phases or
differences in phase of the sources.
2. Because intensity is proportional to the square of the amplitude, I α A2, and because
amplitudes add according to phasor diagrams, label each phasor with the square root of the
intensity of its source when that source alone is active.
3. Add the square roots of the intensities according to the phasor diagram to find the square root
of the intensity with all sources active.
4. Square the result of step 3 to find the intensity with all sources active.
It is this procedure that was used below.
a) Now the same signal is put on both speakers at the same time and in phase. What intensity
does she hear, in terms of I1?
1
I 2
1
I
"In phase" means that the phase difference,
φ
= 0. Thus, the phasors are
parallel, as shown in the diagram. The amplitudes of parallel phasors add
normally.
2
1
1
I
II +=
1
2
1
2
1
191.2
2
1
1
2II
I
II =
⎟
⎠
⎞
⎜
⎝
⎛+=
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛+=
You will always be able to factor out I1
(
)2=I1. The answer will be the initial intensity times a
number.
b) One speaker has an adjustment that permits its phase to be varied with respect to the other
speaker. If she sets this phase difference to –180°, what intensity does she hear?
pf2

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Solution

Physics 214 Problem 1 Week 2 Phasor Addition

We solved a problem similar to this last week. The idea here is to solve it using phasors.

An audiophile is seated in front of her stereo system, equidistant from her two high-end speakers. She is using a test CD that can put a pure harmonic oscillation into either or both channels. The balance (equality of intensity) between the two channels of her system is not correctly adjusted. In the channel-separation test, in which the same signal is first put on one channel and then on the other channel, she first hears intensity I 1 from the right speaker, and then she hears intensity

I 1/2 from the left speaker.

For each of the following questions, first draw the appropriate phasor diagram. Make sure to label the phasor lengths.

To find the intensity due to more than one source, our procedure has four steps.

  1. Draw a phasor diagram, incorporating whatever information is given on the phases or differences in phase of the sources.
  2. Because intensity is proportional to the square of the amplitude, I α A^2 , and because amplitudes add according to phasor diagrams, label each phasor with the square root of the intensity of its source when that source alone is active.
  3. Add the square roots of the intensities according to the phasor diagram to find the square root of the intensity with all sources active.
  4. Square the result of step 3 to find the intensity with all sources active. It is this procedure that was used below.

a) Now the same signal is put on both speakers at the same time and in phase. What intensity does she hear, in terms of I 1?

I (^) (^12)

I 1

"In phase" means that the phase difference, φ = 0. Thus, the phasors are

parallel, as shown in the diagram. The amplitudes of parallel phasors add normally.

1 1

I

I = I +

1

2 1

2 1 1 2.^91 2

I I

I

I I ⎟ =

You will always be able to factor out ( I 1 )

2 = I 1. The answer will be the initial intensity times a

number.

b) One speaker has an adjustment that permits its phase to be varied with respect to the other speaker. If she sets this phase difference to –180°, what intensity does she hear?

I 1

I 1

Here, the phase difference is φ = 180°. The amplitudes of antiparallel phasors subtract.

1 1

I

I = I −

I = I 1 −

I 1

2 = I 1 1 −

2 = 0.0858 I 1

c) This result is not to her liking, and she tries a setting of +90°. Now what intensity does she hear?

I 1

2

I (^) φ I 1

Since φ = 90°, the phasor diagram is a right triangle, and we can use Pythagorean theorem. 2 (^21) 1

2 2

I

I = I +

1

1 (^1 )

I

I

I = I + =

d) Finally, she decides that she wants to hear an intensity of 2 I (^) 1, and she adjusts the phase

shifter accordingly. What phase does she use?

I 1

2

2 I 1 I 1 φ

Since φ is none of the special values given in a) - c), we must use the

law of cosines, A^2 + B^2 +2 AB cos φ = C^2 :

2 1

1 1

2 (^21) 1 cos^2 2

I

I

I

I

I + + φ=

I 1 cancels:

cos 2 2

φ = cos-1( √ 2/4) = 1.209 radians = 69.3 °