Sound and power along the xaxis

1. Sep 11, 2008

xjessica

Two sources of sound are located on the x axis, and each unit emits power uniformly in all directions. There are no reflections. One source is positioned at the origin and the other at x = +164 m. The source at the origin emits four times as much power as the other source. (a) At which location between the two sources on the x axis are the two sounds equal in intensity? (b) At which location to the right of the source at 164 m on the x axis are the two sounds equal in intensity? Describe the locations by giving the distance from the origin.

I attempted to solve this multiple times and it is incorrect each and every time.

I made the origin I1, followed by the +164 I2, and I3 the location to the right

I set I1 and I2 equal to each other by

P1/4pi(164 + d)squared = P2/4pi(164 + d)squared

P2/P1= dsquared/ (164+d)squared

4P2/P1= dsquared/ (164+d)squared

1/4= (164-d)squared/ dsquared

(164+d/d)squared=(1/2)squared

164 + d/d= +-1/2

i solved and got d equal to -290 and -96.667 and these were wrongg so i dont know what i did

2. Sep 11, 2008

gabbagabbahey

I'm not sure I understand what you mean by this. If there are only two sources, why do you need an $$I_3$$? What is the intensity at the point x, due to (a)the source at the origin (b) the source at x=164 (c)both sources combined?

The method of setting $$I_1$$ and $$I_2$$ is correct, but your expressions for $$I_1$$ and $$I_2$$ are not. For $$I_1$$, is the distance from the origin to the point $$x=d$$ really $$164 +d$$? Why wouldn't it just be $${\Delta}x=d-0=d$$? And for $$I_2$$ what is the distance from the point $$x=164$$ to the point $$x=d$$?

This doesn't look right, the proper way to simplify $$\frac{P_2}{P_1}$$ would be to substitute in the equation $$P_1=4P_2$$ as follows: $$\frac{P_2}{P_1}=\frac{P_2}{4P_2}=\frac{1}{4}$$

Once you correct these mistakes, you should get the right answer. Try again and if you still have problems post your new attempt and I will go over it.

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