# A stereo speaker represented by P in the fig- ure emits sound waves with a power output of 103.4 W. What is the intensity of the sound waves at 12.1 m away?

unkyd | High School Teacher | (Level 2) Adjunct Educator

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Intensity = Power/Area

Power (P) is the rate the energy is produced by the sound source in this case 103.4 Watts

Since the sound wave travels in all three directions at once it produces a 3 dimensional wavefront like the surface of a balloon.  As the wavefront travels from the source surface of this wavefront gets bigger just like the surface area of a balloon gets as you fill it with more air.  Also, as the balloon gets bigger the balloon material gets thinner and thinner as it spreads itself out to create the larger and larger surface.  So too, the power of the sound source has to spread itself out over a larger and larger surface area of a sphere.  The surface area of a sphere is calculated:

A = 4*pi*r^2 = 4 * 3.14*(12.1 m)^2 = 1840 m^2

Using this to solve for Intensity you get:

I = P/A = 103.4 Watts/1840 m^2 = 0.0562 W/m^2

Intensity is measured in Watts per square meter (W/m^2)

If you want a decibel level you want to determine what is called Relative Intensity or sometimes loudness.  To be a "Relative" Intensity sound has to be compared to some standard.   A common standard is the threshold of hearing (Io) at around 10^-12 W/m^2.

Relative Intensity = 10 * log(I/Io)

= 10 *log(0.0562/10^-12)

= 107 dB

So the Intensity = .0562 W/m^2

Relative Intensity = 107 dB

AKA VERY LOUD!!!!