The power output of a certain public-address speaker is 6.45 W. Suppose it broad

Question

The power output of a certain public-address speaker is 6.45 W. Suppose it broadcasts equally in all directions. Within what distance from the speaker would the sound be painful to the ear? At what distance from the speaker would the sound be barely audible? Two waves on one string are described by the wave functions y_1= 3.5 cos(4.5x - 1.8t) y_2 = 4.5 sin(6.0x - 2.5t) where x and y are in centimeters and t is in seconds. Find the superposition of the waves y_1 + y_2 at the following points. (Remember that the arguments of the trigonometric functions are in radians.) x = 1.00, t = 1.00 x = 1.00, t = 0.500 x = 0.500, t = 0

Explanation / Answer

solving 11

a) We know that the pain threshold of sound is 1 W/m^2

So Intensity = Power/Area or Area = Power/Intensity = 6.45W/1W/m^2 = 6.45m^2

Now 4*r^2 = 6.45 Soving for r we get r = sqrt(6.45/4) = 0.72 m =>Answer

b) The threshold of sound is 1.0x10^-12W/m^2

Intensity= Power/Area or A = Power/Intensity = 6.45W/(1.0x10^-12W/m^2) = 6.45x10^12m^2

Now 4*r^2 = 6.45x10^12 So r = sqrt(6.45x10^12/4) = 7.2x10^5 m

so sound would be barely available at r=7.2*10^(-12) m

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