A loudspeaker supported well above the ground on a post is supplied with 3.5 W of electrical power. The speaker is able to convert 4.4% of that into sound power. (That number is known as the speaker's efficiency.) If we assume that sound radiates equally in all directions, what is the sound intensity at a distance of 1.9 m?
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Electrical Power supplied = 3.5 Watt.
Efficiency of the speaker = 4.4 %
Power converted to Sound = (3.5)*(4.4) /100 watt = 0.154 watt
Enegy is radiated in all directions uniformly.
Consider a sphere of radius 1.9 m
The surface area of the sphere = 4 π r² = 4*(3.14)*(1.9)² m² =45.34 m²
Power received per unit area of the sphere = Intensity = (0.154) / (45.34)
= 3.4 * 10^( - 3) watt / m² = 3.4 milli watt / m²
Efficiency of the speaker = 4.4 %
Power converted to Sound = (3.5)*(4.4) /100 watt = 0.154 watt
Enegy is radiated in all directions uniformly.
Consider a sphere of radius 1.9 m
The surface area of the sphere = 4 π r² = 4*(3.14)*(1.9)² m² =45.34 m²
Power received per unit area of the sphere = Intensity = (0.154) / (45.34)
= 3.4 * 10^( - 3) watt / m² = 3.4 milli watt / m²
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3.5 x 0.044
---------------- in Watts/m²
4π(1.9)²
---------------- in Watts/m²
4π(1.9)²