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A source emits sound uniformly in all directions. There are no reflections of the sound. At a distance of 12 m from the source, the intensity of the sound is 5.0 × 10−3 W/m2. What is the total sound power P emitted by the source?

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Final answer:

To find the total sound power P emitted by the source, we need to find the intensity at the speaker and the distance at which the intensity is 0.1 W/m². The calculation for the distance is provided using the inverse square law.

Step-by-step explanation:

To find the total sound power P emitted by the source, we need to first find the intensity at the speaker. Since the speaker emits sound uniformly in all directions, the intensity at the speaker is the same as the given intensity of 5.0 × 10−3 W/m² at a distance of 12 m from the source.

Next, we can use the inverse square law to find the distance at which the intensity is 0.1 W/m². The inverse square law states that the intensity of a spherical wave decreases with the square of the distance from the source. So, if the intensity is 0.1 W/m² at a distance of 12 m, the intensity would be 0.1*(12/√(d))² W/m² at a distance d from the source. Solving for d, we find d = 12*√(0.1/5.0 × 10−3) m.

Therefore, the total sound power P emitted by the source can be found using the equation P = IV, where I is the intensity at the speaker and V is the volume of the speaker. Since the intensity at the speaker is 5.0 × 10−3 W/m² and the volume of the speaker depends on its dimensions, we need further information to calculate the exact value of P.

User Samual
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To solve this problem we will apply the concept related to Intensity, that is, the acoustic power transferred by a sound wave per unit of normal area to the direction of propagation:


I = (P)/(A) \rightarrow P= AI

Where,

P = Power

A = Acoustic Area

Our values are given as,

r= 12 m


I = 4.3*10^(-3) W/m^2

The Area then would be


A = 4\pi r^2


A = 4\pi (12)^2


A = 1809.55m^2

Replacing the values we have that


P = (4.3*10^(-3) W/m 2)(1809.55m^2 )


P = 7.781 W

The total sound power P emitted by the source is 7.781W

User War
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