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Intensity and the Inverse Square Law

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Intensity and the Inverse Square Law Intensity Page 1 Tuesday, March 1, 2005 6:56 AM INTENSITY AND THE INVERSE SQUARE LAW The inverse square law for sound applies to small (point) sources of sound that produce sound uniformly in all directions. It assumes that the medium sound source T through which the sound travels is uniform and that the sound does not reflect from, nor is absorbed by, any surfaces or the air. r2 Under these conditions, sound will travel as a spherical wave. It will spread out evenly in all directions and its wavefronts will form the surface of a sphere. r1 The power of the source is therefore spread out over the surface of a sphere. When finding the intensity of a sound at a distance of r m from the source, the sphere will have a surface area of 4 π r2. Since the intensity is the power per unit area, if you divide the power of the source by the area of the sphere, you will calculate the intensity at a distance of r from the source. P I = ----------- Figure 2.3: A sound wave travelling 4πr2 away from a source has a spherical shape. Two compressions Transposing this formula enables you to calculate the power of the (wavefronts) are shown here, with source: 2 radii r1 and r2. P = 4πr I. Sample problem 1 Karen measures the sound intensity at a distance of 5.0 m from a lawn-mower to be 3.0 × 10−2 W m−2. Assuming that the lawn-mower acts as a point sound source and ignoring the effects of reflection and absorption, what is the total power of the sound emitted by the mower? SOLUTION P = 4π r 2I = 4 π (5.0 m)2 × 3.0 × 10−2 W m−2 = 9.4 W Referring back to the formula for the sound intensity produced by a P source, I = ----------- , it can be seen that, for a particular sound source, the 4πr2 sound intensity it produces is inversely proportional to the square of the dis- tance from the source. 1 I ∝ ---- r2 This is the inverse square law which can be restated as: the intensity of sound is inversely proportional to the square of the distance from the source. When comparing the sound intensities at two distances r1 and r2 from a source, it should be remembered that the power of the source is a constant. ==π 2 π 2 Therefore, P 4 r1 I1 4 r2 I2 . This relationship then gives the following useful formula: I r 2 ----2 = -----1- 2 I1 r2 1 Intensity Page 2 Tuesday, March 1, 2005 6:56 AM Sample problem 2 If the sound intensity 3.0 m from a sound source is 4.0 × 10−6 W m−2, what is the intensity at (a) 1.5 m and (b) 12 m from the source? SOLUTION = (a) r1 3.0 m = × −6 −2 I1 4.0 10 W m = r2 1.5 m = I2 ? 2 I2 r 1 ---- = ------- 2 I1 r 2 2 I1r 1 ⇒ I = ------------ 2 2 r 2 4.0× 10–6 W m–2 × ()3.0 m 2 = --------------------------------------------------------------------------- ()1.5 m 2 = 1.6 × 10−5 W m−2 = (b) r1 3.0 m = × −6 −2 I1 4.0 10 W m = r2 12 m = I2 ? 2 I2 r 1 ---- = ------- 2 I1 r 2 2 I1r 1 ⇒ I = ------------ 2 2 r 2 4.0× 10–6 W m–2 × ()3.0 m 2 = -------------------------------------------------------------------------- - ()12 m 2 = 2.5 × 10−7 W m−2 The examples on the previous page show the following general ‘rules of thumb’: if you halve the distance, the intensity is multiplied by 4; if you double the distance, the intensity is divided by 4. QUESTIONS Understanding 1. If the sound intensity 4.0 m from a point sound source is 1.0 × 10–6 W m–2, what will be the sound intensity at each of the following distances from the source? (a) 1.0 m? (b) 2.0 m? (c) 8.0 m? (d) 40 m? (Ans: (a) 1.6 × 10–5 W m–2; (b) 4.0 × 10–6 W m–2; (c) 2.5 × 10–7 W m–2; (d) 1.0 × 10–8 W m–2) Application 2. A siren produces a sound intensity level of 90 dB at a distance of 8.0 m. Assume that the siren acts as a point source of sound. (a) What is the sound intensity at a distance of 8.0 m from the siren? (b) What is the power of the sound emitted by the siren? (c) How much sound energy does the siren produce in 1.0 min? (d) What is the sound intensity level at a distance of 16 m from the siren? (Ans: (a) 1.0 × 10–3 W m–2, (b) 0.80 W; (c) 48.3 J; (d) 84 dB) 2 Intensity Page 3 Tuesday, March 1, 2005 6:56 AM 3. The sound intensity level of a racing car at a distance of 10 m is 130 dB. Assume that the racing car acts like a small point source and ignore the effects of absorption and reflection. (a) What is the power of the sound emitted by the car? (b) What is the sound intensity at a distance of 10 m from the car? (c) What is the sound intensity due to the car at a distance of 1.0 km from the car? (d) At what distance from the car would the sound intensity fall to the threshold of hearing? (Ans: (a) 1.26 × 104 W; (b) 10 W m–2; (c) 1.0 × 10–3 W m–2; (d) 3.2 × 107 m) More of a challenge 4. A mosquito emits a 1000 Hz sound. At a distance of 10.0 m from the mosquito, the sound is at the threshold of hearing. (a) What is the intensity of the sound at the threshold of hearing? (b) What is the acoustical power of the mosquito? (c) What is the intensity and intensity level of the sound when the mos- quito is 10.0 cm from an observer’s ear? (d) How many identical mosquitoes would there have to be at a dis- tance of 10.0 m for them to sound as loud as one mosquito at a dis- tance of 10.0 cm? (Ans: (a) 1.0 × 10–12 W m–2; (b) 1.26 × 10–9 W; (c) I = 1.0 × 10–8 W m–2, L = 40 dB; (d) 1.0 × 104 mosquitoes) SUMMARY Sound from a small source obeys the ‘inverse square law’, namely I α 1/r2. 3.
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