a) What is the displacement amplitude of a sound wave that hasa frequency of 500

Question

a) What is the displacement amplitude of a sound wave that hasa frequency of 500Hz at the pain-threshold pressure amplitude of29.0 Pa? Take densit of air to be 1.29 kg/m3. b)What is the average intensity of the sound wave described inpart (a) at a temperature 20oC. speed of sound: V= 344 m/s at 20oC. a) What is the displacement amplitude of a sound wave that hasa frequency of 500Hz at the pain-threshold pressure amplitude of29.0 Pa? Take densit of air to be 1.29 kg/m3. b)What is the average intensity of the sound wave described inpart (a) at a temperature 20oC. speed of sound: V= 344 m/s at 20oC.

Explanation / Answer

a)The frequency of the sound wave is f = 500 Hz The pain-threshold pressure amplitude is P = 29.0 Pa The density of air is = 1.29 kg/m3 The displacement amplitude of the sound wave is A = V * f Here,V = 344 m/s or A = 344 * 500 = 172000 m b)The average intensity of the sound wave described in part(a) at a temperature 20oC is I = (W/A) Here,W is the power and A is the area ofcrossection. Here,W = F * V and P = (F/A) or F = P * A From equation (1),we get I = (F * V/A) = (P * A * V/A) = P * V or I = 29.0 * 344 = 9976 W/m2 The displacement amplitude of the sound wave is A = V * f Here,V = 344 m/s or A = 344 * 500 = 172000 m b)The average intensity of the sound wave described in part(a) at a temperature 20oC is I = (W/A) Here,W is the power and A is the area ofcrossection. Here,W = F * V and P = (F/A) or F = P * A From equation (1),we get I = (F * V/A) = (P * A * V/A) = P * V or I = 29.0 * 344 = 9976 W/m2

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