A source of sound operates at 2.0 kHz, 20 W emitting sound uniformly in all directions. The speed of sound in air is 340 m s-1 and the density of air is 1.2 kg m-3. (a) What is the intensity at a distance of 6.0 m from the source? (b) What will be the pressure amplitude at this point? (c) What will be the displacement amplitude at this point?

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Navjyot Kalra · Answer

Sol. a) Here given V baseair = 340 m/s., Power = E/t = 20 W f = 2,000 Hz, p = 1.2 kg/m^3 So, intensity I = E/t.A = 20/4πr^2 = 20 4 * π * 6^2 = 44 mw/m^2 (because r = 6m) b) We know that I P^2 base 0/2pV base air⇒ P base 0 = √1 * 2pV base air = √2 * 1.2 * 340 * 44 * 10^-3 = 6.0 N/m^2. C) We know that I = 2π^2 S^2 base 0 v^2 p^V where S base 0 = displacement amplitude ⇒ S base 0 = √l/π^2 p^2 pV base air Putting the value we get S base g = 1.2 * 10^-6 m.

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