The faintest sounds the human ear can detect at a frequency of 1,000 Hz correspo

Question

The faintest sounds the human ear can detect at a frequency of 1,000 Hz correspond to an intensity of about 1.00 times 10^-12 W/m^2, which is called threshold of hearing. The loudest sounds the ear can tolerate at this frequency correspond to an intensity of about 1.00 W/m^2, the threshold of pain. Determine the pressure amplitude and displacement amplitude associated with these two limits. Conceptualize Think about the quietest environment you have ever experienced. It is likely that the intensity of sound in even this quietest environment is higher than the threshold of hearing. Categorize Because we are given intensities and asked to calculate pressure and displacement amplitudes, this problem is an analysis problem requiring the concepts discussed in this section. Analyze To find the pressure amplitude at the threshold of hearing, use the equation, taking the speed of sound waves in air to be v = 343 m/s and the density of air to be rho = 1.20 kg/m^3: DeltaP_max = Squareroot2 pvI = Squareroot2(1.20 kg/m^3)(343 m/s)(1.00 times 10^-12 W/m^2) = .000028691 N/m^2 Calculate the corresponding displacement amplitude. Recall that omega = 2pif: S_max = DeltaP_max/rhov omega = DeltaP_max/(1.20 kg/m^3)(343 m/s)(2pi times 1,000 Hz) In a similar manner, one finds that the loudest sounds the human ear can tolerate (the threshold of pain) correspond to a pressure amplitude of N/m^2 and a displacement amplitude equal to Finalize Because atmospheric pressure is about 10^5 N/m^2, the result for the pressure amplitude tells us that the ear is sensitive to pressure fluctuations as small as 3 parts in 10^10! The displacement amplitude is also a remarkably small number! If we compare this result for s_max to the size of an atom (about 10^-10 m), we see that the ear is an extremely sensitive detector of sound waves. A particular sound wave has a frequency of 1,000 Hz and intensity 1.40 times 10^-12 W/m^2, and is in water at a temperature of 25 degree C. (a) What is the pressure amplitude of this wave? Work through the solution of the example problem with your calculator until you agree with the answers given in example in the text. N/m^2 (b) What is the displacement amplitude of this wave?

Explanation / Answer

from pressure amplitude equation

Pmax^2 = 2 rho V I

here V = 343 m/s

I is 1.4 *10^-12 W/m^2

rho = 1.2 kg/m^3

so

Pmax^2 = 2 * 1.2 * 1.4*10^-12 * 343

Pmax = 0.0000339

---------------------

Displacement S = Pmax/(rho V w)

w = 2pif

S = 0.0000339/(1.2 * 343 * 2*pi * 1000)

S = 1.31*10^-11 m or 0.0000000000131 m

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