just give solution of unsolved & wrong attempted parts The faintest sounds the h

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just give solution of unsolved & wrong attempted parts

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: delta P_max = squareroot 2 rho vI = squareroot 2 (1.20 kg/m^3) (343 m/s) (1.00 times 10^-12 W/m^2) = 0.0000287 N/m^2 Calculate the corresponding displacement amplitude. Recall that omega = 2 pi f: s_max delta P_max/rho v omega = delta P_max/(1.20 kg/m^3) (343 m/s) (2 pi times 1,000 Hz) = 0.0000000001 m 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 2.87 N/m^2 and a displacement amplitude equal to 0.0000000001 m. 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 1.63 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? N/m^2 (b) What is the displacement amplitude of this wave? m

Explanation / Answer

given in the first part
Lowest intensity, Io = 10^-12 W/m^2
Highest intensity, I = 1 W/m^2

now, pressure amplitude is found using the formula
dP = sqroot(2*rho*v*I)
where rho is density of air = 1.2 kg/m^3
and v is sped of sound in air = 343 m/s

so dP min = sqroot(2*1.2*343*10^-12) = 2.869*10^-5 Pa
dP max = sqroot(2*1.2*343*1) = 28.69 Pa

similiarly, displacement amplitude
smin = dPmin/rho*v*w = 2.869*10^-5/1.2*343*(2*pi*1000) = 1.109*10^-11 M

and smax = 28.69/1.2*343*(2*pi*1000) = 1.109*10^-5 m

Now, inside water
rho = 1000 kg/m^3
speed of sound in water, v = 1484 m/s
I = 1.63*10^-12
so, dp = sqroot(2*1000*1484*1.63*10^-12) = 0.0021995 Pa
and ds = 0.0021995/1000*1484*2*pi*(1000) = 2.35*10^-13 m

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