A microphone has an area of 3.0cm^2. It receives during a 6.00s time period a sound energy of 1.0x10-11J. What is the intensity of the sound?

b). Using the intensity in the previous question, what is the variation of pressure in the sound wave if the speed of sound is 343m/s and the density of air is 1.2kg/m^3?

To find the intensity of sound, we can use the formula:

Intensity = Energy / Time

The given energy is 1.0x10^-11 J, and the time is 6.00 s. Plugging in these values:

Intensity = (1.0x10^-11 J) / (6.00 s)
Intensity ≈ 1.67x10^-12 W/m^2

Therefore, the intensity of the sound is approximately 1.67x10^-12 W/m^2.

Now, let's move on to the second part of your question. To find the variation of pressure in the sound wave, we can use the formula:

Intensity = (Density) x (Speed of Sound)^2 x (Amplitude)^2

Here, we need to solve for the amplitude (A), which represents the variation of pressure. Rearranging the equation:

(Amplitude)^2 = Intensity / [(Density) x (Speed of Sound)^2]

The given values are:
Intensity = 1.67x10^-12 W/m^2
Density of air = 1.2 kg/m^3
Speed of sound = 343 m/s

Plugging in these values:

(Amplitude)^2 = (1.67x10^-12 W/m^2) / [(1.2 kg/m^3) x (343 m/s)^2]

Calculating the right-hand side of the equation and taking the square root of the result will give us the amplitude.

(Amplitude) ≈ √[(1.67x10^-12) / (1.2 x 343^2)]

(Amplitude) ≈ 2.69x10^-5 N/m^2

Therefore, the variation of pressure in the sound wave is approximately 2.69x10^-5 N/m^2.