a) A microphone has an area of 5cm^2. It receives during a 4.0 s time period a sound energy of 2.0*10^-11 J. What is the intensity of the sound?
b) using intensity from a) what is the variation in pressure in the sound wave, delta P ? Use T=293 K and density of air=1.2kg/m^3
To find the intensity of the sound, we can use the formula:
Intensity = Energy / Time
For part a):
Given:
Area (A) = 5 cm^2 = 5 * 10^-4 m^2 (since 1 cm^2 = 10^-4 m^2)
Energy (E) = 2.0 * 10^-11 J
Time (t) = 4.0 s
Plug in the values into the formula:
Intensity = Energy / Time
Intensity = (2.0 * 10^-11 J) / (4.0 s)
Intensity = 5.0 * 10^-12 W/m^2
Therefore, the intensity of the sound is 5.0 * 10^-12 W/m^2.
Moving on to part b):
To find the variation in pressure (delta P) in the sound wave, we can use the formula:
Intensity = (delta P)^2 / (2 * density * velocity)
where
Intensity = 5.0 * 10^-12 W/m^2 (from part a)
density of air = 1.2 kg/m^3
velocity = speed of sound = approximately 343 m/s at room temperature (T = 293 K)
Rearranging the formula to solve for delta P:
(delta P)^2 = Intensity * 2 * density * velocity
delta P = √(Intensity * 2 * density * velocity)
Plug in the values into the formula:
delta P = √((5.0 * 10^-12 W/m^2) * (2 * 1.2 kg/m^3) * (343 m/s))
delta P ≈ 3.73 * 10^-6 pa
Therefore, the variation in pressure in the sound wave (delta P) is approximately 3.73 * 10^-6 Pa.