1) A microphone has an area of 6.7 cm2. It receives during a 4.62 s time period a sound energy of 2.5x10-11 J. What is the intensity of the sound?
2) Using the intensity in the previous question, what is the variation of pressure in the sound wave if the speed of sound is 343 m/s and the density of air is 1.2 kg/m3?
To find the intensity of the sound, we can use the formula:
Intensity = Energy / Area * Time
1) Given that the area of the microphone is 6.7 cm2 (or 6.7 * 10^-4 m^2), the time period is 4.62 s, and the sound energy is 2.5x10^-11 J, we can substitute these values into the formula:
Intensity = (2.5x10^-11 J) / (6.7x10^-4 m^2) * (4.62 s)
Simplifying this expression, we get:
Intensity = 2.5x10^-11 J / (6.7x10^-4 m^2 * 4.62 s)
Now we can do the multiplication in the denominator:
Intensity = 2.5x10^-11 J / 3.086x10^-6 m^2s
Dividing the two values, we get:
Intensity = 8.11x10^-6 W/m^2
Therefore, the intensity of the sound is 8.11x10^-6 W/m^2.
2) To find the variation of pressure in the sound wave, we can use the formula:
Pressure = √(2 * Intensity * Density * Speed of Sound)
Given that the intensity is 8.11x10^-6 W/m^2, the density of air is 1.2 kg/m^3, and the speed of sound is 343 m/s, we can substitute these values into the formula:
Pressure = √(2 * (8.11x10^-6 W/m^2) * (1.2 kg/m^3) * (343 m/s))
Simplifying this expression, we get:
Pressure = √(2 * 8.11x10^-6 W/m^2 * 1.2 kg/m^3 * 343 m/s)
Now we can do the multiplication inside the square root:
Pressure = √(2 * 8.11x10^-6 W/m^2 * 1.2 kg/m^3 * 343 m/s)
Pressure = √(1.0108x10^-2 N/m^2)
Taking the square root of 1.0108x10^-2, we get:
Pressure = 0.1005 N/m^2
Therefore, the variation of pressure in the sound wave is 0.1005 N/m^2.