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?
1) I = E/(A*t)= (2.5*10^-11J)/[(6.7*10^-4 m^2)(4.62s)] = 8.1*10^-9 W/m^2
That would be about 39 decibels... not very loud. A soft conversation is about that loud.
2) Use the equation
I = Pmax^2/(2*density*V)
where V is the sound speed. Solve it for Pmax, the pressure amplitude.
I get 2.6*10^-3 N/m^2, which is about 2.5*10^-8 atm
I found that equation in
"Physics Principles and Applications" bu Henry Margenau. It should also be online somewhere.
You could also relate the sound intensity in dB directly to the sound pressure level, but the answer might be different. I get 1.8*10^-3 N/m^2 doing it that way
Thank youuu sooo much i didn't see it that's why i kept posting it :(
P = 2.5*10^-11J./4.62s. = 5.41*10^-12J/s
= 5.41*10^-12 W.
I = 5.41*10^-12W./6.7cm^2 = 8.07*10^-13
W./cm^2. = 8.07*10^-9 W./m^2.
To find the intensity of the sound in question 1, we can use the formula:
Intensity = Sound energy / Time
First, let's convert the given sound energy to SI units. The energy given is 2.5x10^-11 J.
Next, we plug the values into the formula and solve for the intensity:
Intensity = (2.5x10^-11 J) / (4.62 s)
Calculating this will give us the intensity of the sound.
To find the variation of pressure in question 2, we can use the formula:
Intensity = (Peak pressure)^2 / (2 * Density * Speed of Sound)
We have already calculated the intensity from question 1. Now we need to solve for the peak pressure.
Rearranging the formula, we get:
Peak pressure = sqrt(Intensity * 2 * Density * Speed of Sound)
We can plug in the given values for the density of air (1.2 kg/m^3) and the speed of sound (343 m/s) to calculate the variation of pressure.