A baby's mouth is 30.0 cm from her father's ear and 1.50 m from her mother's ear. What is the difference between the sound intensity levels (in dB) heard by her father and by her mother?
(20)^2 = 400 times the sound power (at the mother's ear) = how many decibels?
To find the difference in sound intensity levels heard by the baby's father and mother, we can use the inverse square law for sound.
The inverse square law states that the sound intensity decreases with the square of the distance from the source. Mathematically, it can be expressed as:
I2/I1 = (r1/r2)^2
where I1 and I2 are the sound intensities at distances r1 and r2 from the source, respectively.
Let's assign the following variables:
I1 = sound intensity at the baby's father's ear
I2 = sound intensity at the baby's mother's ear
r1 = distance between the baby's mouth and her father's ear = 30.0 cm = 0.30 m
r2 = distance between the baby's mouth and her mother's ear = 1.50 m
Using the inverse square law equation, we have:
I2/I1 = (r1/r2)^2
Plugging in the values:
I2/I1 = (0.30/1.50)^2
Simplifying:
I2/I1 = (0.2)^2
I2/I1 = 0.04
Therefore, I2 = 0.04 * I1
The difference in the sound intensity levels can be expressed in decibels (dB) using the formula:
ΔdB = 10 * log10(I2/I1)
Plugging in the value of I2/I1 we found earlier:
ΔdB = 10 * log10(0.04)
Using a calculator, we can find that log10(0.04) ≈ -1.39794
Therefore:
ΔdB = 10 * (-1.39794)
ΔdB ≈ -13.9794 dB
So, the difference in sound intensity levels heard by the baby's father and mother is approximately -13.9794 dB.