Question

A baby's mouth is a distance of 35 cm from her father's ear and a distance of 1.60 m from her mother's ear. What is the difference between the sound intensity levels heard by the father and by the mother? Express your answer using two significant figures. βfather−βmother=？

Answer 1

The difference in sound intensity levels heard by the father and mother is approximately -12.11 dB.

Step-by-step explanation:

To find the difference in sound intensity levels heard by the father and mother, we can use the formula:

βfather - βmother = 10 log10(Ifather/Imother)

Given that the distance of the baby's mouth from the father's ear is 35 cm, and the distance from the baby's mouth to the mother's ear is 1.60 m, we can calculate the ratio of the intensities.

Using the inverse square law, the ratio of intensities is:

Ifather/Imother = (distance from father's ear / distance from mother's ear)2

Ifather/Imother = (0.35 / 1.60)2

Ifather/Imother ≈ 0.0614

Substituting this value into the first formula, we can calculate the difference in sound intensity levels:

βfather - βmother = 10 log10(0.0614)

βfather - βmother ≈ 10(-1.211)

βfather - βmother ≈ -12.11 dB

Answer 2

The difference in sound intensity levels heard by the father and the mother can be calculated using the formula βfather - βmother = 10 log10(Ifather/Imother).

Step-by-step explanation:

The difference in sound intensity levels heard by the father and the mother can be calculated using the formula:

βfather - βmother = 10 log10(Ifather/Imother)

where β represents the sound intensity level and I represents the sound intensity. We can convert the given distances to intensities using the inverse square law:

I = k/r2

where I is the intensity, k is a constant, and r is the distance from the source. By substituting the distances into the formula and simplifying the equation, we can find the difference in sound intensity levels between the father and the mother.