Unless indicated otherwise, assume the speed of sound in air to be v = 344 m/s. CP A baby’s mouth is 30 cm from her father’s ear and 1.50 m from her mother’s ear. What is the difference between the sound intensity levels heard by the father and by the mother?

Solution 21E Assume the speed of sound in air to be v = 344 m/s. Given data: The distance between baby’s mouth and mother’s ear r = 1.5 m. 1 The distance between baby’s mouth and father’s ear r = 30 cm. 2 The difference between sound intensity levels = 10 dB log(I /I ) 2 1 2 1 Intensity drops as the square of the distance from the source.the distance ratio squared is (1.5/0.30) = 25 times greater at father’s ear than mothers.in decibels it can expressed as 10 log(25).