Intensity and Power of Sound The intensity of the sound from a certain sourceis
ID: 1727653 • Letter: I
Question
Intensity and Power of SoundThe intensity of the sound from a certain sourceis measured by two observers located at different positions along aline from the source. The observers are located on the same side ofthe source and are separated by 180.1 m. The observer that isclosest to the source hears the sound with an intensity of59.6 dB. The intensity of the sound heard by the more distantobserver is 50 dB.
What is the distance from the source to the closestobserver? What is the power output of thesource? Intensity and Power of Sound
The intensity of the sound from a certain sourceis measured by two observers located at different positions along aline from the source. The observers are located on the same side ofthe source and are separated by 180.1 m. The observer that isclosest to the source hears the sound with an intensity of59.6 dB. The intensity of the sound heard by the more distantobserver is 50 dB.
What is the distance from the source to the closestobserver? What is the power output of thesource?
Explanation / Answer
ratio of intensities can be found by . diff in sound level = 10 log (ratioof intensities) . 9.6 = 10 log (ratio of I) . ratio of I = 9.120 . The intensities are proportional to the distance from thesource squared, so the ratio of the distance is 9.120 = 3.020 . This means: closest distanceis x further distanceis y and . y -x = 180.1 and y = 3.020 x . So... 3.020x - x = 180.1 2.020 x = 180.1 . x = 89.16 meters is the distance from the source to the closerobserver. . Now... the intensity at the closer observer is . sound level = 10 log (intensity / Io) where Io = 1 x10-12 W/m2 . 59.6 = 10 log(intensity / Io ) . intensity = 912011x Io = 9.120 x10-7 W/m2 . Then... power output = intensity * 4r2 = 9.120 x 10-7 * 4 *89.162 = 0.09111 Watts . Then... power output = intensity * 4r2 = 9.120 x 10-7 * 4 *89.162 = 0.09111 WattsRelated Questions
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