A common commercial jet aircraft can give off \'edge noise\", asound radially ou

Question

A common commercial jet aircraft can give off 'edge noise", asound radially outwards in all directions. The intensity of thesounds is I1=10^-2 W/m^2 at a distance, r1, from the aircraft andis I2=10^-4 W/m^2 at a distance, r2, from the aircraft. a.) If we assume a constant output sound source, the distancer2 is (choose one) greater or lesser than r1 by a factor of (fillin the blank) _______. b.) On a cold day (0deg C) the speed of the sound from a jetaircraft is 331 m/s and has a low frequency of 200 Hz. What is thechange in sound wavelength, , when the temperature of the airraises to 20 deg C? A common commercial jet aircraft can give off 'edge noise", asound radially outwards in all directions. The intensity of thesounds is I1=10^-2 W/m^2 at a distance, r1, from the aircraft andis I2=10^-4 W/m^2 at a distance, r2, from the aircraft. a.) If we assume a constant output sound source, the distancer2 is (choose one) greater or lesser than r1 by a factor of (fillin the blank) _______. b.) On a cold day (0deg C) the speed of the sound from a jetaircraft is 331 m/s and has a low frequency of 200 Hz. What is thechange in sound wavelength, , when the temperature of the airraises to 20 deg C?

Explanation / Answer

(a) you are given intensity of   0.01   at   r1   and    0.0001 ar r2. . Since the sound is weaker at r2, we know that   r2is further than r1. Also, intensity varies inversely by the squareof the distance. Since the sound at r2 is 100 times weaker, we knowthat... .    r2 is   10 times greater than r1 . (b) The initial wavelength is .     wavelength = speed / freq = 331/ 200 = 1.655 meters . The new speed of sound, at 20 degrees, is   331 + 0.6 * 20 =   343 . So the new wavelength is    343 / 200 =    1.715 . And the change in wavelength is    171.5 -1.655 =    0.060 meters = 6.00 cm

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