Question

At 10:02 am local time on August 27, 1883, the Krakatoa volcano exploded. This volcanic eruption supposedly generated the loudest sound ever heard on Earth. Measurements at the time placed the peak sound level of 165 dB at a distance of 40 miles from the eruption! This large transient sound level will break your ear drums and corresponds to an increase in air pressure of 2.5 inches of mercury. (a) What is the intensity (I) of the eruption sound at a measured distance of 40 miles (4 points)? (b) The approximate threshold for pain in the human ear is 120 dB. How much does the intensity have to be reduced by in order for the eruption sound to be at the pain threshold at 40 miles? I imagine the involved spectators would have to use some type of earplug to achieve this reduction (4 points). 3 (c) Residents of a small town some distance away reported hearing artillery firing in the distance at a sound level of 130 dB. How far was this small town from the Krakatoa volcano (4 points)?

Answer 1

i) Date what is intensity (165 db) at 40 miles Intensity (db) = 10 log, / (Intensity at A) Inverse square law (1/2) for sound intensity threshold of hearing vs distance lit Il be I (ab= lo logo intensity unit distance volcano I away from T ( where I is in watt/m2) Not in db here ) Io = 10-12 22 W/m2 165 = 10 logio I 10-12 16.5 10 -T2 * 10 >> 3.168 LO' But it is easier to use 6 db rule at 80 miles 165-6 x 159 db at 160 miles 189-6 ~ 153 db 320 miles 153-6 147 db 640 147-6 141 db 1280 135db 2560 miles 1356 129 db So about would be Bodb 104.5 네 141 - 6 Watts m2 2300 miles away intensity 120 10 o log [ Ireduced By formula! То 3.16 x 10 4 W/m² 10 W/m² I reduced = 165 db = 130 db 1 watt me 2 3.168104 r = 40 miles 10 22 reduction by a factor of 3.16810 31 thousand time less } ii) Inverse distance Law (in db) for every doubling of distance, The Sound Leved reduce by 6 db. 3.16 8 1008 40 2250 miles