You are hiking in Eastern Montana about 400 miles Fast of Yellowstone National P

Question

You are hiking in Eastern Montana about 400 miles Fast of Yellowstone National Park (a super volcano that you thought was dominant). However, as you are taking pictures the volcano erupts. A man near to you yells "Vancouver, this is it!" and runs off at a velocity of 6.7 m/s in the direction opposite to the explosion (East). The power of the eruption is 4.2 *. In this problem assume the speed of sound is still 340 m/s. If you are located a distance o $640 km from the center of Yellowstone how long will it take the sound of the explosion to get to you? What will the intensity of sound of the explosion be when it reaches you AND what will the decibel level of the sound be? 190 decibels over a long period of time would kill a person. How far will you need to be from the volcano so that the concussion (the sound) from the eruption does NOT kill you? If the rest frequency of the sound from the explosion is 120 Hz then what frequency will the man who ran away hear from you hear when it gets to him?

Explanation / Answer

A) time taken = (640 * 1000)/340

                         =   31.37 min

B)   intensity of sound = (4.2 * 1021)/(4 * 3.14 * 640000 * 640000)

                                  =   8.1639 * 108 W/m2

=> decibel level =    10 * log(8.1639 * 108/10-12)

                            = 209.12 dB

C)    Here,    For 190 dB intensity = 107 W/m2

=> far will you need to be = 5782.68 km

D)    frequency man will hear = 120 * (340-6.7)/340

                                               = 117.63 Hz

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