The ability to hear a \"pin drop\" is the sin of sensitivehearing. Suppose a 0.5

Question

The ability to hear a "pin drop" is the sin of sensitivehearing. Suppose a 0.55-g pin is dropped from a height of 28cm, and that the pin emits sound for 1.5 s when it lands. Assuming all of the mechanical energy of the pin is converted tosound energy, and that the sound radiates uniformly in alldirections, find the maximum distance from which a person can hearthe pin drop. (This is the ideal maximum distance butatmospheric absorption and other factors will make the actualmaximum distance considerably smaller.)

Explanation / Answer

Mass of pin is(m) = 0.55g                           = 0.55 x 10-3 kg The pin dropped form a height (h) =28cm                                                    = 0.28m Acceleration due to gravity (g) =9.81m/s2 The reference intensity (I)=10-12W/m2 The reference intensity (I)=10-12W/m2 All the potential energy is converted tosound energy here.          U =mgh             =(0.55 x 10-3 kg)(9.81 m/s2)(0.28m)            = 1.511 x 10-3 J We have to find thepower          P =E/t             =(1.511 x 10-3 J) / (1.5 s)            = 1.01 x 10-3 W And we know that sound intensity isgiven by          I = P /4r2   4r2 =1.01 x 10-3W / 10-12 W /m2          =1.01 x 109 m2 The maximum distance from which theperson can hear the pin drop is The maximum distance from which theperson can hear the pin drop is      r   = ( 1.01 x 109m2)/4(3.14)           = (0.804*108)            =0.896*104m           = 8.96*103m          =9.0km

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