At point A, 3.00 m from a small source of sound that is emitting uniformly in al

Question

At point A, 3.00 m from a small source of sound that is emitting uniformly in all directions, the sound level is 53.0 dB. A: What is the intensity of the sound at point A? B: How far from the source would the intensity of the sound be one-fourth the intensity at point A? C: How far from the source would the sound level of the second be one-fourth the sound level at point A? how far must you go so that the sound intensity level is one-fourth of what it was at A d) dos intensity obey the inverse-square law? What about sound intensity level? Please don't answer(this answer is wrong): 55 + (20log(3) + (10log4pi)) = 75.5345 dB @ source. 55/4 = 13.75 dB 13.75 - 75.5345 + 10log(4pi) = - 50.7924 - 20log(x) = - 50.7924 -50.7924/-20 = log(x) 10^x = 346.434 m from source. check: 75.5345 - (20log(346.434) + 10log(4pi)) = 13.75 dB

Explanation / Answer

beta = ( 10 dB) log _10 ( I/ I0)

log_10 ( I/ 10^-12) = 53/10

I = 10 ^-5.3-12

= 10^-6.7 W/m^2

(b)

I1/I2 = (r2/r1)^2

r2 = sqrt ( 3)^2 * I1/ I1/4= 6 m

(c)

difference in sound intensiy level

Beta_2- Beta1 = 10 dB log_10 ( I2/I1)=10 dB log_10 ( r1/r2)^2

log_10 ( r1/r2)^2 = 53dB- 1/4 ( 53 dB)/ 10 dB

( r1/r2)^2= 10^3.975

r1 = sqrt r2^2 ( 10 ^3.975)

= sqrt 3^2 ( 10 ^3.975)

= 291.488 m

(d) and E

intensity of sound obeys inverse square law, where intensity levle does not depend on it

because decible scale is logarithemic

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