Find the ratio of sound intensity heard?

Question

A skyrosket explodes 90m above the ground. Three observers are spaced 82. apart, with observer A directly under the point of the explosion. Find the ratio of the sound intensity heard by observer A to that heard by observer B. Find ratio of intensity heard by observer A to that heard by observer C.

Answer 1

The information in this problem is confusing. it is implied but not stated that the observers are in a line. Only if that is the case can a definite answer be obtained. Sound intensity will decrease in proportion to the square of the distance from the source. Observer A is 90m from the source. Observer B's dlstance from the source is the hypotenuse of a right triange with legs 90m and 82m, and it's length is √[90^2 + 82^2]. The ratio (A's intensity to B's) will then be [B's distance}^2 / [A's distance]^2. This is [90^2 + 82^2] / 90^2, or

[82/90]^2 + 1.

If the observers are in a line, then C's distance is 2*82m from A, and the same approach is used to get the result.

Answer 2

Sound intensity point SIL is a logarithmic degree of the sound intensity I (measured in W/m2), in assessment to a reference point I0. The degree L of a ratio of two sound intensities in decibels is: (a million) L = 10 * log (I / I0) dB [see source a million] enable l. a. be the backside SIL and Lb be the better SIL; then: (2) 30.3 = Lb - l. a. [given] increasing (2) utilising (a million): (3) 30.3 = ( 10 * log (Ib / I0) ) - ( 10 * log (Ia / I0) ) = 10 * ( log (Ib / I0) - log (Ia / I0) ) = 10 * log [ (Ib / I0) / (Ia / I0) ] = 10 * log (Ib / Ia ) => (4) log (Ib / Ia ) = 30.3 / 10 => (5) Ib / Ia = 10 ^ (30.3 / 10) = 10 ^ 3.03 ~= 1072. <<<=== ratio of the better intensity to the unique (unaided) intensity. .