Question

A trumpet creates a sound intensity level of 52.1 dB at a distance of 3.6 m. What is the sound intensity level (in dB) of 7 such trumpets at a distance of 10.9 m?

Answer 1

Intensity level of sound and distance can be related by the equation, - 82 where, L, intensity level at - distancer, 7 = - 2 L₂ > intensity level at distance rz > 10 - 10.9 3.6 [no] - 3.0277 20 2,c2z = log (3.0277) = 3 0.4811 2,-L₂ = 20x0.4811 = 9.622 Given, L=52.1dB 52.1_L₂=9.622 L₂ = 52.1-9.622 L2 = 42.5dB Ans: Intensity level of t toumpets at a distance 10.9m is 42.5dB

Answer 2

SOLUTION :

I₂ 

= intensity level of 1 trumpet at a distance of  r2.

= I₁ * r₁^2 / r₂^2    

= 52.1 * 3.6^2 / 10.9^2.

 (I₁ is intensity of a trumpet at r₁ (3.6 m) away = 52.1 DB, r₂ = 10.9 m)

= 5.6832 DB 

So,

Intensity level of 7 trumpets 

= 7 * 5.6832

= 39.78 DB (ANSWER).