Sound 1 has an intensity of 38 W/m2. Sound 2 has an intensity level that is 2.5 dB greater than the intensity level of sound 1. What is the intensity of sound 2?

The intensity of sound 2 is 67.6 W/m²

Explanation:

First we convert the intensity of sound 1 to the intensity level in db:

For this we use the formula:

L₁ = 10 log₁₀[I₁/I₀]

where,

L₁ = intensity level of sound 1

I₁ = Intensity of sound 1 = 38 W/m²

I₀ = Minimum Audible Intensity = 10⁻¹² W/m²

Therefore:

L₁ = 10 log₁₀ [38/10⁻¹²]

L₁ = 135.8 dB

Since, the intensity level of sound 2 is 2.5 dB greater than intensity level of sound 1. Therefore,

L₂ = 135.8 dB + 2.5 dB

L₂ = 138.3 dB

Now, we calculate intensity of sound by the same formula:

L₂ = 10 log₁₀[I₂/I₀]

10^(L₂/10) = [I₂/I₀]

(I₀)[10^(L₂/10)] = I₂

where,

L₂ = intensity level of sound 2 = 138.3 dB

I₂ = Intensity of sound 2 = ?

I₀ = Minimum Audible Intensity = 10⁻¹² W/m²

Therefore:

I₂ = (10⁻¹²)(10^13.83)

I₂ = 10^1.83

I₂ = 67.6 W/m²

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