Question

The sound level at a point 5.0 m away from a point source is 40 dB. What will be the level at a point 50 m away from the source?

Answer 1

Answer ⇒ 40 db.

Explanation ⇒ We know that Sound level is given by the formula,

τ = 10 log₁₀ [I/I₀]

Let at the point 5 m away from the source, sound level be τ and 50 m away be τ' and intensity be I and I'.

∴ τ' = 10 log₁₀ [I'/I₀]

Now,

τ' - τ = 10log₁₀ [I'/I₀] - 10log₁₀ [I/I₀]

∴ τ' - τ = 10 log[I'/I]

we know that intensity is inversely proportional to the square of the distance, thus,

I'/I = 5²/50²

∴ I'/I = 1/100

∴  τ' - τ = 10 log[1/100]

∴ τ' - τ = 10[0 - 2]

∴ τ' = τ - 20

Now,

τ = 40 db,

thus, τ' = 40 - 20 = 20 db.

Hope it helps.

Answer 2

Answer:

40 db

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