Question

If the sound level in a room is increased from 50 dB to 60 dB, by what factor is the pressure amplitude increased?

Answer 1

Answer ⇒  p₀' = p₀√10

Explanation ⇒

Let the intensity of sound at 50 dB = I and at 60 dB = I'.

Since, sound level = 10 log[I/I₀]

∴ 10 log[I/I₀] = 50 and

∴ 10 log[I'/I₀] = 60

then, 10 log[I'/I₀] - 10 log[I/I₀] = 60-50

log[I'/I] = 1

I'/I = 10    [Since log10 = 1.]

Now, we know that pressure amplitude is directly proportional to the square of under root of amplitude.

∴ I'/I = p₀'²/p₀²

∴ p₀'/p₀ = √10

∴ p₀' = p₀√10

Hope it helps.

Answer 2

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