Question

If the intensity of sound is doubled, by how many decibels does the sound level increase?

Answer 1

Explanation ⇒ We know that,

For the intensity I, sound level  

 τ = 10 log₁₀ [I/I₀],

when the intensity will changes to 2I,

then,

 τ' =  10 log₁₀ [2I/I₀]

On subtracting both the equation,

τ' - τ =  10 log₁₀ [2I/I₀] -  10 log₁₀ [I/I₀]

τ' - τ = 10(log₁₀ [2I/I₀] - log₁₀ [I/I₀])

τ' - τ = 10(log2)    [Since, logm - logn = log(m/n)]

∴ τ' - τ = 10 × 0.3

∴ τ' - τ = 3

∴ τ' = τ + 3

Thus, when intensity is doubled, then sound level increases by 3 db.

Hope it helps.

Answer 2

Explanation:

The sound intensity is given by :;

$dB=10\ log(\dfrac{I}{I_o})$

It is given that, if the sound intensity is doubled i.e I = 2I₀

$dB=10\ log(\dfrac{2I}{I})$

$dB=10\ log(2)$

Since, log 2 = 0.301

$dB=3.01\ dB$

So, if the sound intensity is doubled, sound level increases by a factor of 3.01 decibels. Hence, this is the required solution.