Get the solution to "If the intensity of sound is doubled, by..."

odigranogi5

Answered question

2023-01-01

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

Dorian Beasley

Beginner2023-01-02Added 13 answers

Step 1: We have
Let the intensity of the sound be

I

.
According to the question,

I' = 2 I

Step 2: Formula used
Let the intensity of the sound be

I

and it's sound level is

β 1

The sound level

β

provided by,

β = 10 \log 10 (\frac{I}{I})

Where the constant reference intensity is

I 0

Step 3: The sound level increased
The sound level

β 1

provided by,

β 1 = 10 \log 10 (\frac{I}{I}) . . . (1)

When the intensity doubles, the sound level is given by,

β 2 = 10 \log 10 (\frac{2 I}{I}) . . . (2)

By subtracting equations (2) and (1), the sound level increased is,

β 2 - β 1 = (10 \log 10 (\frac{2 I}{I})) - (10 \log 10 (\frac{I}{I})) β 2 - β 1 = 10 (\log 10 (\frac{2 I}{I}) - \log 10 (\frac{I}{I})) . . . (3)

Since

\log a (x) - \log a (y) = \log a (\frac{x}{y})

equation 3 becomes,

β 2 - β 1 = 10 (\log 10 (\frac{2 I}{I}) - \log 10 (\frac{I}{I})) β 2 - β 1 = 10 \log 10 (\frac{2 I}{I}) β 2 - β 1 = 10 \log 10 2 β 2 - β 1 = 10 \times 0.3010 β 2 - β 1 = 3 dB

Hence, the volume of the sound rises by

3 dB

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