odigranogi5

Answered

2023-01-01

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

Answer & Explanation

Dorian Beasley

Expert

2023-01-02Added 11 answers

Step 1: We have

Let the intensity of the sound be $I$.

According to the question, $I\text{'}=2I$

Step 2: Formula used

Let the intensity of the sound be $I$ and it's sound level is $\beta 1$

The sound level $\beta $ provided by, $\beta =10\mathrm{log}10\left(\frac{I}{I}\right)$

Where the constant reference intensity is $I0$

Step 3: The sound level increased

The sound level $\beta 1$ provided by, $\beta 1=10\mathrm{log}10\left(\frac{I}{I}\right)...\left(1\right)$

When the intensity doubles, the sound level is given by, $\beta 2=10\mathrm{log}10\left(\frac{2I}{I}\right)...\left(2\right)$

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

$\beta 2-\beta 1=\left(10\mathrm{log}10\left(\frac{2I}{I}\right)\right)-\left(10\mathrm{log}10\left(\frac{I}{I}\right)\right)\beta 2-\beta 1=10\left(\mathrm{log}10\left(\frac{2I}{I}\right)-\mathrm{log}10\left(\frac{I}{I}\right)\right)...\left(3\right)$

Since $\mathrm{log}a\left(x\right)-\mathrm{log}a\left(y\right)=\mathrm{log}a\left(\frac{x}{y}\right)$ equation 3 becomes,

$\beta 2-\beta 1=10\left(\mathrm{log}10\left(\frac{2I}{I}\right)-\mathrm{log}10\left(\frac{I}{I}\right)\right)\beta 2-\beta 1=10\mathrm{log}10\left(\frac{2I}{I}\right)\beta 2-\beta 1=10\mathrm{log}102\beta 2-\beta 1=10\times 0.3010\beta 2-\beta 1=3\mathrm{dB}$

Hence, the volume of the sound rises by $3\mathrm{dB}$.

Let the intensity of the sound be $I$.

According to the question, $I\text{'}=2I$

Step 2: Formula used

Let the intensity of the sound be $I$ and it's sound level is $\beta 1$

The sound level $\beta $ provided by, $\beta =10\mathrm{log}10\left(\frac{I}{I}\right)$

Where the constant reference intensity is $I0$

Step 3: The sound level increased

The sound level $\beta 1$ provided by, $\beta 1=10\mathrm{log}10\left(\frac{I}{I}\right)...\left(1\right)$

When the intensity doubles, the sound level is given by, $\beta 2=10\mathrm{log}10\left(\frac{2I}{I}\right)...\left(2\right)$

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

$\beta 2-\beta 1=\left(10\mathrm{log}10\left(\frac{2I}{I}\right)\right)-\left(10\mathrm{log}10\left(\frac{I}{I}\right)\right)\beta 2-\beta 1=10\left(\mathrm{log}10\left(\frac{2I}{I}\right)-\mathrm{log}10\left(\frac{I}{I}\right)\right)...\left(3\right)$

Since $\mathrm{log}a\left(x\right)-\mathrm{log}a\left(y\right)=\mathrm{log}a\left(\frac{x}{y}\right)$ equation 3 becomes,

$\beta 2-\beta 1=10\left(\mathrm{log}10\left(\frac{2I}{I}\right)-\mathrm{log}10\left(\frac{I}{I}\right)\right)\beta 2-\beta 1=10\mathrm{log}10\left(\frac{2I}{I}\right)\beta 2-\beta 1=10\mathrm{log}102\beta 2-\beta 1=10\times 0.3010\beta 2-\beta 1=3\mathrm{dB}$

Hence, the volume of the sound rises by $3\mathrm{dB}$.

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