Chapter 2: Q24E (page 539)

(a) If two sounds differ by 5.00 dB, find the ratio of the intensity of the louder sound to that of the softer one. (b) If one sound is 100 times as intense as another, by how much do they differ in sound intensity level (in decibels)? (c) If you increase the volume of your stereo so that the intensity doubles, by how much does the sound intensity level increase?

Short Answer

Expert verified

a) The ratio of the intensity is 3.16 b) $Δβ = 20 db$ c) $Δβ = 3 db$

Step by step solution

STEP 1 The difference between the sound intensity levels for two different intensities

The formula is given by $β_{2} - β_{1} = 10 \log (\frac{I_{2}}{I_{1}})$ where, $l_{2}$ is the new intensity and $l_{1}$ is the previous intensity

STEP 2 Find the ratio of these two intensities

Substitute the values in the equation $β_{2} - β_{1} = 10 \log (\frac{I_{2}}{I_{1}})$ we get,

$5 db = 10 \log (\frac{l_{2}}{l_{1}}) 0.5 = \log (\frac{l_{2}}{l_{1}}) 10^{0.5} = \frac{l_{2}}{l_{1}} \frac{l_{2}}{l_{1}} = 3.16$

Step 3 Find the difference in sound intensity level between the two sounds

If the intensity of the first sound is $l_{1}$ , then the intensity of the second sound will be $l_{2}$ = $100 l_{1}$ , to find the difference in sound intensity level between the two sounds $∆ β$ , we need to substitute into (l) for $l_{1}$ and $l_{2}$

$∆ β = 10 \log (\frac{100 l_{1}}{l_{1}}) = 10 \log (100) = 20 db$

Now the intensity of the second sound will be $l_{2} = 2 l_{1}$ . Therefore, we get,

$∆ β = 10 \log (\frac{2 l_{1}}{l_{1}}) = 10 \log (2) = 3 db$

Thus, the difference in sound intensity level between the two sounds is 3db

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Short Answer

Step by step solution

STEP 1 The difference between the sound intensity levels for two different intensities

STEP 2 Find the ratio of these two intensities

Step 3 Find the difference in sound intensity level between the two sounds

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