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Chapter 3: Problem 24

Suppose a rock band is playing loudly at a sound level of 100 decibels. What decibel level would make the band sound three-fifths as loud?

Short Answer

Expert verified
The new decibel level that would make the band sound three-fifths as loud is approximately 94.77 dB.

Step by step solution

01

Understand the Decibel Scale

Decibel (dB) is a logarithmic unit used to measure the intensity of sound. The intensity or sound level L1 is related to the reference intensity (L0) through the following formula: \[ L_1 = 10\log_{10}\left(\frac{I_1}{I_0}\right) \] Where L1 is the sound level in decibels, I1 is the intensity of the sound, and I0 is the reference intensity level. In this problem, we are given the sound level (L1 = 100 dB) and need to find the decibel level that would make the band sound three-fifths (3/5) as loud.
02

Calculate the Intensity Ratio

To find the intensity ratio, we first need to remove the logarithmic scale from the loudness level. We can do this by using the inverse operation, which is finding the antilogarithm: \[ \frac{I_1}{I_0} = 10^{\frac{L_1}{10}} \] Now, we can calculate the intensity ratio of the sound level at 100 dB: \[ \frac{I_1}{I_0} = 10^{\frac{100}{10}} = 10^{10} \]
03

Find the New Intensity

Now that we have the intensity ratio for the loudest sound, we want to find the intensity ratio for the new sound level that is three-fifths (3/5) as loud. Multiply the original intensity ratio with 3/5: \[ \frac{I_2}{I_0} = \frac{3}{5} \times \frac{I_1}{I_0} = \frac{3}{5} \times 10^{10} \]
04

Calculate the New Sound Level

Now that we have the new intensity ratio, we can calculate the new sound level L2 using the decibel formula: \[ L_2 = 10\log_{10}\left(\frac{I_2}{I_0}\right) \] \[ L_2 = 10\log_{10}\left(\frac{3}{5} \times 10^{10}\right) \] By using the properties of logarithms, we can simplify the expression: \[ L_2 = 10\log_{10}\left(\frac{3}{5}\right)+ 10\log_{10}\left(10^{10}\right) \] Now, we can find the decibel level that makes the band sound three-fifths as loud: \[ L_2 = 10\log_{10}\left(\frac{3}{5}\right)+ 100 = 10(-0.5229) + 100 \approx 94.77 \] So, the new decibel level that would make the band sound three-fifths as loud is approximately 94.77 dB.

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