if a sound intensity changes by 1/10 of the original level, its dB changes by

The decibel (dB) scale is logarithmic, and the relationship between sound intensity levels in dB and the ratio of intensities is given by the following formula:

\[
\Delta L = 10 \cdot \log_{10} \left( \frac{I_2}{I_1} \right)
\]

where:
- \(\Delta L\) is the change in sound level (in decibels, dB),
- \(I_1\) is the original intensity,
- \(I_2\) is the new intensity.

If the sound intensity decreases to \( \frac{1}{10} \) of the original level, we plug this into the formula:

\[
\Delta L = 10 \cdot \log_{10} \left( \frac{1}{10} \right)
\]

The logarithm of \( \frac{1}{10} \) is:

\[
\log_{10} \left( \frac{1}{10} \right) = -1
\]

So:

\[
\Delta L = 10 \cdot (-1) = -10 \text{ dB}
\]

Thus, if the sound intensity decreases by a factor of \( \frac{1}{10} \), the sound level decreases by 10 dB.

Would you like more details or have any questions?

Here are 5 related questions:
1. How is sound intensity level measured in decibels?
2. What is the formula to calculate sound level changes for different intensity ratios?
3. What does a negative change in dB indicate about the sound intensity?
4. How would the dB change if the sound intensity decreases to \( \frac{1}{100} \)?
5. How does the logarithmic nature of the dB scale affect perception of sound?

**Tip**: When comparing sound intensities, small changes in decibels can represent large differences in actual sound intensity.

Learn how to calculate the decibel (dB) change when sound intensity decreases by 1/10 of the original level. This problem involves logarithmic relationships and the formula ΔL = 10 * log10(I2 / I1), suitable for students in Grades 9-12.

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