We are given the formula for loudness in decibels (dB):

$L = 10 \log \left( \frac{I}{I_0} \right)$

where:

• $L$ is the loudness in decibels,
• $I$ is the intensity of the sound in watts per square meter ($W/m^2$),
• $I_0 = 10^{-12}$ $W/m^2$ is the reference intensity.

Step 1: Solve for $I$

Rearrange the formula to solve for $I$:

$I = I_0 \times 10^{L/10}$

Step 2: Calculate $I$ for the top up (80 dB)

$I_{\text{up}} = (10^{-12}) \times 10^{80/10}$

$I_{\text{up}} = (10^{-12}) \times 10^8$

$I_{\text{up}} = 10^{-4} \text{ W/m}^2$

Step 3: Calculate $I$ for the top down (96 dB)

$I_{\text{down}} = (10^{-12}) \times 10^{96/10}$

$I_{\text{down}} = (10^{-12}) \times 10^9.6$

$I_{\text{down}} = 10^{-2.4} \text{ W/m}^2$

Approximating:

$I_{\text{down}} \approx 3.98 \times 10^{-3} \text{ W/m}^2$

Final Answers:

• The intensity of the sound with the top up is $10^{-4}$ W/m².
• The intensity of the sound with the top down is $3.98 \times 10^{-3}$ W/m².

Would you like a breakdown of logarithmic calculations? Let me know!

Related Questions:

1. How does doubling the intensity affect the loudness in decibels?
2. If the intensity increases by a factor of 100, by how many dB does the loudness increase?
3. What is the intensity level at 50 dB?
4. How does sound intensity relate to human perception of loudness?
5. What is the threshold of pain in decibels, and what is its intensity?

Tip:

Every 10 dB increase represents a 10-fold increase in intensity, but only about a doubling in perceived loudness.