To solve this problem, we'll break it into two parts.

Part (a) - Sound Intensity Level of 5 Trombones

The sound intensity level $L$ in decibels (dB) is related to the sound intensity $I$ by the formula: $L = 10 \log_{10} \left( \frac{I}{I_0} \right)$ where:

• $L$ is the sound intensity level in dB,
• $I$ is the sound intensity in watts per square meter (W/m²),
• $I_0$ is the reference intensity, typically $1 \times 10^{-12} \, \text{W/m²}$.

When multiple sound sources of the same intensity are combined, the total intensity is multiplied by the number of sources. For 5 trombones, the total intensity would be 5 times the intensity of a single trombone.

If the intensity of a single trombone is $I$, then for 5 trombones, the total intensity becomes $5I$.

The new sound intensity level $L_5$ for 5 trombones is: $L_5 = 10 \log_{10} \left( \frac{5I}{I_0} \right)$

We can rewrite this using logarithmic properties: $L_5 = 10 \log_{10} \left( 5 \right) + 10 \log_{10} \left( \frac{I}{I_0} \right)$ $L_5 = 10 \log_{10} \left( 5 \right) + L$

Given that $\log_{10} 5 \approx 0.699$ and the sound intensity level $L$ for one trombone is 70 dB: $L_5 = 10 \times 0.699 + 70 = 6.99 + 70 = 76.99 \, \text{dB}$

So, the sound intensity level for 5 trombones is approximately 77 dB.

Part (b) - Barely Audible Sound from a Single Trombone

The threshold of hearing corresponds to a sound intensity level of 0 dB, which is equivalent to the reference intensity $I_0 = 1 \times 10^{-12} \, \text{W/m²}$.

Sound intensity decreases with the square of the distance from the source. If $I_1$ is the intensity at a distance $r_1 = 1 \, \text{m}$ (which corresponds to 70 dB), the intensity at a farther distance $r_2$ is: $I_2 = I_1 \left( \frac{r_1}{r_2} \right)^2$

Since sound intensity levels in dB are given by: $L_2 = 10 \log_{10} \left( \frac{I_2}{I_0} \right)$

At the threshold of hearing, $L_2 = 0 \, \text{dB}$, so: $0 = 10 \log_{10} \left( \frac{I_1 \left( \frac{1}{r_2} \right)^2}{I_0} \right)$

Simplifying this equation: $\log_{10} \left( \frac{I_1}{I_0} \right) + 2 \log_{10} \left( \frac{1}{r_2} \right) = 0$ $\log_{10} \left( \frac{I_1}{I_0} \right) = -2 \log_{10} \left( r_2 \right)$ Since $\log_{10} \left( \frac{I_1}{I_0} \right) = \frac{70}{10} = 7$: $7 = -2 \log_{10} \left( r_2 \right)$ $\log_{10} \left( r_2 \right) = \frac{-7}{2} = -3.5$

Thus: $r_2 = 10^{-3.5} \approx 0.000316 \, \text{km} = 316 \, \text{m}$

So, you must be approximately 316 meters away from the single trombone to barely hear it.

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Would you like more detailed explanations, or do you have any other questions?

Related Questions:

1. How does the intensity of sound change when the number of sound sources doubles?
2. What is the formula to calculate the sound level in dB from sound intensity?
3. How does distance affect the intensity of sound from a point source?
4. What factors other than distance can influence sound intensity levels in open spaces?
5. How do different frequencies (like 523 Hz for this C note) affect sound propagation?

Tip: Sound intensity diminishes following an inverse square law, meaning doubling the distance from a sound source reduces the intensity to a quarter!