Let's solve the given problem using the formula provided:

$L_2 = L_1 + 20 \log \left( \frac{r_1}{r_2} \right)$

where:

• $L_1$ is the initial loudness at distance $r_1$.
• $L_2$ is the loudness at a different distance $r_2$.
• $r_1$ is the initial distance.
• $r_2$ is the new distance.

Given Data:

• $r_1 = 10$ meters
• $L_1 = 84$ dB
• $r_2 = 30$ meters

Part (a): Find $L_2$ at 30 meters.

$L_2 = 84 + 20 \log \left( \frac{10}{30} \right)$ $L_2 = 84 + 20 \log \left( \frac{1}{3} \right)$ $L_2 = 84 + 20 \times (-0.4771)$ $L_2 = 84 - 9.542$ $L_2 \approx 74.46 \, \text{dB}$

So, the sound level at 30 meters from the speaker is approximately $74.46 \, \text{dB}$.

Part (b): Find the distance $r_2$ for the volume to be halved.

A halving of the volume in decibels means a reduction of 3 dB, so $L_2 = 84 - 3 = 81$ dB.

$81 = 84 + 20 \log \left( \frac{10}{r_2} \right)$ $81 - 84 = 20 \log \left( \frac{10}{r_2} \right)$ $-3 = 20 \log \left( \frac{10}{r_2} \right)$ $\log \left( \frac{10}{r_2} \right) = \frac{-3}{20}$ $\log \left( \frac{10}{r_2} \right) = -0.15$ $\frac{10}{r_2} = 10^{-0.15}$ $\frac{10}{r_2} \approx 0.7079$ $r_2 \approx \frac{10}{0.7079} \approx 14.12 \, \text{meters}$

So, you have to move approximately 14.12 meters away from the speaker for the volume to be halved.

Do you want more details or have any other questions?

Related Questions:

1. What would be the sound level at 50 meters from the speaker?
2. If the initial loudness was 90 dB at 5 meters, what would be the loudness at 20 meters?
3. How does the sound level change if you move from 10 meters to 5 meters?
4. What distance is required to reduce the loudness by 10 dB?
5. How does the formula change if we consider sound intensity instead of loudness?

Tip: Remember, sound intensity decreases with the square of the distance, while sound level in dB decreases logarithmically.