A certain sound level is increased by an additional 30 dB, Find the factor by which
(a) its intensity increases and
(b) its pressure amplitude increases.

Explanation

To solve the problem, we need to find the factors by which the intensity and pressure amplitude of a sound increase when the sound level is increased by 30 dB.

### Step-by-Step Solution:

**Step 1: Understand the relationship between sound level (in dB) and intensity.**
The relationship between the change in sound level (ΔL) in decibels and the intensities (I1 and I2) is given by the formula:
\[ \Delta L = 10 \log_{10} \left(\frac{I_2}{I_1}\right) \]
where:
- \(I_1\) is the initial intensity,
- \(I_2\) is the final intensity after the increase.

**Step 2: Set up the equation for the increase in sound level.**
Given that the increase in sound level is 30 dB, we can substitute this into the equation:
\[ 30 = 10 \log_{10} \left(\frac{I_2}{I_1}\right) \]

**Step 3: Simplify the equation.**
Dividing both sides by 10 gives:
\[ 3 = \log_{10} \left(\frac{I_2}{I_1}\right) \]

**Step 4: Exponentiate to eliminate the logarithm.**
To eliminate the logarithm, we exponentiate both sides:
\[ 10^3 = \frac{I_2}{I_1} \]
This simplifies to:
\[ \frac{I_2}{I_1} = 1000 \]

**Step 5: Determine the factor by which intensity increases.**
From the equation above, we find that:
\[ I_2 = 1000 I_1 \]
Thus, the intensity increases by a factor of 1000.

**Step 6: Relate intensity to pressure amplitude.**
The intensity (I) is related to the pressure amplitude (p) by the equation:
\[ I \propto p^2 \]
This means that:
\[ I = k p^2 \]
for some constant \(k\).

**Step 7: Set up the relationship for pressure amplitudes.**
Let \(p_1\) be the initial pressure amplitude and \(p_2\) be the final pressure amplitude. We can write:
\[ I_1 = k p_1^2 \quad \text{and} \quad I_2 = k p_2^2 \]

**Step 8: Form the ratio of the pressure amplitudes.**
Taking the ratio of the two intensities gives:
\[ \frac{I_2}{I_1} = \frac{k p_2^2}{k p_1^2} = \frac{p_2^2}{p_1^2} \]
Substituting the factor we found for intensity:
\[ 1000 = \frac{p_2^2}{p_1^2} \]

**Step 9: Solve for the ratio of pressure amplitudes.**
Taking the square root of both sides:
\[ \frac{p_2}{p_1} = \sqrt{1000} = 31.62 \]

**Step 10: Conclusion.**
Thus, the pressure amplitude increases by a factor of approximately 31.62.

### Final Answers:
(a) The intensity increases by a factor of **1000**.
(b) The pressure amplitude increases by a factor of approximately **31.62**.