Problem 69 - Physics

Question

Suppose that when a certain sound intensity level (in dB) triples, the sound intensity (in \(\mathrm{W} / \mathrm{m}^{2}\) ) also triples. Determine this sound intensity level.

Step-by-Step Solution

Answer

The sound intensity level is approximately 2.39 dB.

Step 1: Understand Sound Intensity Level in dB

The sound intensity level in decibels (dB) is given by the formula: \( L = 10 \log_{10}\left(\frac{I}{I_0}\right) \), where \( L \) is the sound level in dB, \( I \) is the sound intensity in watts per square meter (\( \,\mathrm{W/m^2} \, \)), and \( I_0 \) is the reference intensity, typically \( 10^{-12} \, \mathrm{W/m^2} \).

Step 2: Define the Problem

Let's assume the initial sound intensity level is \( L \) dB with sound intensity \( I \, \mathrm{W/m^2} \). We know that tripling the sound intensity level triples the sound intensity. We need to find \( L \) using the defined relationship.

Step 3: Relate Intensity and dB Level Change

When sound intensity triples from \( I \) to \( 3I \), the new sound intensity level \( L' \) is given by: \( L' = 10 \log_{10}\left(\frac{3I}{I_0}\right) \).

Step 4: Express Sound Intensity Level Change

The change in sound intensity level due to tripling intensity is: \( L' - L = 10 \log_{10}(3) \). Given that this change is also equal to tripling the initial sound intensity level, we can establish that: \( 3L = L + 10 \log_{10}(3) \).

Step 5: Solve for Initial Sound Intensity Level (L)

Rearranging the previous equation: \( 2L = 10 \log_{10}(3) \). Calculate \( \log_{10}(3) \approx 0.4771 \), then \( 2L = 10 \times 0.4771 \) which gives \( 2L = 4.771 \). Solving for \( L \) gives: \( L = \frac{4.771}{2} = 2.3855 \) dB.

Key Concepts

Decibels

The term decibels (dB) is often encountered in the world of sound, measuring how intense or powerful a sound is. Decibels offer a convenient way to express sound intensity levels, especially since they compress a wide range of sound intensities into a manageable scale.
This makes it easier to compare different sound levels without dealing with large or cumbersome numbers. One key aspect of decibels is that they represent a relative measure; that is, decibels compare a given sound level to a standard reference point, instead of stating an absolute value.
This concept allows us to easily express large ratios of sound intensities using simple numbers.

Logarithmic Scale

A logarithmic scale is essential when dealing with measurements like sound intensity because it turns multiplicative relationships into additive ones. This means that if the sound intensity is increased by a factor, the intensity level measured in decibels will only increase by a relatively small number.
For example, doubling or tripling a sound intensity does not double or triple the decibel measure. Instead, it adds to it by a fixed number of decibels. In this way, a logarithmic scale allows us to express sound intensities ranging from very soft whispers to loud noises on the same scale, without large, unwieldy numbers.
In sound measurement, a tenfold increase in intensity equals a 10 dB increase in sound intensity level, due to the properties of the logarithmic scale used for decibels.

Reference Intensity

Reference intensity is a foundational concept when calculating sound intensity levels, particularly in decibels. It is the baseline against which we measure other sound intensities. The standard reference intensity is typically set at \(10^{-12} \, \mathrm{W/m^2}\), which is widely considered the threshold of human hearing.
This is the lowest sound intensity that the average human ear can perceive. Using this reference intensity ensures consistency and allows for universally comparable sound intensity measurements, since it effectively standardizes the way sound levels are calculated.
It provides the baseline denominator in the decibel formula: \( L = 10 \log_{10}\left(\frac{I}{I_0}\right) \). By using a uniform reference intensity, sound measurements become meaningful and useful across different contexts and scales.