Question

At a distance of \(20.0 \mathrm{~m}\) from a sound source, the intensity of the sound is \(60.0 \mathrm{~dB}\). What is the intensity (in \(\mathrm{dB}\) ) at a point \(2.00 \mathrm{~m}\) from the source? Assume that the sound radiates equally in all directions from the source.

Short answer

Answer: The intensity at a point 2.00 m from the source is 100 dB.

Step-by-step solution

Define the inverse square law for intensity with respect to distance

The inverse square law dictates that the ratio of the intensities is proportional to the inverse square of the ratio of the distances. We can define this mathematically as follows: \(I_1 / I_2 = (d_2 / d_1)^2\)

Convert decibels to intensity

Given the intensity in decibels, we need to convert it to Watts per square meter (W/m²). The formula for converting decibels to intensity is: \(I = 10^{L/10} \times 10^{-12} \mathrm{~W/m²}\) First, use the provided intensity level \(60.0 \mathrm{~dB}\) and convert it to intensity (W/m²): \(I_1 = 10^{60/10} \times 10^{-12} = 10^6 \times 10^{-12} = 10^{-6} \mathrm{~W/m²}\)

Use the inverse square law formula to find the new intensity

Now that we have the initial intensity \(I_1\) and the distances \(d_1 = 20.0 \mathrm{~m}\) and \(d_2 = 2.00 \mathrm{~m}\), we can plug in the values to the inverse square law formula: \(I_2 = I_1 \times (d_1 / d_2)^2 = 10^{-6} \mathrm{~W/m²} \times (20.0 \mathrm{~m} / 2.00 \mathrm{~m})^2\)

Solve for the new intensity

From the previous step, we get: \(I_2 = 10^{-6} \mathrm{~W/m²} \times (10)^2 = 10^{-6} \mathrm{~W/m²} \times 100 = 10^{-4} \mathrm{~W/m²}\)

Convert back to decibels

Now, we need to convert the new intensity (in W/m²) back to decibels. The formula for converting intensity to decibels is: \(L = 10 \log_{10}(I / 10^{-12})\) Plug in the calculated intensity and find the new intensity level: \(L_2 = 10 \log_{10}(10^{-4} \mathrm{~W/m²} / 10^{-12}) = 100 \mathrm{~dB}\) The intensity at a point 2.00 m from the source is \(100 \mathrm{~dB}\).

Key concepts

Understanding the Inverse Square Law

The inverse square law is a physical principle that describes how certain quantities decrease in strength as they spread out from a source. When it comes to sound intensity, this law tells us that as you move away from a sound source, the intensity of the sound diminishes rapidly. Specifically, the intensity is inversely proportional to the square of the distance from the source. To put it in simpler terms, if you double the distance from the source, the sound intensity becomes one-fourth of its original value.

This relationship is crucial for understanding how sound behaves in a three-dimensional space. In the context of the exercise, by using the inverse square law, we can calculate how much quieter a sound will be at a certain distance compared to another point. The formula expressed mathematically is \(I_1 / I_2 = (d_2 / d_1)^2\), where \(I_1\) and \(I_2\) are the sound intensities at two different distances, \(d_1\) and \(d_2\), from the sound source.

Intensity Conversion Between Decibels and Watts per Square Meter

Sound intensity can be measured in two different units: decibels (dB), which is a logarithmic unit, and watts per square meter (W/m²), a linear unit. Converting between these units is a common step in solving problems related to sound intensity.

To convert from decibels to watts per square meter, we use the formula \(I = 10^{L/10} \times 10^{-12} \mathrm{~W/m^2}\), where \(L\) is the sound intensity level in decibels. Conversely, to convert from watts per square meter to decibels, you would use \(L = 10 \log_{10}(I / 10^{-12})\). These formulas account for the fact that the decibel scale is logarithmic, meaning each step on the scale represents a factor of change, and the applicable baseline intensity for decibels in acoustics is typically \(10^{-12} \mathrm{~W/m^2}\), which corresponds to the threshold of human hearing.

Using these formulas, students can easily switch between a linear representation of sound intensity (which makes calculating changes over distance straightforward) and the more commonly used decibel scale.

The Logarithmic Scale of Decibels

The concept of a logarithmic scale is essential when discussing sound intensity levels measured in decibels. A logarithmic scale is used to represent a wide range of values in a more compact and manageable way. In the case of sound, the range of human hearing spans from the quietest sound at 0 dB (the threshold of hearing) to sounds that are damagingly loud, which can be over 120 dB. A logarithmic scale is helpful here because it allows us to handle these large variations with smaller, more intuitive numbers.

The decibel system is a relative scale that compares the measured intensity to a reference value. It's based on powers of 10, following the formula \(L = 10 \log_{10}(I / I_0)\), where \(L\) is the sound level in decibels, \(I\) is the sound intensity, and \(I_0\) is the reference intensity. Every 10 dB increase represents a tenfold increase in the sound intensity, and every 20 dB increase represents a hundredfold increase. This is why in our exercise, a relatively small change in distance results in a large change in decibel level—because the decibel scale compresses changes exponentially.

Appreciating the logarithmic nature of the decibel scale helps students understand why distances impact sound intensity levels so dramatically and why such calculations are fundamental in fields concerned with noise management and audio engineering.