Problem 23 - University Physics with Modern Physics

Question

CP At point \(A, 3.0\) m from a small source of sound that is emitting uniformly in all directions, the sound intensity level is 53 dB. (a) What is the intensity of the sound at \(A ?\) (b) How far from the source must you go so that the intensity is one-fourth of what it was at \(A ?\) (c) How far must you go so that the sound intensity level is one-fourth of what it was at \(A ?\) (d) Does intensity obey the inverse-square law? What about sound intensity level?

Step-by-Step Solution

Answer

(a) Intensity at A is \(2.0 \times 10^{-7} \text{ W/m}^2\). (b) 6.0 m. (c) 163.2 m. (d) Intensity obeys the inverse-square law; intensity level does not.

Step 1: Convert Sound Intensity Level to Intensity

The sound intensity level in decibels (dB) is given by: \[ \beta = 10 \log_{10} \left( \frac{I}{I_0} \right) \]where \( \beta = 53 \text{ dB} \) and \( I_0 = 1.0 \times 10^{-12} \text{ W/m}^2 \).We can rearrange this formula to find \( I \):\[ I = I_0 \times 10^{\frac{\beta}{10}} \]Substitute the known values:\[ I = 1.0 \times 10^{-12} \times 10^{\frac{53}{10}} = 2.0 \times 10^{-7} \text{ W/m}^2 \]

Step 2: Calculate Distance for Intensity to be One-Fourth

Intensity follows the inverse-square law: \[ I \propto \frac{1}{r^2} \]If the intensity is one-fourth of its initial value, \( I' = \frac{I}{4} \):\[ \frac{I'}{I} = \frac{r^2}{r_0^2} = \frac{1}{4} \]Given \( r_0 = 3.0 \, \text{m} \), solve for \( r \):\[ r = r_0 \times 2 = 6.0 \, \text{m} \]

Step 3: Calculate Distance for Intensity Level One-Fourth

Since intensity level \( \beta \) in dB scales logarithmically, dividing \( \beta \) by 4 corresponds to a quarter of the difference between \( \beta \) and 0:\[ \beta' = \frac{53}{4} = 13.25 \, \text{dB} \]Find the intensity corresponding to this new \( \beta' \):\[ I'' = I_0 \times 10^{\frac{13.25}{10}} = 1.86 \times 10^{-10} \, \text{W/m}^2 \]Use the inverse-square law: \[ \frac{I''}{I} = \frac{r^2}{r_0^2} \Rightarrow \frac{1.86 \times 10^{-10}}{2.0 \times 10^{-7}} = \frac{r^2}{3^2} \]\[ r = 3 \times \sqrt{\frac{2.0 \times 10^{-7}}{1.86 \times 10^{-10}}} \approx 163.2 \, \text{m} \]

Step 4: Discuss Intensity and Intensity Level

The sound intensity \( I \) obeys the inverse-square law, as shown in the calculations where doubling the distance decreased the intensity by a factor of four. However, the intensity level \( \beta \) does not follow the inverse-square law directly due to its logarithmic nature; it decreases logarithmically instead.

Key Concepts

Inverse-Square Law

The inverse-square law is a principle that describes how a physical quantity diminishes with distance. When applied to sound intensity, it states that the intensity of sound waves reduces as the square of the distance from the source. If you move twice as far from the source, the intensity drops to one-quarter; if you move three times farther, it drops to one-ninth.

This happens because the sound energy spreads out over the surface of a sphere as it moves away from the source. Thus, mathematically, we express this as \( I \propto \frac{1}{r^2} \), where \( I \) is intensity and \( r \) is the distance. This law is crucial in understanding how energy dissipates in a medium, affecting how we perceive sound over varying distances.

Decibels (dB)

The decibel is a unit used to quantify sound intensity levels. It's a logarithmic unit that expresses the ratio of two values, helpful in dealing with very large or very small numbers. The formula to calculate sound intensity level in decibels is \( \beta = 10 \log_{10} \left( \frac{I}{I_0} \right) \), where \( I \) is the intensity of sound, and \( I_0 \) is the reference intensity, typically \( 1.0 \times 10^{-12} \text{ W/m}^2 \).

This means a small change in decibels corresponds to a significant change in intensity. For example, an increase of 10 dB represents a tenfold increase in intensity. This scale is used because it aligns more closely with the human ear’s response to changes in intensity, which is also logarithmic in nature.

Logarithmic Scale

A logarithmic scale is used to represent data that spans multiple orders of magnitude. In the context of sound, it means that increments on the decibel scale are not linear but exponential. This scale is particularly useful in sound because sound intensity levels can vary widely, and using a logarithmic scale makes these levels more manageable.

For example, on a linear scale, moving from an intensity of \(1\) to \(100\) would involve increments of \(1\), totaling \(99\) steps. On a logarithmic scale, this same change might only involve a few steps because each step represents a multiplication rather than an addition. This scale helps in compressing a wide dynamic range into a more comprehensible format, making it easier to compare different sound levels.

Sound Propagation

Sound propagation refers to how sound waves travel through a medium. The medium can be air, water, or any material through which sound can move. When sound is emitted from a source, it propagates outward in all directions. This propagation is affected by factors such as the medium's density and elasticity.

As sound travels, its energy spreads out, causing a decrease in intensity, which is well described by the inverse-square law. Moreover, the frequency of the sound can affect how it propagates; high-frequency sounds tend to dissipate more quickly compared to low-frequency sounds. Understanding sound propagation is essential in fields such as acoustics and audio engineering, where the dispersion and absorption of sound waves play a critical role in design and analysis.