Problem 24 - University Physics with Modern Physics

Question

(a) If two sounds differ by 5.00 \(\mathrm{dB}\) , find the ratio of the intensity of the louder sound to that of the softer one. (b) If one sound is 100 times as intense as another, by how much do they differ in sound intensity level (in decibels)? (c) If you increase the volume of your stereo so that the intensity doubles, by how much does the sound intensity level increase?

Step-by-Step Solution

Answer

(a) Ratio is \(3.16\). (b) Difference is \(20\) dB. (c) Increase is \(3.01\) dB.

Step 1: Understand the decibel formula

The sound intensity level in decibels (dB) is given by the formula \( L = 10 \, \log_{10}\left(\frac{I}{I_0}\right) \), where \( I \) is the intensity of the sound and \( I_0 \) is the reference intensity.

Step 2: Find the intensity ratio for part (a)

Given two sounds differing by 5 dB, use the logarithmic relationship: \( \Delta L = 10 \, \log_{10}\left(\frac{I_2}{I_1}\right) \), where \( \Delta L = 5 \) dB. Solving for the intensity ratio yields: \[ \frac{I_2}{I_1} = 10^{\frac{5}{10}} = 10^{0.5} = \sqrt{10} \approx 3.16 \].

Step 3: Calculate the decibel difference for part (b)

If one sound intensity \( I_2 \) is 100 times another \( I_1 \), then \( \frac{I_2}{I_1} = 100 \). Use the formula \( \Delta L = 10 \, \log_{10}(\frac{I_2}{I_1}) \). This gives \[ \Delta L = 10 \times \log_{10}(100) = 10 \times 2 = 20 \, \text{dB} \].

Step 4: Find intensity level increase for part (c)

If the intensity doubles \( \frac{I_2}{I_1} = 2 \), use \( \Delta L = 10 \, \log_{10}(2) \). Thus \[ \Delta L = 10 \times \log_{10}(2) \approx 10 \times 0.301 = 3.01 \] dB.

Key Concepts

Intensity

Intensity describes the amount of energy a sound wave carries per unit area per unit time in a specified direction. It is typically measured in watts per square meter (W/m²). When dealing with sound, this concept is essential because it determines how strong or weak a sound is. Several factors affect intensity:

• The power of the sound source: More power usually means greater intensity.
• The distance from the sound: Intensity decreases as distance from the source increases due to the spreading out of sound waves.
• The medium through which sound travels: Different materials can absorb sound to varying degrees.

Understanding intensity is crucial in fields such as acoustics, where it helps engineers design effective soundproofing and audio equipment. The challenge often lies in measuring and comparing intensities, which is where mathematical tools like logarithms come into play.

Sound Intensity Level

The sound intensity level is a logarithmic measure of the intensity of sound relative to a fixed reference level, typically threshold of hearing, denoted as \(I_0\). It is expressed in decibels (dB) and is calculated using the equation: \[ L = 10 \, \log_{10}\left(\frac{I}{I_0}\right) \]where \(L\) is the sound intensity level in dB, \(I\) is the intensity of the sound, and \(I_0\) is the reference intensity of \(10^{-12} \text{ W/m}^2 \).This concept is particularly useful because:

• It allows for easier comparison between vastly different sound intensities due to its logarithmic nature.
• It reflects how humans perceive changes in sound intensity. A 10 dB increase typically sounds about twice as loud.
• It is widely used in environmental noise assessments, audio engineering, and more.

By converting physical sound intensity into a logarithmic scale, the sound intensity level provides a more manageable way to represent and work with the wide range of sounds we encounter daily.

Logarithm

A logarithm is an exponent or power to which a base must be raised to produce a given number. In mathematical terms, for numbers \(a\), \(b\), and \(c\), the equation \(a^b = c\) translates to \(\log_a c = b\). In sound calculations, the base 10 logarithm (common logarithm) is preferred, represented as \(\log_{10}\). Logarithms are particularly helpful for simplifying multiplication into addition and division into subtraction, making calculations more manageable.Here is why logarithms are vital in sound intensity calculations:

• They accommodate the vast range of human hearing, from the quietest sound we can detect to the loudest we can tolerate.
• They simplify the comparison of sound intensities, enabling us to express small changes in intensity as manageable dB values.
• The logarithmic nature corresponds with our perception of sound's non-linear character, meaning small intensity changes can lead to significant changes in perceived loudness.

By converting linear quantities like sound intensity into logarithmic scales, we align our calculations with human hearing more effectively.