Problem 17 - University Physics with Modern Physics

Question

A sound wave in air at \(20^{\circ} \mathrm{C}\) has a frequency of 150 \(\mathrm{Hz}\) and a displacement amplitude of \(5.00 \times 10^{-3} \mathrm{mm} .\) For this sound wave calculate the (a) pressure amplitude (in \(\mathrm{Pa} ) ;\) (b) intensity (in \(\mathrm{W} / \mathrm{m}^{2} ) ;(\mathrm{c})\) sound intensity level (in decibels).

Step-by-Step Solution

Answer

(a) 1.95 Pa; (b) 0.0046 W/m²; (c) 96.6 dB

Step 1: Calculate Speed of Sound

The speed of sound in air at a given temperature can be calculated using the formula:\[ v = 331.4 + 0.6 imes T \]where \( T \) is the temperature in degrees Celsius. For \( T = 20^{\circ} \),\[ v = 331.4 + 0.6 \times 20 = 343.4 \text{ m/s} \]

Step 2: Calculate Pressure Amplitude

Pressure amplitude \( P_0 \) can be calculated using the formula:\[ P_0 = B k A \]where:- \( A = 5.00 \times 10^{-3} \times 10^{-3} \text{ m} = 5.00 \times 10^{-6} \text{ m} \) (convert mm to m)- Bulk modulus \( B = 1.42 \times 10^5 \text{ Pa} \) for dry air at \( 20^{\circ}C \).Wave number \( k \) is:\[ k = \frac{2\pi f}{v} = \frac{2\pi \times 150}{343.4} \approx 2.74 \text{ rad/m} \]Calculate \( P_0 \):\[ P_0 = 1.42 \times 10^5 \times 2.74 \times 5.00 \times 10^{-6} \approx 1.95 \text{ Pa} \]

Step 3: Calculate Intensity

Intensity \( I \) is given by:\[ I = \frac{(P_0)^2}{2 \rho v} \]For air, assume \( \rho = 1.21 \text{ kg/m}^3 \) (density of air).Substitute the values:\[ I = \frac{(1.95)^2}{2 \times 1.21 \times 343.4} \approx 0.0046 \text{ W/m}^2 \]

Step 4: Calculate Sound Intensity Level

Sound intensity level \( L \) in decibels is given by:\[ L = 10 \log_{10} \left( \frac{I}{I_0} \right) \]where \( I_0 = 1.0 \times 10^{-12} \text{ W/m}^2 \) is the reference intensity. Substitute:\[ L = 10 \log_{10} \left( \frac{0.0046}{1.0 \times 10^{-12}} \right) \approx 96.6 \text{ dB} \]

Key Concepts

Pressure Amplitude

Pressure amplitude is a crucial concept when discussing sound waves. It refers to the maximum change in pressure from its equilibrium value, caused by a sound wave through a medium.

• The formula to find pressure amplitude is: \[ P_0 = B k A \] where:
  • \( B \) is the bulk modulus of the air, representing how the medium resists compression.
  • \( k \) represents the wave number, calculated as \( k = \frac{2\pi f}{v} \), with \( f \) being the frequency and \( v \) the speed of sound.
  • \( A \) stands for displacement amplitude, essentially the degree of motion within the medium.

Pressure amplitude is directly influenced by the wave properties (frequency and amplitude) and the medium's resistance (bulk modulus). In practice, calculating pressure amplitude helps in understanding the perceived loudness and energy of the sound.

Intensity of Sound

The intensity of sound is an important measure of the energy carried by a sound wave through a unit area per unit time. It's often referred to as the "loudness" of the sound, but technically offers a more precise measurement.

• Intensity \( I \) can be calculated using: \[ I = \frac{(P_0)^2}{2 \rho v} \] where:
  • \( P_0 \) is the pressure amplitude.
  • \( \rho \) is the density of the air (typical value is 1.21 \( \text{ kg/m}^3 \) for dry air).
  • \( v \) is the speed of sound in the medium. For air at 20°C, it's about 343.4 \( \text{ m/s} \).

Sound intensity is usually expressed in \( \text{ W/m}^2 \), which provides a good representation of how much power a sound wave emits in a particular area. It helps in distinguishing between sounds of different volumes and is vital in applications like soundproofing or speaker design.

Sound Intensity Level

Sound intensity level offers a way to quantify sound intensity using a logarithmic scale. This is expressed in decibels (dB), which gives a more intuitive understanding of sound volume differences.

• The formula used is: \[ L = 10 \log_{10} \left( \frac{I}{I_0} \right) \] where:
  • \( I \) is the sound intensity.
  • \( I_0 \) is a reference intensity, typically \( 1.0 \times 10^{-12} \text{ W/m}^2 \), which represents the threshold of human hearing.

Using sound intensity level, we can compare different sound sources or environments effectively. For instance, an increase of 10 dB represents a sound that is 10 times more intense. This logarithmic nature helps in managing large ranges of sound intensities in a practical and meaningful way.