Problem 19 - University Physics with Modern Physics

Question

For a person with normal hearing, the faintest sound that can be heard at a frequency of 400 \(\mathrm{Hz}\) has a pressure amplitude of about \(6.0 \times 10^{-5} \mathrm{Pa}\) . Calculate the (a) intensity; (b) sound intensity level; (c) displacement amplitude of this sound wave at \(20^{\circ} \mathrm{C}\) .

Step-by-Step Solution

Answer

(a) Intensity is \(2.94 \times 10^{-12} \, \text{W/m}^2\); (b) Sound intensity level is \(4.68\, \text{dB}\); (c) Displacement amplitude is \(4.01 \times 10^{-11} \, \text{m}\).

Step 1: Calculate the Intensity

The intensity of a sound wave can be calculated using the formula:\[ I = \frac{(p_0)^2}{2\rho v} \]where \( p_0 = 6.0 \times 10^{-5} \, \text{Pa} \) is the pressure amplitude, \( \rho \) is the density of air, and \( v \) is the speed of sound in air.At \( 20^\circ \text{C} \), \( \rho = 1.21 \, \text{kg/m}^3 \) and \( v = 343 \, \text{m/s} \). Substituting these values, we have:\[ I = \frac{(6.0 \times 10^{-5})^2}{2 \times 1.21 \times 343} \approx 2.94 \times 10^{-12} \, \text{W/m}^2 \]

Step 2: Calculate the Sound Intensity Level

The sound intensity level \( L \) in decibels (dB) 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 level.Using the intensity calculated in Step 1, substitute into the formula:\[ L = 10 \log_{10}\left(\frac{2.94 \times 10^{-12}}{1.0 \times 10^{-12}}\right) \approx 4.68 \, \text{dB} \]

Step 3: Calculate the Displacement Amplitude

The displacement amplitude \( s_0 \) can be calculated using:\[ s_0 = \frac{p_0}{\omega \cdot \rho \cdot v} \]where \( \omega = 2\pi f \) (angular frequency) and \( f = 400 \, \text{Hz} \).First, calculate the angular frequency:\[ \omega = 2 \pi \times 400 = 800\pi \, \text{rad/s} \]Then, substitute into the formula for \( s_0 \):\[ s_0 = \frac{6.0 \times 10^{-5}}{800\pi \times 1.21 \times 343} \approx 4.01 \times 10^{-11} \, \text{m} \]

Key Concepts

Sound Intensity

Sound intensity refers to the power per unit area carried by a sound wave, and it's important for understanding how much energy is received by a surface from a sound source. Sound intensity is typically measured in watts per square meter (W/m\(^2\)). Unlike loudness, which is how we perceive sound, intensity is an objective measurement. The formula to calculate sound intensity is:\[ I = \frac{(p_0)^2}{2\rho v} \]Here, \( p_0 \) is the pressure amplitude, \( \rho \) is the air density, and \( v \) is the speed of sound. At 20°C, air density is usually about 1.21 kg/m\(^3\), and the speed of sound is approximately 343 m/s.

• To calculate sound intensity, square the pressure amplitude.
• Divide by 2, multiply with air density and sound speed.
• This gives the acoustic power per unit area, essential in acoustics for studying sound distribution.

Sound Intensity Level

Sound intensity level is a logarithmic measure of the intensity of a sound relative to a reference intensity. It is expressed in decibels (dB). This measurement helps us understand how loud a sound is perceived in relation to a baseline, which is usually the threshold of hearing.The formula for sound intensity level \( L \) is:\[ L = 10 \log_{10}\left(\frac{I}{I_0}\right) \]where \( I \) is the sound intensity, and \( I_0 = 1.0 \times 10^{-12} \, \text{W/m}^2 \) is the reference intensity often used for the threshold of hearing in air.

• Decibel scale is logarithmic; small changes in intensity can mean large changes in dB.
• This scale helps visualize how much louder one sound is compared to another.
• Useful in various applications such as audio engineering and environmental noise monitoring.

Displacement Amplitude

Displacement amplitude in a sound wave measures the maximum distance that particles in the medium are displaced from their rest position due to a passing wave. It's a fundamental concept because it directly links to how energetic a sound wave is in terms of particle movement.The formula to calculate displacement amplitude \( s_0 \) is:\[ s_0 = \frac{p_0}{\omega \cdot \rho \cdot v} \]where \( p_0 \) is the pressure amplitude, \( \omega \) is the angular frequency calculated as \( 2\pi \times f \) with \( f \) being the frequency, \( \rho \) is the density of the medium, and \( v \) is the speed of sound.

• Angular frequency \( \omega \) connects the wave's frequency with a circular motion concept, making calculations easier.
• Displacement amplitude is directly influenced by both the frequency and medium properties.
• Knowing this value helps in understanding how intense a sound feels and its potential impact on surroundings.