Problem 59 - University Physics with Modern Physics

Question

A soprano und a bass are singing a duet. While the soprano sings an \(A\) -sharp at 932 \(\mathrm{Hz}\) , the bass sings an \(A\) -sharp but three octaves lower. In this concert hall, the density of air is 1.20 \(\mathrm{kg} / \mathrm{m}^{3}\) and its bulk modulus is \(1.42 \times 10^{5} \mathrm{Pa.}\) In order for their notes to have the same sound intensity level, what must be (a) the ratio of the pressure amplitude of the bass to that of the soprano and (b) the ratio of the displacement amplitude of the bass to that of the soprano? (c) What displacement amplitude (in \(m\) and in nm \()\) does the soprano produce to sing her A-sharp at 72.0 \(\mathrm{dB} ?\)

Step-by-Step Solution

Answer

(a) 1, (b) 8, (c) 1.58 x 10^{-8} m or 15.8 nm.

Step 1: Understand Frequency of Bass Note

The frequency of the bass note is three octaves lower than the soprano. Since an octave is a doubling of frequency, three octaves lower means dividing the soprano frequency by 2 three times: \( \frac{932}{2^3} = \frac{932}{8} = 116.5 \, \text{Hz} \).

Step 2: Calculate Sound Intensity Level

The sound intensity level in decibels (dB) is given by the formula \( L = 10 \log_{10} \left( \frac{I}{I_0} \right) \), where \( I_0 = 10^{-12} \, \text{W/m}^2 \) is the reference intensity. For a level of 72 dB, \( L = 72 = 10 \log_{10} \left( \frac{I}{10^{-12}} \right) \). Thus, \( I = 10^{-12} \times 10^{7.2} = 1.585 \times 10^{-5} \, \text{W/m}^2 \).

Step 3: Find Pressure Amplitude Ratio

Intensity \( I \) is related to pressure amplitude \( p_0 \) by the equation \( I = \frac{p_0^2}{2 \rho v} \). Since intensity is the same for both, equating \( \frac{p_{0,\text{soprano}}^2}{2 \rho v_{\text{soprano}}} = \frac{p_{0,\text{bass}}^2}{2 \rho v_{\text{bass}}} \), the pressure amplitude ratio \( \frac{p_{0,\text{bass}}}{p_{0,\text{soprano}}} = \sqrt{\frac{v_{\text{bass}}}{v_{\text{soprano}}}} \).

Step 4: Calculate Speed of Sound in Air

The speed of sound \( v \) is given by \( v = \sqrt{\frac{B}{\rho}} \) where \( B \) is the bulk modulus and \( \rho \) is the density. So, \( v = \sqrt{\frac{1.42 \times 10^5}{1.20}} = 345.6 \, \text{m/s} \). Hence \( v_{\text{bass}} = v_{\text{soprano}} = 345.6 \, \text{m/s} \) so the pressure amplitude ratio \( \frac{p_{0,\text{bass}}}{p_{0,\text{soprano}}} = 1 \).

Step 5: Find Displacement Amplitude Ratio

Displacement amplitude \( s_0 \) is related to pressure amplitude by \( p_0 = \rho v \omega s_0 \) and \( \omega = 2\pi f \). Thus \( \frac{s_{0,\text{bass}}}{s_{0,\text{soprano}}} = \frac{p_{0,\text{bass}} \cdot f_{\text{soprano}}}{p_{0,\text{soprano}} \cdot f_{\text{bass}}} \). With \( p_{0,\text{bass}} = p_{0,\text{soprano}} \), we have \( \frac{s_{0,\text{bass}}}{s_{0,\text{soprano}}} = \frac{932}{116.5} = 8 \).

Step 6: Calculate Soprano Displacement Amplitude

Using \( s_0 = \frac{p_0}{\rho v \omega} \) with intensity \( I = 1.585 \times 10^{-5} \, \text{W/m}^2 \) and \( f = 932 \, \text{Hz} \), \( p_0 = \sqrt{2 \rho v I} = \sqrt{2 \times 1.20 \times 345.6 \times 1.585 \times 10^{-5}} = 0.109 \, \text{Pa} \). Thus, \( s_{0,\text{soprano}} = \frac{0.109}{1.20 \times 345.6 \times 2\pi \times 932} = 1.58 \times 10^{-8} \, \text{m} \) or \( 15.8 \, \text{nm} \).

Key Concepts

Pressure Amplitude

Pressure amplitude in sound waves is the measure of the maximum pressure variation from the average atmospheric pressure caused by the sound wave. It is an indicator of how strong a sound is, often linked to the sound intensity which determines how loud we perceive the sound to be. In the context of sound waves:

• Pressure amplitude is denoted as \( p_0 \).
• It relates to sound intensity \( I \) by the formula \( I = \frac{p_0^2}{2 \rho v} \), where \( \rho \) is the density of the medium and \( v \) is the speed of sound in that medium.
• In the exercise, since the soprano and bass sing with the same sound intensity, their pressure amplitudes must have a specific ratio to maintain this balance.

Understanding and calculating the pressure amplitude is crucial because it helps in determining how different notes blend harmoniously in music or how different sound levels are measured in acoustics.

Displacement Amplitude

Displacement amplitude refers to the maximum displacement of particles in the medium as the sound wave passes through it. It tells us how far the particles move from their rest position when a sound wave travels through:

• Displacement amplitude is represented as \( s_0 \).
• There is a direct relationship between pressure amplitude and displacement amplitude, expressed as \( p_0 = \rho v \omega s_0 \), where \( \omega = 2\pi f \) is the angular frequency.
• In the solution, the displacement amplitude of the bass is compared to that of the soprano to determine how different frequencies affect particle movement in the air.

While pressure amplitude highlights the change in pressure, displacement amplitude focuses on the physical movement of particles. Together, they provide a comprehensive picture of how sound energy propagates through a medium.

Frequency of Sound

Frequency of sound is one of the fundamental properties of sound waves, indicating how many wave cycles pass a certain point per unit of time. It's directly related to the perceived pitch of the sound:

• Frequency is denoted in Hertz (Hz).
• A high frequency results in a high-pitched sound, whereas a low frequency corresponds to a low-pitched sound.
• In this exercise, the soprano sings at 932 Hz, and the bass sings three octaves lower, which geometrically reduces the frequency to 116.5 Hz.

Frequency influences how we perceive music and plays a crucial role in various applications, from tuning instruments to analyzing acoustic signals. Knowing how frequency relates to other properties like pressure and displacement amplitude is vital in understanding and manipulating sound.