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Chapter 23: Problem 6

What is the speed of compression waves (sound waves) in water? The bulk modulus for water is \(2.2 \times 10^{9} \mathrm{~N} / \mathrm{m}^{2}\). $$ v=\sqrt{\frac{\text { Bulk modulus }}{\text { Density }}}=\sqrt{\frac{2.2 \times 10^{9} \mathrm{~N} / \mathrm{m}^{2}}{1000 \mathrm{~kg} / \mathrm{m}^{3}}}=1.5 \mathrm{~km} / \mathrm{s} $$

Short Answer

Expert verified
The speed of sound in water is approximately 1.5 km/s.

Step by step solution

01

Understand the Formula

The speed of sound waves in a medium can be calculated using the formula \( v = \sqrt{\frac{B}{\rho}} \), where \( v \) is the speed of sound, \( B \) is the bulk modulus of the medium, and \( \rho \) is the density of the medium.
02

Identify the Given Values

From the problem, we know the bulk modulus \( B \) for water is given as \( 2.2 \times 10^{9} \mathrm{~N} / \mathrm{m}^{2} \) and the density \( \rho \) is \( 1000 \mathrm{~kg} / \mathrm{m}^{3} \).
03

Plug Values into the Formula

Substitute the given values into the formula: \( v = \sqrt{\frac{2.2 \times 10^{9} \mathrm{~N} / \mathrm{m}^{2}}{1000 \mathrm{~kg} / \mathrm{m}^{3}}} \).
04

Simplify the Calculation

Simplify the fraction inside the square root: \( \frac{2.2 \times 10^{9}}{1000} = 2.2 \times 10^{6} \). Thus, \( v = \sqrt{2.2 \times 10^{6}} \).
05

Calculate the Speed

Calculate the square root to find the speed: \( \sqrt{2.2 \times 10^{6}} \approx 1483 \mathrm{~m/s} \), which can be approximated to \( 1.5 \mathrm{~km/s} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Bulk Modulus
The bulk modulus is a key concept in understanding how sound travels through different materials. It measures how resistant a material is to uniform compression, or how much it can "handle" pressure without changing its volume.
The formula for bulk modulus, denoted as \( B \), is \( B = -V \frac{\Delta P}{\Delta V} \) where \( V \) is the volume, \( \Delta P \) is the change in pressure, and \( \Delta V \) is the change in volume as a result of that pressure.
  • Units: Bulk modulus is expressed in pascals \( (\text{Pa}) \), which is equivalent to \( \text{N}/\text{m}^2 \).
  • Physical Insight: A higher bulk modulus indicates that the material is difficult to compress; it responds less to pressure changes, which is a crucial factor in wave propagation.
In water, the bulk modulus is \( 2.2 \times 10^9 \text{ N/m}^2 \). This significant value means water is relatively incompressible, facilitating the efficient transmission of sound waves.
Density of Water
Density is another fundamental concept when considering the behavior of sound waves in water. It refers to how much mass is contained within a certain volume of a substance. For water, the density is typically \( 1000 \text{ kg/m}^3 \) at standard temperature and pressure.
  • Role in Sound Transmission: Density directly affects the speed of sound in a medium because sound waves move through mediums by vibrating particles; the more closely packed these particles are, the faster sound can travel.
  • Impact on Calculation: Having a consistent density allows for predicting how sound waves behave in water, which is essential when calculating the wave speed using the formula \( v = \sqrt{\frac{B}{\rho}} \).
This stable density enables us to accurately use predefined formulas to determine the speed of sound in water, which is crucial for various applications, including marine navigation and sonar.
Physics Problem Solving
Approaching physics problems systematically makes the process less intimidating and more manageable. Understanding each component of a problem is vital. Here are some tips:
  • Identify Given Variables: Write down what you know and what needs to be calculated. Look for units and convert them if necessary to ensure consistency across the calculation.
  • Use Relevant Formulas: Every aspect of physics has specific formulas designed to interrelate variables. Here, we're focusing on the formula for wave speed in a medium, \( v = \sqrt{\frac{B}{\rho}} \), where solving involves algebraic manipulation and logical thinking.
  • Step-by-step Calculation: Plug known values into the formula and simplify step-by-step, as demonstrated in the detailed solution.
Approaching problems methodically enhances your understanding and ability to tackle similar equations independently.
Wave Speed Calculation
Calculating the wave speed, particularly the speed of sound in water, involves integrating our understanding of bulk modulus and density. The formula \( v = \sqrt{\frac{B}{\rho}} \) captures how these properties influence wave propagation speed.
  • Insert Known Values: The bulk modulus \( B \) and density \( \rho \) of water have predefined values: \( B = 2.2 \times 10^9 \text{ N/m}^2 \) and \( \rho = 1000 \text{ kg/m}^3 \).
  • Perform the Calculation: Substituting, we find \( v = \sqrt{\frac{2.2 \times 10^9}{1000}} \). First, perform the division inside the square root, giving \( 2.2 \times 10^6 \). Then, the square root of this value gives us approximately \( 1483 \mathrm{~m/s} \), or converted, \( 1.5 \mathrm{~km/s} \).
This calculation is crucial in fields like acoustics and oceanography, where understanding wave dynamics enhances technological and scientific applications.

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Most popular questions from this chapter

Determine the speed of sound in carbon dioxide \((M=44 \mathrm{~kg} / \mathrm{kmo}\) \(y=1.30\) ) at a pressure of \(0.50\) atm and a temperature of \(400^{\circ} \mathrm{C}\).

A noise-level meter reads the sound level in a room to be \(85.0 \mathrm{~dB}\). What is the sound intensity in the room? Sound level \((\beta)\), in \(\mathrm{dB}\), is given by \(\beta=10 \log _{10}\left(I / I_{0}\right)\) and here it equals \(85.0 \mathrm{~dB}\). Accordingly, $$ \begin{array}{l} \text { and }\\\ \begin{aligned} \beta &=10 \log _{10}\left(\frac{1}{1.00 \times 10^{-12} \mathrm{w} / \mathrm{m}^{2}}\right)=85.0 \mathrm{~dB} \\ \log _{10}\left(\frac{l}{1.00 \times 10^{-12} \mathrm{~W} / \mathrm{m}^{2}}\right) &=\frac{85.0}{10}=8.50 \\ \frac{I}{1.00 \times 10^{-12} \mathrm{~W} / \mathrm{m}^{2}} &=\text { antilog }_{10} 8.50=3.16 \times 10^{\mathrm{k}} \\ I &=\left(1.00 \times 10^{-12} \mathrm{~W} / \mathrm{m}^{2}\right)\left(3.16 \times 10^{8}\right)=3.16 \times 10^{-4} \mathrm{~W} / \mathrm{m}^{2} \end{aligned} \end{array} $$

Back in the days before computers, a single typist typing furiously could generate an average sound level nearby of \(60.0 \mathrm{~dB}\). What would be the decibel level in the vicinity if three equally noisy typists were working close to one another? If each typist emits the same amount of sound energy, then the final sound intensity \(I_{f}\) should be three times the initial intensity \(I_{i}\) We have $$ \begin{array}{l} \beta_{f}=10 \log _{10}\left(\frac{I_{f}}{I_{0}}\right)=10 \log _{10} I_{f}-10 \log _{10} I_{0} \\ \text { and } \beta_{i}=10 \log _{10} I_{i}-10 \log _{10} I_{0} \end{array} $$ Subtracting these yields the change in sound level in going from \(I_{i}\) $$ \begin{array}{l} \text { to } I_{f}=3 I_{i}, \\ \qquad \begin{aligned} \beta_{f}-\beta_{i}=10 \log _{10} I_{f}-10 \log _{10} I_{i} \\ \quad \beta_{f}=\beta_{i}+10 \log _{10}\left(\frac{I_{I}}{I_{i}}\right)=60.0 \mathrm{~dB}+10 \log _{10} 3=64.8 \mathrm{~dB} \end{aligned} \end{array} $$ The sound level, being a logarithmic measure, rises very slowly with the number of sources.

At S.T.P., the speed of sound in air is \(331 \mathrm{~m} / \mathrm{s}\). Determine the speed of sound in hydrogen at S.T.P. if the specific gravity of hydrogen relative to air is \(0.0690\) and if \(y=1.40\) for both gases.

One sound has an intensity level of \(75.0 \mathrm{~dB}\), while a second has an intensity level of \(72.0 \mathrm{~dB}\). What is the intensity level when the two sounds are combined?
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