Problem 15 - University Physics with Modern Physics

Question

Longitudinal Waves in Different Fluids. (a) A longitudinal wave propagating in a water-filled pipe has intensity \(3.00 \times 10^{-6} \mathrm{W} / \mathrm{m}^{2}\) and frequency 3400 \(\mathrm{Hz}\) . Find the amplitude \(A\) and wavelength \(\lambda\) of the wave. Water has density 1000 \(\mathrm{kg} / \mathrm{m}^{3}\) and bulk modulus \(2.18 \times 10^{9} \mathrm{Pa}\) . (b) If the pipe is filled with air at pressure \(1.00 \times 10^{5} \mathrm{Pa}\) and density \(1.20 \mathrm{kg} / \mathrm{m}^{3},\) what will be the amplitude \(A\) and wavelength \(\lambda\) of a longitudinal wave with the same intensity and frequency as in part (a)? (c) In which fluid is the amplitude larger, water or air? What is the ratio of the two amplitudes? Why is this ratio so different from 1.00\(?\)

Step-by-Step Solution

Answer

The amplitude in air is greater; the ratio is 2330.6. This is due to air's lower density compared to water.

Step 1: Calculate the Speed of Sound in Water

The speed of sound in a medium can be calculated using the formula \( v = \sqrt{\frac{B}{\rho}} \) where \( B \) is the bulk modulus and \( \rho \) is the density. For water, \( B = 2.18 \times 10^{9} \ \text{Pa} \) and \( \rho = 1000 \ \text{kg/m}^3 \). Thus, the speed of sound in water is given by:\[ v = \sqrt{\frac{2.18 \times 10^{9}}{1000}} = 1478 \ \text{m/s} \].

Step 2: Determine the Wavelength in Water

Knowing the speed of sound \( v \) and the frequency \( f \), the wavelength \( \lambda \) is found using the relationship \( \lambda = \frac{v}{f} \). For the given frequency, \( f = 3400 \ \text{Hz} \), the wavelength in water is:\[ \lambda = \frac{1478}{3400} = 0.4347 \ \text{m} \].

Step 3: Calculate the Amplitude in Water

The intensity \( I \) of a wave is given by \( I = \frac{1}{2} \rho v \omega^2 A^2 \) where \( \omega = 2\pi f \). We rearrange to solve for \( A \):\[ A = \sqrt{\frac{2I}{\rho v \omega^2}} \].Substituting \( I = 3.00 \times 10^{-6} \ \text{W/m}^2 \), \( \rho = 1000 \ \text{kg/m}^3 \), \( v = 1478 \ \text{m/s} \), and \( \omega = 2\pi \times 3400 \):\[ A = \sqrt{\frac{2 \times 3.00 \times 10^{-6}}{1000 \times 1478 \times (2\pi \times 3400)^2}} = 1.24 \times 10^{-7} \ \text{m} \].

Step 4: Calculate the Speed of Sound in Air

For air, use the formula \( v = \sqrt{\frac{B}{\rho}} \) where the effective bulk modulus \( B \) is approximated by the air pressure: \( B = 1.00 \times 10^{5} \ \text{Pa} \) and \( \rho = 1.20 \ \text{kg/m}^3 \). Calculate the speed of sound:\[ v = \sqrt{\frac{1.00 \times 10^{5}}{1.20}} = 290 \ \text{m/s} \].

Step 5: Determine the Wavelength in Air

Using the same relationship as with water, \( \lambda = \frac{v}{f} \) with the speed of sound in air, \( v = 290 \ \text{m/s} \), we find:\[ \lambda = \frac{290}{3400} = 0.0853 \ \text{m} \].

Step 6: Calculate the Amplitude in Air

Use the same formula for amplitude as in Step 3 but replace water parameters with air parameters: \( \rho = 1.20 \ \text{kg/m}^3 \), and \( v = 290 \ \text{m/s} \):\[ A = \sqrt{\frac{2 \times 3.00 \times 10^{-6}}{1.20 \times 290 \times (2\pi \times 3400)^2}} = 2.89 \times 10^{-4} \ \text{m} \].

Step 7: Compare Amplitudes and Calculate Ratio

The amplitude in air is \( 2.89 \times 10^{-4} \ \text{m} \), while in water it is \( 1.24 \times 10^{-7} \ \text{m} \). The ratio of amplitudes \( \frac{A_{\text{air}}}{A_{\text{water}}} \) is:\[ \frac{2.89 \times 10^{-4}}{1.24 \times 10^{-7}} = 2330.6 \].The ratio is much greater than 1 because air is less dense than water, making it easier for the same energy to create larger displacements.

Key Concepts

Wave Intensity

Wave intensity is a measure of the energy carried by a wave per unit area per unit time. It is expressed as a power per area, usually in watts per square meter (W/m²). Intensity helps us understand how much energy is passing through a particular area in a certain amount of time, which is crucial when looking at how waves propagate through different media.

In many physical situations, intensity is used to quantify how strong a wave is. For longitudinal waves, such as sound waves moving through fluids, the intensity can be calculated using the relation:

• \( I = \frac{1}{2} \rho v \omega^2 A^2 \)

where \( \rho \) is the density of the medium, \( v \) is the wave speed, \( \omega \) is the angular frequency, and \( A \) is the amplitude of the wave.

In the context of the problem, the intensity remains constant whether the wave travels through water or air, helping us explore how the wave behaves in each medium by keeping one parameter the same.

Wave Amplitude

The amplitude of a wave is a measure of how large the oscillations of the wave are, which translates to the maximum displacement of particles from their mean position. It is these oscillations that carry energy through the medium. In essence, the larger the amplitude, the more energy the wave is transferring.

For longitudinal waves, amplitude is particularly important because it directly relates to the perceived loudness in sound waves. When we calculated the amplitude of the wave in both water and air, we saw that the amplitude was vastly different due to the differing densities of the mediums.

• In water, the calculated amplitude was very small (\(1.24 \times 10^{-7} \ \text{m}\)).
• In air, the amplitude was significantly larger (\(2.89 \times 10^{-4} \ \text{m}\)).

This is because less dense air allows greater displacement for the same energy input, resulting in a much higher wave amplitude. This property of amplitude is why a sound wave tends to be louder in air compared to denser mediums.

Wave Speed

Wave speed is the rate at which a wave travels through a medium. It's determined by the properties of the medium and can be calculated by the formula:

• \( v = \sqrt{\frac{B}{\rho}} \)

where \( B \) is the bulk modulus of the medium, indicating its compressibility, and \( \rho \) is the density.

For longitudinal waves in fluids, wave speed plays a crucial role as it determines how quickly the wave can propagate through the medium. In our exercise, the speed of sound was higher in water (1478 m/s) compared to air (290 m/s). This is primarily due to water's higher density and bulk modulus compared to air.

Understanding wave speed is essential in predicting how waves behave as they move from one medium to another, affecting other properties like wavelength. Wave speed also influences wave interference and refraction, which are vital in many applications, from sonar technology to musical acoustics.

Fluid Dynamics

Fluid dynamics is the study of how fluids (liquids and gases) move and the forces acting on them. It comprises understanding the properties of waves as they traverse different fluids, giving insights into phenomena such as wave propagation, reflection, and refraction.

In relation to longitudinal waves, fluid dynamics help explore how changes in the medium affect wave characteristics like speed, amplitude, and intensity. Different properties such as density and pressure directly influence how waves behave in a fluid.

• Higher density fluids typically result in smaller amplitude waves due to increased resistance.
• The speed of sound changes drastically with variations in fluid's density and compressibility, like air compared to water.

By applying principles of fluid dynamics, we understand why a wave moves faster in water than in air and why the amplitude of that wave is larger in air, given the same frequency and intensity. This explains how energy transfer varies across mediums, influencing how sound waves are experienced depending on the environment they travel through.