91Ó°ÊÓ

Tap a metal spoon inside a mug of water and note the frequency \(f_{i}\) you hear. Then add a spoonful of powder (say, chocolate mix or instant coffee) and tap again as you stir the powder. The frequency you hear has a lower value \(f_{s}\) because the tiny air bubbles released by the powder change the water's bulk modulus. As the bubbles reach the water surface and disappear, the frequency gradually shifts back to its initial value. During the effect, the bubbles don't appreciably change the water's density or volume or the sound's wavelength. Rather, they change the value of \(d V / d p-\) that is, the differential change in volume due to the differential change in the pressure caused by the sound wave in the water. If \(f_{s} / f_{i}=0.333\), what is the ratio \((d V / d p)_{s} /(d V / d p)_{i} ?\)

Step by step solution

01

Understand the Bulk Modulus

The bulk modulus of a fluid, represented by \( K \), is related to the compressibility, which is \( \frac{dV}{dP} \), the fractional volume change per unit pressure change. The presence of air bubbles affects this compressibility. Hence, \( (dV/dP) \) changes with bubbles.

02

Relate Frequency to Bulk Modulus

The speed of sound \( v \) in a medium is expressed as \( v = \sqrt{\frac{K}{\rho}} \), where \( \rho \) is the density of the medium. Since \( \rho \) and wavelength \( \lambda \) remain constant, the relationship between frequency \( f \) and bulk modulus is \( f \propto \sqrt{K} \).

03

Express Ratios of Frequencies and Bulk Moduli

Given that \( \frac{f_s}{f_i} = 0.333 \), and knowing \( f \propto \sqrt{K} \), we can relate this to the bulk modulus as \( \sqrt{\frac{K_s}{K_i}} = 0.333 \).

04

Solve for Bulk Modulus Ratio

Since \( K = \frac{1}{dV/dP} \), the ratio of derivatives \( \frac{(dV/dP)_s}{(dV/dP)_i} \) is the inverse of the squared frequency ratio due to \( f \propto \sqrt{K} \). Thus, \( \frac{1}{\frac{K_s}{K_i}} = (\frac{dV/dP)_s}{(dV/dP)_i} = \left(\frac{f_i}{f_s}\right)^2 \).

05

Calculate the Final Result

Substitute \( \frac{f_i}{f_s} = \frac{1}{0.333} \approx 3 \) into the equation. Therefore, \( \left(\frac{f_i}{f_s}\right)^2 \approx 3^2 = 9 \).

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Key Concepts

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

Understanding Bulk Modulus

The bulk modulus is a measure of a fluid's resistance to uniform compression. It's mathematically expressed as the ratio of pressure change to the relative change in volume:\[ K = -V \left( \frac{dP}{dV} \right) \]Here, a high bulk modulus indicates that a substance is incompressible, meaning it doesn't change volume easily under pressure.* Because sound waves involve pressure changes, the bulk modulus is a key determinant for sound propagation in fluids.* When air bubbles are introduced, as in our exercise, the bulk modulus decreases. This happens because the bubbles make the fluid more compressible.Comprehending this concept is essential as it closely ties to how sound travels through different mediums.

Compressibility of Fluids

Compressibility is the reciprocal of the bulk modulus and is concerned with how much a fluid can be compressed. It can be expressed as:\[ \text{Compressibility} = \frac{1}{K} = \frac{dV}{dP} \]For practical understanding:

• Higher compressibility means the fluid changes volume easily when pressure changes are applied.
• Air bubbles increase the fluid's compressibility significantly, affecting sound propagation within the medium.

When mixing substances, like in our spoon-tapping experiment, bubbles reduce the frequency of sound because they affect fluid compressibility, altering how vibrations travel through the liquid.

The Speed of Sound in Fluids

The speed of sound in a medium is defined by the formula:\[ v = \sqrt{\frac{K}{\rho}} \]where:* \( v \) is the speed of sound,* \( K \) is the bulk modulus,* \( \rho \) is the density of the medium.In the context of our exercise, the bubbles introduced by stirring a powdered substance into the water drastically affect the value of \( K \), thereby impacting the speed of sound. An increased compressibility (decreased \( K \)) results in a slower speed of sound, explaining the observed lowering of sound frequency.

Recognizing Density's Role

Density, denoted \( \rho \), is the mass per unit volume of a substance and is crucial in determining the speed of sound. However, in our experiment, the introduction of bubbles does not alter the water's density notably. Therefore, while \( \rho \) is a factor in the speed of sound equation, it's considered constant in this specific scenario.

• Sound travels at different speeds in mediums with different densities.
• For this exercise, focusing on bulk modulus and compressibility is more pertinent, since density remains unaffected and constant.

This emphasizes the concept that, although density is a fundamental factor in acoustics, it sometimes plays a lesser role in dynamic experiments like this one.

Frequency Ratio and Its Implications

The frequency ratio \( \left( \frac{f_s}{f_i} \right) \) is a critical concept in this experiment to understand how alterations in physical properties affect sound. The frequency \( f \) is directly proportional to the square root of the bulk modulus:\[ f \propto \sqrt{K} \]This means any change in \( K \) directly causes a change in the frequency of the sound. Our exercise shows:

• The frequency decreases as compressibility increases (due to bubbles).
• The ratio of initial to final frequency gives insight into how much the bulk modulus has changed.

For the presented problem, knowing \( \left( \frac{f_s}{f_i} = 0.333 \right) \) allows us to use the relationship between frequency and bulk modulus to solve for the ratio of compressibility before and after the change, confirming the impact bubbles have on sound frequency.