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Chapter 17: Problem 8

Hot chocolate effect. 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 \((d V / d p)_{s} /(d V / d p)_{i} ?\)

Short Answer

Expert verified
The ratio is approximately 9.02.

Step by step solution

01

Understanding the Given Information

We are given the initial frequency \( f_i \) when tapping the spoon in water. After adding powder, the frequency changes to \( f_s \). The ratio of these frequencies is provided as \( \frac{f_s}{f_i} = 0.333 \). We need to find the ratio of \( \left(\frac{dV}{dp}\right)_s \) to \( \left(\frac{dV}{dp}\right)_i \).
02

Frequency and Bulk Modulus Relationship

The speed of sound \( v \) in a medium is related to the bulk modulus \( B \) and density \( \rho \) by the formula: \( v = \sqrt{\frac{B}{\rho}} \). Since \( f = \frac{v}{\lambda} \) (where \( \lambda \) is wavelength), and wavelength remains constant, \( f \propto \sqrt{B} \). Thus, \( \frac{f_s}{f_i} = \sqrt{\frac{B_s}{B_i}} \).
03

Relating Bulk Modulus and Volume Change

The bulk modulus \( B \) is defined as \( B = -V \frac{dP}{dV} \). The volume change with pressure, \( \frac{dV}{dP} \), is inversely related to the bulk modulus: \( \frac{dV}{dP} = -\frac{1}{B} \).
04

Equation of Ratios

Since \( \frac{dV}{dP} \) is inversely related to \( B \), the ratios are related by: \( \frac{\left(\frac{dV}{dP}\right)_s}{\left(\frac{dV}{dP}\right)_i} = \frac{B_i}{B_s} \). From the frequency relationship, \( \frac{B_s}{B_i} = \left(\frac{f_s}{f_i}\right)^2 \).
05

Substitute Given Frequency Ratio

Substitute the frequency ratio: \( \frac{B_s}{B_i} = 0.333^2 = 0.110889 \). Therefore, \( \frac{B_i}{B_s} = \frac{1}{0.110889} \).
06

Final Calculation

Calculate \( \frac{B_i}{B_s} = \frac{1}{0.110889} \approx 9.0206 \). This value represents the ratio \( \frac{(dV/dP)_s}{(dV/dP)_i} \).

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Key Concepts

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

Bulk Modulus
Bulk modulus is a fundamental property of materials that describes how incompressible a substance is. It is defined mathematically as: \[ B = -V \frac{dP}{dV} \] where:
  • \( B \) is the bulk modulus,
  • \( V \) is the initial volume,
  • \( dP \) is the differential change in pressure,
  • \( dV \) is the differential change in volume.
Bulk modulus essentially quantifies the material's resistance to uniform compression.
The negative sign indicates that an increase in pressure results typically in a decrease in volume.
When air bubbles from the chocolate powder mix with the water, they reduce the effective bulk modulus of the liquid. Decreased bulk modulus results in lower sound speed through the medium, affecting the frequency you hear.
Speed of Sound
The speed of sound in a medium is a fascinating concept that depends on both the medium鈥檚 bulk modulus and its density. The relationship is given by:\[ v = \sqrt{\frac{B}{\rho}} \] where:
  • \( v \) is the speed of sound,
  • \( B \) is the bulk modulus,
  • \( \rho \) is the density of the medium.
In simpler terms, a larger bulk modulus indicates a stiffer medium where sound can travel faster, while a greater density slows it down.
In the hot chocolate effect, when the air bubbles lower the bulk modulus, the speed of sound in the water decreases. This change in sound speed reduces the frequency, which is why tapping the spoon creates a different tone after the powder mix is added.
Frequency and Wavelength
Frequency and wavelength are key characteristics of sound waves. The frequency determines the pitch of the sound, while the wavelength is the spatial length between two consecutive points of a wave in phase.
The formula relating frequency \( f \), speed of sound \( v \), and wavelength \( \lambda \) is given by:\[ f = \frac{v}{\lambda} \] In the hot chocolate effect, the addition of air bubbles affects the speed of sound without altering the wavelength. This is why you observe changes in frequency (or pitch). Since the wavelength remains constant, the frequency is directly affected by any variations in the speed of sound.
Thus, the changes in frequency, as heard through the mug of water, are directly tied to variations in the bulk modulus caused by the air bubbles.
Pressure and Volume Change
The relationship between pressure and volume change in a medium is an important aspect of understanding sound wave behavior. This relationship is captured by the bulk modulus, and changes in this property indicate how a medium reacts to pressure changes.
Visualize that when the pressure exerted on a medium increases, its volume generally decreases, provided the material is compressible.
In the context of the hot chocolate effect, when air bubbles are introduced, they alter this natural resistance to compression. This leads to a change in water's ability to compress under pressure from sound waves. \[ \frac{dV}{dP} \] This expression describes how much the volume \( V \) changes for a given change in pressure \( P \). When the bulk modulus decreases due to added bubbles, the medium becomes easier to compress, affecting the propagation of sound waves through the liquid.
Understanding this dynamic is essential for comprehending why the frequency changes occur during the hot chocolate effect.

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Most popular questions from this chapter

A state trooper chases a speeder along a straight road; both vehicles move at \(160 \mathrm{~km} / \mathrm{h}\). The siren on the trooper's vehicle produces sound at a frequency of \(500 \mathrm{~Hz}\). What is the Doppler shift in the frequency heard by the speeder?

When the door of the Chapel of the Mausoleum in Hamilton, Scotland, is slammed shut, the last echo heard by someone standing just inside the door reportedly comes 15 s later. (a) If that echo were due to a single reflection off a wall opposite the door, how far from the door is the wall? (b) If, instead, the wall is \(25.7 \mathrm{~m}\) away, how many reflections (back and forth) occur?

On July \(10,1996,\) a granite block broke away from a wall in Yosemite Valley and, as it began to slide down the wall, was launched into projectile motion. Seismic waves produced by its impact with the ground triggered seismographs as far away as \(200 \mathrm{~km} .\) Later measurements indicated that the block had a mass between \(7.3 \times 10^{7} \mathrm{~kg}\) and \(1.7 \times 10^{8} \mathrm{~kg}\) and that it landed \(500 \mathrm{~m}\) vertically below the launch point and \(30 \mathrm{~m}\) horizontally from it. (The launch angle is not known.) (a) Estimate the block's kinetic energy just before it landed. Consider two types of seismic waves that spread from the impact point - a hemispherical body wave traveled through the ground in an expanding hemisphere and a cylindrical surface wave traveled along the ground in an expanding shallow vertical cylinder (Fig. 17-49). Assume that the impact lasted \(0.50 \mathrm{~s}\), the vertical cylinder had a depth \(d\) of \(5.0 \mathrm{~m},\) and each wave type received \(20 \%\) of the energy the block had just before impact. Neglecting any mechanical energy loss the waves experienced as they traveled, determine the intensities of (b) the body wave and (c) the surface wave when they reached a seismograph \(200 \mathrm{~km}\) away. (d) On the basis of these results, which wave is more easily detected on a distant seismograph?

An avalanche of sand along some rare desert sand dunes can produce a booming that is loud enough to be heard \(10 \mathrm{~km}\) away. The booming apparently results from a periodic oscillation of the sliding layer of sand \(-\) the layer's thickness expands and contracts. If the emitted frequency is \(90 \mathrm{~Hz}\), what are (a) the period of the thickness oscillation and (b) the wavelength of the sound?

In pipe \(A\), the ratio of a particular harmonic frequency to the next lower harmonic frequency is \(1.2 .\) In pipe \(B,\) the ratio of a particular harmonic frequency to the next lower harmonic frequency is 1.4. How many open ends are in (a) pipe \(A\) and (b) pipe \(B ?\)
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