17.3 Sound Intensity

In a quiet forest, you can sometimes hear a single leaf fall to the ground. But when a passing motorist has their stereo turned up, you cannot even hear what the person next to you in your car is saying (Figure 17.12). We are all very familiar with the loudness of sounds and are aware that loudness is related to how energetically the source is vibrating. High noise exposure is hazardous to hearing, which is why it is important for people working in industrial settings to wear ear protection. The relevant physical quantity is sound intensity, a concept that is valid for all sounds whether or not they are in the audible range.

Photograph shows a roadway crowded with cars and motorcycles in Delhi. — **Figure 17.12** Noise on crowded roadways, like this one in Delhi, makes it hard to hear others unless they shout.Noise on crowded roadways, like this one in Delhi, makes it hard to hear others unless they shout. (credit: “Lingaraj G J”/Flickr)

In Waves, we defined intensity as the power per unit area carried by a wave. Power is the rate at which energy is transferred by the wave. In equation form, intensity I is

$I = \frac{P}{A},$

where P is the power through an area A. The SI unit for I is $W/m squared.$ If we assume that the sound wave is spherical, and that no energy is lost to thermal processes, the energy of the sound wave is spread over a larger area as distance increases, so the intensity decreases. The area of a sphere is $A equals 4 π r squared.$ As the wave spreads out from $r sub 1$ to $r sub 2,$ the energy also spreads out over a larger area:

$\begin{matrix} P_{1} & = & P_{2} \\ I_{1} 4 π r_{1}^{2} & = & I_{2} 4 π r_{2}^{2}; \end{matrix}$

$I_{2} = I_{1} {(\frac{r_{1}}{r_{2}})}^{2} .$

The intensity decreases as the wave moves out from the source. In an inverse square relationship, such as the intensity, when you double the distance, the intensity decreases to one quarter,

$I_{2} = I_{1} {(\frac{r_{1}}{r_{2}})}^{2} = I_{1} {(\frac{r_{1}}{2 r_{1}})}^{2} = \frac{1}{4} I_{1} .$

Generally, when considering the intensity of a sound wave, we take the intensity to be the time-averaged value of the power, denoted by $〈 P 〉,$ divided by the area,

$I = \frac{〈 P 〉}{A} .$

The intensity of a sound wave is proportional to the change in the pressure squared and inversely proportional to the density and the speed. Consider a parcel of a medium initially undisturbed and then influenced by a sound wave at time t, as shown in Figure 17.13.

Picture is a drawing of a parcel of a medium initially undisturbed and then influenced by a sound wave. A sound wave moves through the medium at time t, and the parcel is displaced and expands in the displacement direction. — **Figure 17.13** An undisturbed parcel of a medium with a volume $V = A Δ x$ shown in blue. A sound wave moves through the medium at time t, and the parcel is displaced and expands, as shown by dotted lines. The change in volume is $Δ V = A Δ s = A (s_{2} - s_{1})$ , where $s sub 1$ is the displacement of the leading edge of the parcel and $s sub 2$ is the displacement of the trailing edge of the parcel. In the figure, $s sub 2 greater than s sub 1$ and the parcel expands, but the parcel can either expand or compress $open parenthesis s sub 2 less than s sub 1 close parenthesis$ , depending on which part of the sound wave (compression or rarefaction) is moving through the parcel.

As the sound wave moves through the parcel, the parcel is displaced and may expand or contract. If $s sub 2 greater than s sub 1$ , the volume has increased and the pressure decreases. If $s sub 2 less than s sub 1,$ the volume has decreased and the pressure increases. The change in the volume is

$Δ V = A Δ s = A (s_{2} - s_{1}) = A (s (x + Δ x, t) - s (x, t)) .$

The fractional change in the volume is the change in volume divided by the original volume:

$\frac{d V}{V} = lim_{Δ x \to 0} \frac{A [s (x + Δ x, t) - s (x, t)]}{A Δ x} = \frac{\partial s (x, t)}{\partial x} .$

The fractional change in volume is related to the pressure fluctuation by the bulk modulus $β = - \frac{Δ p (x, t)}{d V / V} .$ Recall that the minus sign is required because the volume is inversely related to the pressure. (We use lowercase p for pressure to distinguish it from power, denoted by P.) The change in pressure is therefore $Δ p (x, t) = − β \frac{d V}{V} = − β \frac{\partial s (x, t)}{\partial x} .$ If the sound wave is sinusoidal, then the displacement as shown in Equation 17.2 is $s (x, t) = s_{max} cos (k x \mp ω t + ϕ)$ and the pressure is found to be

$Δ p (x, t) = − β \frac{d V}{V} = − β \frac{\partial s (x, t)}{\partial x} = β k s_{max} sin (k x - ω t + ϕ) = Δ p_{max} sin (k x - ω t + ϕ) .$

The intensity of the sound wave is the power per unit area, and the power is the force times the velocity, $I = \frac{P}{A} = \frac{F v}{A} = p v .$ Here, the velocity is the velocity of the oscillations of the medium, and not the velocity of the sound wave. The velocity of the medium is the time rate of change in the displacement:

$v (x, t) = \frac{\partial}{\partial t} s (x, t) = \frac{\partial}{\partial t} (s_{max} cos (k x - ω t + ϕ)) = s_{max} ω sin (k x - ω t + ϕ) .$

Thus, the intensity becomes

$\begin{matrix} I & = Δ p (x, t) v (x, t) \\ = β k s_{max} sin (k x - ω t + ϕ) [s_{max} ω sin (k x - ω t + ϕ)] \\ = β k ω s_{max}^{2} {sin}^{2} (k x - ω t + ϕ) . \end{matrix}$

To find the time-averaged intensity over one period $T = \frac{2 π}{ω}$ for a position x, we integrate over the period, $I = \frac{β k ω s_{max}^{2}}{2} .$ Using $Δ p_{max} = β k s_{max},$ $v = \sqrt{\frac{β}{ρ}},$ and $v = \frac{ω}{k},$ we obtain

$I = \frac{β k ω s_{max}^{2}}{2} = \frac{β^{2} k^{2} ω s_{max}^{2}}{2 β k} = \frac{ω {(Δ p_{max})}^{2}}{2 (ρ v^{2}) k} = \frac{v {(Δ p_{max})}^{2}}{2 (ρ v^{2})} = \frac{{(Δ p_{max})}^{2}}{2 ρ v} .$

That is, the intensity of a sound wave is related to its amplitude squared by

$I = \frac{{(Δ p_{max})}^{2}}{2 ρ v} .$

Here, $Δ p_{max}$ is the pressure variation or pressure amplitude in units of pascals (Pa) or $N/m squared$ . The energy (as kinetic energy $\frac{1}{2} m v^{2}$ ) of an oscillating element of air due to a traveling sound wave is proportional to its amplitude squared. In this equation, $ρ$ is the density of the material in which the sound wave travels, in units of $kg/m cubed,$ and v is the speed of sound in the medium, in units of m/s. The pressure variation is proportional to the amplitude of the oscillation, so I varies as ${(Δ p)}^{2} .$ This relationship is consistent with the fact that the sound wave is produced by some vibration; the greater its pressure amplitude, the more the air is compressed in the sound it creates.

Human Hearing and Sound Intensity Levels

As stated earlier in this chapter, hearing is the perception of sound. The hearing mechanism involves some interesting physics. The sound wave that impinges upon our ear is a pressure wave. The ear is a transducer that converts sound waves into electrical nerve impulses in a manner much more sophisticated than, but analogous to, a microphone. Figure 17.14 shows the anatomy of the ear.

Picture is a drawing of an ear. It shows the ear canal finishing with the eardrum. Hammer connected to the anvil is in the in the contact with the eardrum. Behind the eardrum is the hammer and the anvil. The anvil is connected to the stirrup which is attached to the oval window. Cochlea, cochlear nerve and vestibular nerve are in contact with the stirrup. — **Figure 17.14** The anatomy of the human ear.

The outer ear, or ear canal, carries sound to the recessed, protected eardrum. The air column in the ear canal resonates and is partially responsible for the sensitivity of the ear to sounds in the 2000–5000-Hz range. The middle ear converts sound into mechanical vibrations and applies these vibrations to the cochlea.

The range of intensities that the human ear can hear depends on the frequency of the sound, but, in general, the range is quite large. The minimum threshold intensity that can be heard is $I_{0} = 10^{−12} {W/m}^{2} .$ Pain is experienced at intensities of $I sub pain equals 1 W/m squared.$ Measurements of sound intensity (in units of $W/m squared$ ) are very cumbersome due to this large range in values. For this reason, as well as for other reasons, the concept of sound intensity level was proposed.

The sound intensity level $β$ of a sound, measured in decibels, having an intensity I in watts per meter squared, is defined as

$β (dB) = 10 {log}_{10} (\frac{I}{I_{0}}),$

(17.12)

where $I_{0} = 10^{−12} {W/m}^{2}$ is a reference intensity, corresponding to the threshold intensity of sound that a person with normal hearing can perceive at a frequency of 1.00 kHz. It is more common to consider sound intensity levels in dB than in $W/m squared.$ How human ears perceive sound can be more accurately described by the logarithm of the intensity rather than directly by the intensity. Because $β$ is defined in terms of a ratio, it is a unitless quantity, telling you the level of the sound relative to a fixed standard ( $10 to the power −12 W/m squared$ ). The units of decibels (dB) are used to indicate this ratio is multiplied by 10 in its definition. The bel, upon which the decibel is based, is named for Alexander Graham Bell, the inventor of the telephone.

The decibel level of a sound having the threshold intensity of $10 to the power −12 W/m squared$ is $β equals 0 dB,$ because $log sub 10 1 equals 0.$ Table 17.2 gives levels in decibels and intensities in watts per meter squared for some familiar sounds. The ear is sensitive to as little as a trillionth of a watt per meter squared—even more impressive when you realize that the area of the eardrum is only about $1 cm squared,$ so that only $10 to the power −16 W$ falls on it at the threshold of hearing. Air molecules in a sound wave of this intensity vibrate over a distance of less than one molecular diameter, and the gauge pressures involved are less than $10 to the power −9 atm.$

**Table 17.2** Sound Intensity Levels and Intensities
Sound intensity level $β$ (dB)	Intensity I $open parenthesis W/m squared close parenthesis$	Example/effect
0	$1 \times 10^{- 12}$	Threshold of hearing at 1000 Hz
10	$1 \times 10^{- 11}$	Rustle of leaves
20	$1 \times 10^{- 10}$	Whisper at 1-m distance
30	$1 \times 10^{- 9}$	Quiet home
40	$1 \times 10^{- 8}$	Average home
50	$1 \times 10^{- 7}$	Average office, soft music
60	$1 \times 10^{- 6}$	Normal conversation
70	$1 \times 10^{- 5}$	Noisy office, busy traffic
80	$1 \times 10^{- 4}$	Loud radio, classroom lecture
90	$1 \times 10^{- 3}$	Inside a heavy truck; damage from prolonged exposure¹
100	$1 \times 10^{- 2}$	Noisy factory, siren at 30 m; damage from 8 h per day exposure
110	$1 \times 10^{- 1}$	Damage from 30 min per day exposure
120	1	Loud rock concert; pneumatic chipper at 2 m; threshold of pain
140	$1 times 10 squared$	Jet airplane at 30 m; severe pain, damage in seconds
160	$1 \times 10^{4}$	Bursting of eardrums

An observation readily verified by examining Table 17.2 or by using Equation 17.12 is that each factor of 10 in intensity corresponds to 10 dB. For example, a 90-dB sound compared with a 60-dB sound is 30 dB greater, or three factors of 10 (that is, $10 cubed$ times) as intense. Another example is that if one sound is $10 to the power 7$ as intense as another, it is 70 dB higher (Table 17.3).

**Table 17.3** Ratios of Intensities and Corresponding Differences in Sound Intensity Levels
$I sub 2 / I sub 1$	$β sub 2 minus β sub 1$
2.0	3.0 dB
5.0	7.0 dB
10.0	10.0 dB
100.0	20.0 dB
1000.0	30.0 dB

Example 1 Calculating Sound Intensity Levels

Calculate the sound intensity level in decibels for a sound wave traveling in air at $0 °C$ and having a pressure amplitude of 0.656 Pa.

Strategy We are given $Δ p$ , so we can calculate I using the equation $I = \frac{{(Δ p)}^{2}}{2 ρ v_{w}} .$ Using I, we can calculate $β$ straight from its definition in $β (d B) = 10 {log}_{10} (\frac{I}{I_{0}}) .$

Solution

Identify knowns:
Sound travels at 331 m/s in air at $0 °C.$
Air has a density of $1.29 kg/m cubed$ at atmospheric pressure and $0 °C.$
Enter these values and the pressure amplitude into $I = \frac{{(Δ p)}^{2}}{2 ρ v} .$
$I = \frac{{(Δ p)}^{2}}{2 ρ v} = \frac{{(0.656 Pa)}^{2}}{2 (1.29 {kg/m}^{3}) (331 m/s)} = 5.04 \times 10^{−4} {W/m}^{2} .$
Enter the value for I and the known value for $I sub 0$ into $β open parenthesis dB close parenthesis equals 10 log sub 10 open parenthesis I / I sub 0 close parenthesis.$ Calculate to find the sound intensity level in decibels:
$10 {log}_{10} (5.04 \times 10^{8}) = 10 (8.70) dB = 87 dB .$

Significance This 87-dB sound has an intensity five times as great as an 80-dB sound. So a factor of five in intensity corresponds to a difference of 7 dB in sound intensity level. This value is true for any intensities differing by a factor of five.

Example 2 Changing Intensity Levels of a Sound

Show that if one sound is twice as intense as another, it has a sound level about 3 dB higher.

Strategy We are given that the ratio of two intensities is 2 to 1, and are then asked to find the difference in their sound levels in decibels. We can solve this problem by using of the properties of logarithms.

Solution

Identify knowns:
The ratio of the two intensities is 2 to 1, or
$\frac{I_{2}}{I_{1}} = 2.00 .$
We wish to show that the difference in sound levels is about 3 dB. That is, we want to show:
$β_{2} - β_{1} = 3 dB.$
Note that

${log}_{10} b - {log}_{10} a = {log}_{10} (\frac{b}{a}) .$
Use the definition of $β$ to obtain
$β_{2} - β_{1} = 10 {log}_{10} (\frac{I_{2}}{I_{1}}) = 10 {log}_{10} 2.00 = 10 (0.301) dB .$
Thus,
$β_{2} - β_{1} = 3.01 dB .$

Significance This means that the two sound intensity levels differ by 3.01 dB, or about 3 dB, as advertised. Note that because only the ratio $I sub 2 / I sub 1$ is given (and not the actual intensities), this result is true for any intensities that differ by a factor of two. For example, a 56.0-dB sound is twice as intense as a 53.0-dB sound, a 97.0-dB sound is half as intense as a 100-dB sound, and so on.

Another decibel scale is also in use, called the sound pressure level, based on the ratio of the pressure amplitude to a reference pressure. This scale is used particularly in applications where sound travels in water. It is beyond the scope of this text to treat this scale because it is not commonly used for sounds in air, but it is important to note that very different decibel levels may be encountered when sound pressure levels are quoted.

Hearing and Pitch

The human ear has a tremendous range and sensitivity. It can give us a wealth of simple information—such as pitch, loudness, and direction.

The perception of frequency is called pitch. Typically, humans have excellent relative pitch and can discriminate between two sounds if their frequencies differ by 0.3% or more. For example, 500.0 and 501.5 Hz are noticeably different. Musical notes are sounds of a particular frequency that can be produced by most instruments and in Western music have particular names, such as A-sharp, C, or E-flat.

The perception of intensity is called loudness. At a given frequency, it is possible to discern differences of about 1 dB, and a change of 3 dB is easily noticed. But loudness is not related to intensity alone. Frequency has a major effect on how loud a sound seems. Sounds near the high- and low-frequency extremes of the hearing range seem even less loud, because the ear is less sensitive at those frequencies. When a violin plays middle C, there is no mistaking it for a piano playing the same note. The reason is that each instrument produces a distinctive set of frequencies and intensities. We call our perception of these combinations of frequencies and intensities tone quality or, more commonly, the timbre of the sound. Timbre is the shape of the wave that arises from the many reflections, resonances, and superposition in an instrument.

A unit called a phon is used to express loudness numerically. Phons differ from decibels because the phon is a unit of loudness perception, whereas the decibel is a unit of physical intensity. Figure 17.15 shows the relationship of loudness to intensity (or intensity level) and frequency for persons with normal hearing. The curved lines are equal-loudness curves. Each curve is labeled with its loudness in phons. Any sound along a given curve is perceived as equally loud by the average person. The curves were determined by having large numbers of people compare the loudness of sounds at different frequencies and sound intensity levels. At a frequency of 1000 Hz, phons are taken to be numerically equal to decibels.

The graph is the plot of sound level in decibels versus frequency in Herz. Data for 0, 10, 20, 30, 40, 50, 60, 70, 80, 90, 100, 110, and 120 phons is plotted. Data is plotted as curved lines stacked one a top of other. — **Figure 17.15** The relationship of loudness in phons to intensity level (in decibels) and intensity (in watts per meter squared) for persons with normal hearing. The curved lines are equal-loudness curves—all sounds on a given curve are perceived as equally loud. Phons and decibels are defined to be the same at 1000 Hz.

Example 3 Measuring Loudness

(a) What is the loudness in phons of a 100-Hz sound that has an intensity level of 80 dB? (b) What is the intensity level in decibels of a 4000-Hz sound having a loudness of 70 phons? (c) At what intensity level will an 8000-Hz sound have the same loudness as a 200-Hz sound at 60 dB?

Strategy The graph in Figure 17.15 should be referenced to solve this example. To find the loudness of a given sound, you must know its frequency and intensity level, locate that point on the square grid, and then interpolate between loudness curves to get the loudness in phons. Once that point is located, the intensity level can be determined from the vertical axis.

Solution

Identify knowns: The square grid of the graph relating phons and decibels is a plot of intensity level versus frequency—both physical quantities: 100 Hz at 80 dB lies halfway between the curves marked 70 and 80 phons.
Find the loudness: 75 phons.
Identify knowns: Values are given to be 4000 Hz at 70 phons.
Follow the 70-phon curve until it reaches 4000 Hz. At that point, it is below the 70 dB line at about 67 dB.
Find the intensity level: 67 dB.
Locate the point for a 200 Hz and 60 dB sound.
Find the loudness: This point lies just slightly above the 50-phon curve, and so its loudness is 51 phons.
Look for the 51-phon level is at 8000 Hz: 63 dB.

Significance These answers, like all information extracted from Figure 17.15, have uncertainties of several phons or several decibels, partly due to difficulties in interpolation, but mostly related to uncertainties in the equal-loudness curves.

In this section, we discussed the characteristics of sound and how we hear, but how are the sounds we hear produced? Interesting sources of sound are musical instruments and the human voice, and we will discuss these sources. But before we can understand how musical instruments produce sound, we need to look at the basic mechanisms behind these instruments. The theories behind the mechanisms used by musical instruments involve interference, superposition, and standing waves, which we discuss in the next section.

Summary

Intensity $I equals P / A$ is the same for a sound wave as was defined for all waves, where P is the power crossing area A. The SI unit for I is watts per meter squared. The intensity of a sound wave is also related to the pressure amplitude $Δ p :$
$I = \frac{{(Δ p)}^{2}}{2 ρ v},$
where $ρ$ is the density of the medium in which the sound wave travels and $v_{w}$ is the speed of sound in the medium.
Sound intensity level in units of decibels (dB) is
$β (dB) = 10 {log}_{10} (\frac{I}{I_{0}}),$
where $I_{0} = 10^{−12} {W/m}^{2}$ is the threshold intensity of hearing.
The perception of frequency is pitch. The perception of intensity is loudness and loudness has units of phons.

Conceptual Questions

Six members of a synchronized swim team wear earplugs to protect themselves against water pressure at depths, but they can still hear the music and perform the combinations in the water perfectly. One day, they were asked to leave the pool so the dive team could practice a few dives, and they tried to practice on a mat, but seemed to have a lot more difficulty. Why might this be?

The ear plugs reduce the intensity of the sound both in water and on land, but Navy researchers have found that sound under water is heard through vibrations mastoid, which is the bone behind the ear.

A community is concerned about a plan to bring train service to their downtown from the town’s outskirts. The current sound intensity level, even though the rail yard is blocks away, is 70 dB downtown. The mayor assures the public that there will be a difference of only 30 dB in sound in the downtown area. Should the townspeople be concerned? Why?

Problems

What is the intensity in watts per meter squared of a 85.0-dB sound?

The warning tag on a lawn mower states that it produces noise at a level of 91.0 dB. What is the intensity of this sound in watts per meter squared?

$1.26 \times 10^{− 3} {W/m}^{2}$

A sound wave traveling in air has a pressure amplitude of 0.5 Pa. What is the intensity of the wave?

What intensity level does the sound in the preceding problem correspond to?

85 dB

What sound intensity level in dB is produced by earphones that create an intensity of $4.00 \times 10^{−2} {W/m}^{2}$ ?

What is the decibel level of a sound that is twice as intense as a 90.0-dB sound? (b) What is the decibel level of a sound that is one-fifth as intense as a 90.0-dB sound?

a. 93 dB; b. 83 dB

What is the intensity of a sound that has a level 7.00 dB lower than a $4.00 \times 10^{−9} {-W/m}^{2}$ sound? (b) What is the intensity of a sound that is 3.00 dB higher than a $4.00 \times 10^{−9} {-W/m}^{2}$ sound?

People with good hearing can perceive sounds as low as −8.00 dB at a frequency of 3000 Hz. What is the intensity of this sound in watts per meter squared?

$1.58 \times 10^{− 13} {W/m}^{2}$

If a large housefly 3.0 m away from you makes a noise of 40.0 dB, what is the noise level of 1000 flies at that distance, assuming interference has a negligible effect?

Ten cars in a circle at a boom box competition produce a 120-dB sound intensity level at the center of the circle. What is the average sound intensity level produced there by each stereo, assuming interference effects can be neglected?

A decrease of a factor of 10 in intensity corresponds to a reduction of 10 dB in sound level: $120 dB - 10 dB = 110 dB .$

The amplitude of a sound wave is measured in terms of its maximum gauge pressure. By what factor does the amplitude of a sound wave increase if the sound intensity level goes up by 40.0 dB?

If a sound intensity level of 0 dB at 1000 Hz corresponds to a maximum gauge pressure (sound amplitude) of $10 to the power −9 atm$ , what is the maximum gauge pressure in a 60-dB sound? What is the maximum gauge pressure in a 120-dB sound?

We know that 60 dB corresponds to a factor of $10 to the power 6$ increase in intensity. Therefore,
$I \propto X^{2} \Rightarrow \frac{I_{2}}{I_{1}} = {(\frac{X_{2}}{X_{1}})}^{2}, so that X_{2} = 10^{− 6} atm .$
120 dB corresponds to a factor of $10 to the power 12$ increase $\Rightarrow 10^{− 9} atm {(10^{12})}^{1/2} = 10^{− 3} atm .$

An 8-hour exposure to a sound intensity level of 90.0 dB may cause hearing damage. What energy in joules falls on a 0.800-cm-diameter eardrum so exposed?

Sound is more effectively transmitted into a stethoscope by direct contact rather than through the air, and it is further intensified by being concentrated on the smaller area of the eardrum. It is reasonable to assume that sound is transmitted into a stethoscope 100 times as effectively compared with transmission though the air. What, then, is the gain in decibels produced by a stethoscope that has a sound gathering area of $15.0 cm squared$ , and concentrates the sound onto two eardrums with a total area of $0.900 cm squared$ with an efficiency of $40.0 %$ ?

28.2 dB

Loudspeakers can produce intense sounds with surprisingly small energy input in spite of their low efficiencies. Calculate the power input needed to produce a 90.0-dB sound intensity level for a 12.0-cm-diameter speaker that has an efficiency of $1.00 %$ . (This value is the sound intensity level right at the speaker.)

The factor of 10^-12 in the range of intensities to which the ear can respond, from threshold to that causing damage after brief exposure, is truly remarkable. If you could measure distances over the same range with a single instrument and the smallest distance you could measure was 1 mm, what would the largest be?

$1 \times 10^{6} km$

What are the closest frequencies to 500 Hz that an average person can clearly distinguish as being different in frequency from 500 Hz? The sounds are not present simultaneously.

Can you tell that your roommate turned up the sound on the TV if its average sound intensity level goes from 70 to 73 dB?

$73 dB - 70 dB = 3 dB;$ Such a change in sound level is easily noticed.

If a woman needs an amplification of $5.0 \times 10^{5}$ times the threshold intensity to enable her to hear at all frequencies, what is her overall hearing loss in dB? Note that smaller amplification is appropriate for more intense sounds to avoid further damage to her hearing from levels above 90 dB.

A person has a hearing threshold 10 dB above normal at 100 Hz and 50 dB above normal at 4000 Hz. How much more intense must a 100-Hz tone be than a 4000-Hz tone if they are both barely audible to this person?

2.5; The 100-Hz tone must be 2.5 times more intense than the 4000-Hz sound to be audible by this person.