problem solving on the speed of sound

17.2 Speed of Sound

Learning objectives.

By the end of this section, you will be able to:

Explain the relationship between wavelength and frequency of sound
Determine the speed of sound in different media
Derive the equation for the speed of sound in air
Determine the speed of sound in air for a given temperature

Sound, like all waves, travels at a certain speed and has the properties of frequency and wavelength. You can observe direct evidence of the speed of sound while watching a fireworks display ( (Figure) ). You see the flash of an explosion well before you hear its sound and possibly feel the pressure wave, implying both that sound travels at a finite speed and that it is much slower than light.

Picture shows a photograph of colorful fireworks illuminating night sky.

Figure 17.4 When a firework shell explodes, we perceive the light energy before the sound energy because sound travels more slowly than light does.

The difference between the speed of light and the speed of sound can also be experienced during an electrical storm. The flash of lighting is often seen before the clap of thunder. You may have heard that if you count the number of seconds between the flash and the sound, you can estimate the distance to the source. Every five seconds converts to about one mile. The velocity of any wave is related to its frequency and wavelength by

where v is the speed of the wave, f is its frequency, and [latex] \lambda [/latex] is its wavelength. Recall from Waves that the wavelength is the length of the wave as measured between sequential identical points. For example, for a surface water wave or sinusoidal wave on a string, the wavelength can be measured between any two convenient sequential points with the same height and slope, such as between two sequential crests or two sequential troughs. Similarly, the wavelength of a sound wave is the distance between sequential identical parts of a wave—for example, between sequential compressions ( (Figure) ). The frequency is the same as that of the source and is the number of waves that pass a point per unit time.

Picture is a schematic drawing of a tuning fork emanating sound waves.

Figure 17.5 A sound wave emanates from a source, such as a tuning fork, vibrating at a frequency f. It propagates at speed v and has a wavelength [latex] \lambda [/latex].

Speed of Sound in Various Media

(Figure) shows that the speed of sound varies greatly in different media. The speed of sound in a medium depends on how quickly vibrational energy can be transferred through the medium. For this reason, the derivation of the speed of sound in a medium depends on the medium and on the state of the medium. In general, the equation for the speed of a mechanical wave in a medium depends on the square root of the restoring force, or the elastic property , divided by the inertial property ,

Also, sound waves satisfy the wave equation derived in Waves ,

Recall from Waves that the speed of a wave on a string is equal to [latex] v=\sqrt{\frac{{F}_{T}}{\mu }}, [/latex] where the restoring force is the tension in the string [latex] {F}_{T} [/latex] and the linear density [latex] \mu [/latex] is the inertial property. In a fluid, the speed of sound depends on the bulk modulus and the density,

The speed of sound in a solid the depends on the Young’s modulus of the medium and the density,

In an ideal gas (see The Kinetic Theory of Gases in the second volume of this text), the equation for the speed of sound is

where [latex] \gamma [/latex] is the adiabatic index, [latex] R=8.31\,\text{J/mol}·\text{K} [/latex] is the gas constant, [latex] {T}_{\text{K}} [/latex] is the absolute temperature in kelvins, and M is the molecular mass. In general, the more rigid (or less compressible) the medium, the faster the speed of sound. This observation is analogous to the fact that the frequency of simple harmonic motion is directly proportional to the stiffness of the oscillating object as measured by k , the spring constant. The greater the density of a medium, the slower the speed of sound. This observation is analogous to the fact that the frequency of a simple harmonic motion is inversely proportional to m , the mass of the oscillating object. The speed of sound in air is low, because air is easily compressible. Because liquids and solids are relatively rigid and very difficult to compress, the speed of sound in such media is generally greater than in gases.

Because the speed of sound depends on the density of the material, and the density depends on the temperature, there is a relationship between the temperature in a given medium and the speed of sound in the medium. For air at sea level, the speed of sound is given by

where the temperature in the first equation (denoted as [latex] {T}_{\text{C}} [/latex]) is in degrees Celsius and the temperature in the second equation (denoted as [latex] {T}_{\text{K}} [/latex]) is in kelvins. The speed of sound in gases is related to the average speed of particles in the gas, [latex] {v}_{\text{rms}}=\sqrt{\frac{3{k}_{\text{B}}T}{m}}, [/latex] where [latex] {k}_{\text{B}} [/latex] is the Boltzmann constant [latex] (1.38\,×\,{10}^{-23}\,\text{J/K}) [/latex] and m is the mass of each (identical) particle in the gas. Note that v refers to the speed of the coherent propagation of a disturbance (the wave), whereas [latex] {v}_{\text{rms}} [/latex] describes the speeds of particles in random directions. Thus, it is reasonable that the speed of sound in air and other gases should depend on the square root of temperature. While not negligible, this is not a strong dependence. At [latex] 0\text{°C} [/latex], the speed of sound is 331 m/s, whereas at [latex] 20.0\text{°C} [/latex], it is 343 m/s, less than a [latex] 4\text{%} [/latex] increase. (Figure) shows how a bat uses the speed of sound to sense distances.

Picture is a drawing of a flying bat that emits sound waves. Waves are reflected from the flying insect and are returned to the bat.

Figure 17.6 A bat uses sound echoes to find its way about and to catch prey. The time for the echo to return is directly proportional to the distance.

Derivation of the Speed of Sound in Air

As stated earlier, the speed of sound in a medium depends on the medium and the state of the medium. The derivation of the equation for the speed of sound in air starts with the mass flow rate and continuity equation discussed in Fluid Mechanics .

Consider fluid flow through a pipe with cross-sectional area A ( (Figure) ). The mass in a small volume of length x of the pipe is equal to the density times the volume, or [latex] m=\rho V=\rho Ax. [/latex] The mass flow rate is

The continuity equation from Fluid Mechanics states that the mass flow rate into a volume has to equal the mass flow rate out of the volume, [latex] {\rho }_{\text{in}}{A}_{\text{in}}{v}_{\text{in}}={\rho }_{\text{out}}{A}_{\text{out}}{v}_{\text{out}}. [/latex]

Picture is a schematic drawing of a mass flowing through with the speed v for the distance x through the cylinder with the cross-sectional area A.

Figure 17.7 The mass of a fluid in a volume is equal to the density times the volume, [latex] m=\rho V=\rho Ax. [/latex] The mass flow rate is the time derivative of the mass.

Now consider a sound wave moving through a parcel of air. A parcel of air is a small volume of air with imaginary boundaries ( (Figure) ). The density, temperature, and velocity on one side of the volume of the fluid are given as [latex] \rho ,T,v, [/latex] and on the other side are [latex] \rho +d\rho ,T+dT,v+dv. [/latex]

Picture is a schematic drawing of a sound wave moving through a volume of fluid. The density, temperature, and velocity of the fluid change from one side to the other.

Figure 17.8 A sound wave moves through a volume of fluid. The density, temperature, and velocity of the fluid change from one side to the other.

The continuity equation states that the mass flow rate entering the volume is equal to the mass flow rate leaving the volume, so

This equation can be simplified, noting that the area cancels and considering that the multiplication of two infinitesimals is approximately equal to zero: [latex] d\rho (dv)\approx 0, [/latex]

The net force on the volume of fluid ( (Figure) ) equals the sum of the forces on the left face and the right face:

Picture is a schematic drawing of a sound wave moving through a volume of fluid with the sides of dimensions dx, dy, and dz. The pressure is different on the opposite sides.

Figure 17.9 A sound wave moves through a volume of fluid. The force on each face can be found by the pressure times the area.

The acceleration is the force divided by the mass and the mass is equal to the density times the volume, [latex] m=\rho V=\rho \,dx\,dy\,dz. [/latex] We have

From the continuity equation [latex] \rho \,dv=\text{−}vd\rho [/latex], we obtain

Consider a sound wave moving through air. During the process of compression and expansion of the gas, no heat is added or removed from the system. A process where heat is not added or removed from the system is known as an adiabatic system. Adiabatic processes are covered in detail in The First Law of Thermodynamics , but for now it is sufficient to say that for an adiabatic process,[latex] p{V}^{\gamma }=\text{constant,} [/latex] where p is the pressure, V is the volume, and gamma [latex] (\gamma ) [/latex] is a constant that depends on the gas. For air, [latex] \gamma =1.40 [/latex]. The density equals the number of moles times the molar mass divided by the volume, so the volume is equal to [latex] V=\frac{nM}{\rho }. [/latex] The number of moles and the molar mass are constant and can be absorbed into the constant [latex] p{(\frac{1}{\rho })}^{\gamma }=\text{constant}\text{.} [/latex] Taking the natural logarithm of both sides yields [latex] \text{ln}\,p-\gamma \,\text{ln}\,\rho =\text{constant}\text{.} [/latex] Differentiating with respect to the density, the equation becomes

If the air can be considered an ideal gas, we can use the ideal gas law:

Here M is the molar mass of air:

Since the speed of sound is equal to [latex] v=\sqrt{\frac{dp}{d\rho }} [/latex], the speed is equal to

Note that the velocity is faster at higher temperatures and slower for heavier gases. For air, [latex] \gamma =1.4, [/latex] [latex] M=0.02897\frac{\text{kg}}{\text{mol}}, [/latex] and [latex] R=8.31\frac{\text{J}}{\text{mol}·\text{K}}. [/latex] If the temperature is [latex] {T}_{\text{C}}=20\text{°}\text{C}(T=293\,\text{K}), [/latex] the speed of sound is [latex] v=343\,\text{m/s}\text{.} [/latex]

The equation for the speed of sound in air [latex] v=\sqrt{\frac{\gamma RT}{M}} [/latex] can be simplified to give the equation for the speed of sound in air as a function of absolute temperature:

One of the more important properties of sound is that its speed is nearly independent of the frequency. This independence is certainly true in open air for sounds in the audible range. If this independence were not true, you would certainly notice it for music played by a marching band in a football stadium, for example. Suppose that high-frequency sounds traveled faster—then the farther you were from the band, the more the sound from the low-pitch instruments would lag that from the high-pitch ones. But the music from all instruments arrives in cadence independent of distance, so all frequencies must travel at nearly the same speed. Recall that

In a given medium under fixed conditions, v is constant, so there is a relationship between f and [latex] \lambda ; [/latex] the higher the frequency, the smaller the wavelength ( (Figure) ).

Picture is a schematic drawing of a speaker system emanating sound waves. The lower-frequency sounds are emitted by the bottom large speaker; the higher-frequency sounds are emitted by the top small speaker.

Figure 17.10 Because they travel at the same speed in a given medium, low-frequency sounds must have a greater wavelength than high-frequency sounds. Here, the lower-frequency sounds are emitted by the large speaker, called a woofer, whereas the higher-frequency sounds are emitted by the small speaker, called a tweeter.

Calculating Wavelengths

Calculate the wavelengths of sounds at the extremes of the audible range, 20 and 20,000 Hz, in [latex] 30.0\text{°C} [/latex] air. (Assume that the frequency values are accurate to two significant figures.)

To find wavelength from frequency, we can use [latex] v=f\lambda . [/latex]

Identify knowns. The value for v is given by [latex] v=(331\,\text{m/s})\sqrt{\frac{T}{273\,\text{K}}}. [/latex]
Convert the temperature into kelvins and then enter the temperature into the equation [latex] v=(331\,\text{m/s})\sqrt{\frac{303\,\text{K}}{273\,\text{K}}}=348.7\,\text{m/s}\text{.} [/latex]
Solve the relationship between speed and wavelength for λ : [latex] \lambda =\frac{v}{f}. [/latex]
Enter the speed and the minimum frequency to give the maximum wavelength: [latex] {\lambda }_{\text{max}}=\text{}\frac{348.7\,\text{m/s}}{20\,\text{Hz}}=17\,\text{m}\text{.} [/latex]
Enter the speed and the maximum frequency to give the minimum wavelength: [latex] {\lambda }_{\text{min}}=\frac{348.7\,\text{m/s}}{20,000\,\text{Hz}}=0.017\,\text{m}=1.7\,\text{cm}\text{.} [/latex]

Significance

Because the product of f multiplied by [latex] \lambda [/latex] equals a constant, the smaller f is, the larger [latex] \lambda [/latex] must be, and vice versa.

The speed of sound can change when sound travels from one medium to another, but the frequency usually remains the same. This is similar to the frequency of a wave on a string being equal to the frequency of the force oscillating the string. If v changes and f remains the same, then the wavelength [latex] \lambda [/latex] must change. That is, because [latex] v=f\lambda [/latex], the higher the speed of a sound, the greater its wavelength for a given frequency.

Check Your Understanding

Imagine you observe two firework shells explode. You hear the explosion of one as soon as you see it. However, you see the other shell for several milliseconds before you hear the explosion. Explain why this is so.

Sound and light both travel at definite speeds, and the speed of sound is slower than the speed of light. The first shell is probably very close by, so the speed difference is not noticeable. The second shell is farther away, so the light arrives at your eyes noticeably sooner than the sound wave arrives at your ears.

Although sound waves in a fluid are longitudinal, sound waves in a solid travel both as longitudinal waves and transverse waves. Seismic waves , which are essentially sound waves in Earth’s crust produced by earthquakes, are an interesting example of how the speed of sound depends on the rigidity of the medium. Earthquakes produce both longitudinal and transverse waves, and these travel at different speeds. The bulk modulus of granite is greater than its shear modulus. For that reason, the speed of longitudinal or pressure waves (P-waves) in earthquakes in granite is significantly higher than the speed of transverse or shear waves (S-waves). Both types of earthquake waves travel slower in less rigid material, such as sediments. P-waves have speeds of 4 to 7 km/s, and S-waves range in speed from 2 to 5 km/s, both being faster in more rigid material. The P-wave gets progressively farther ahead of the S-wave as they travel through Earth’s crust. The time between the P- and S-waves is routinely used to determine the distance to their source, the epicenter of the earthquake. Because S-waves do not pass through the liquid core, two shadow regions are produced ( (Figure) ).

Picture is a drawing of P and S waves that travel from a source. Shadow regions, where S-waves are absent, is also indicated. There is color coded labeling for Crust, Mantle, Liquid outer core, and Solid inner core.

Figure 17.11 Earthquakes produce both longitudinal waves (P-waves) and transverse waves (S-waves), and these travel at different speeds. Both waves travel at different speeds in the different regions of Earth, but in general, P-waves travel faster than S-waves. S-waves cannot be supported by the liquid core, producing shadow regions.

As sound waves move away from a speaker, or away from the epicenter of an earthquake, their power per unit area decreases. This is why the sound is very loud near a speaker and becomes less loud as you move away from the speaker. This also explains why there can be an extreme amount of damage at the epicenter of an earthquake but only tremors are felt in areas far from the epicenter. The power per unit area is known as the intensity, and in the next section, we will discuss how the intensity depends on the distance from the source.

The speed of sound depends on the medium and the state of the medium.
In a fluid, because the absence of shear forces, sound waves are longitudinal. A solid can support both longitudinal and transverse sound waves.
In air, the speed of sound is related to air temperature T by[latex] v=\,331\frac{\text{m}}{\text{s}}\sqrt{\frac{{T}_{\text{K}}}{273\,\text{K}}}=331\frac{\text{m}}{\text{s}}\sqrt{1+\frac{{T}_{\text{C}}}{273\text{°}\text{C}}}. [/latex]
v is the same for all frequencies and wavelengths of sound in air.

Conceptual Questions

How do sound vibrations of atoms differ from thermal motion?

When sound passes from one medium to another where its propagation speed is different, does its frequency or wavelength change? Explain your answer briefly.

The frequency does not change as the sound wave moves from one medium to another. Since the speed changes and the frequency does not, the wavelength must change. This is similar to the driving force of a harmonic oscillator or a wave on the string.

A popular party trick is to inhale helium and speak in a high-frequency, funny voice. Explain this phenomenon.

You may have used a sonic range finder in lab to measure the distance of an object using a clicking sound from a sound transducer. What is the principle used in this device?

The transducer sends out a sound wave, which reflects off the object in question and measures the time it takes for the sound wave to return. Since the speed of sound is constant, the distance to the object can found by multiplying the velocity of sound by half the time interval measured.

The sonic range finder discussed in the preceding question often needs to be calibrated. During the calibration, the software asks for the room temperature. Why do you suppose the room temperature is required?

When poked by a spear, an operatic soprano lets out a 1200-Hz shriek. What is its wavelength if the speed of sound is 345 m/s?

What frequency sound has a 0.10-m wavelength when the speed of sound is 340 m/s?

[latex] f=3400\,\text{Hz} [/latex]

Calculate the speed of sound on a day when a 1500-Hz frequency has a wavelength of 0.221 m.

(a) What is the speed of sound in a medium where a 100-kHz frequency produces a 5.96-cm wavelength? (b) Which substance in (Figure) is this likely to be?

a. [latex] v=5.96\,×\,{10}^{3}\,\text{m/s} [/latex]; b. steel (from value in (Figure) )

Show that the speed of sound in [latex] 20.0\text{°}\text{C} [/latex] air is [latex] 343\,\text{m/s}, [/latex] as claimed in the text.

Air temperature in the Sahara Desert can reach [latex] 56.0\text{°}\text{C} [/latex] (about [latex] 134\text{°}\text{F} [/latex]). What is the speed of sound in air at that temperature?

Show Answer

Dolphins make sounds in air and water. What is the ratio of the wavelength of a sound in air to its wavelength in seawater? Assume air temperature is [latex] 20.0\text{°}\text{C}\text{.} [/latex]

A sonar echo returns to a submarine 1.20 s after being emitted. What is the distance to the object creating the echo? (Assume that the submarine is in the ocean, not in fresh water.)

[latex] \text{Δ}x=924\,\text{m} [/latex]

(a) If a submarine’s sonar can measure echo times with a precision of 0.0100 s, what is the smallest difference in distances it can detect? (Assume that the submarine is in the ocean, not in fresh water.) (b) Discuss the limits this time resolution imposes on the ability of the sonar system to detect the size and shape of the object creating the echo.

Ultrasonic sound waves are often used in methods of nondestructive testing. For example, this method can be used to find structural faults in a steel I-beams used in building. Consider a 10.00 meter long, steel I-beam with a cross-section shown below. The weight of the I-beam is 3846.50 N. What would be the speed of sound through in the I-beam? [latex] ({Y}_{\text{steel}}=200\,\text{GPa},{\beta }_{\text{steel}}=159\,\text{GPa}) [/latex].

Picture is a drawing of an I beam. The central rod is 10 cm long and 2.5 cm thick. Two parallel rods, 5 cm wide and 2.5 cm thick, are connected to the opposite sides of the center rod.

A physicist at a fireworks display times the lag between seeing an explosion and hearing its sound, and finds it to be 0.400 s. (a) How far away is the explosion if air temperature is [latex] 24.0\text{°}\text{C} [/latex] and if you neglect the time taken for light to reach the physicist? (b) Calculate the distance to the explosion taking the speed of light into account. Note that this distance is negligibly greater.

During a 4th of July celebration, an M80 firework explodes on the ground, producing a bright flash and a loud bang. The air temperature of the night air is [latex] {T}_{\text{F}}=90.00\text{°}\text{F}. [/latex] Two observers see the flash and hear the bang. The first observer notes the time between the flash and the bang as 1.00 second. The second observer notes the difference as 3.00 seconds. The line of sight between the two observers meet at a right angle as shown below. What is the distance [latex] \text{Δ}x [/latex] between the two observers?

Picture is a drawing of a triangle formed by the source of fireworks and two observers. Distance between two observers is delta x. Line of sight from the first observer to the source of fireworks is delta x1. Line of sight from the second observer to the source of fireworks is delta x2.

The density of a sample of water is [latex] \rho =998.00\,{\text{kg/m}}^{3} [/latex] and the bulk modulus is [latex] \beta =2.15\,\text{GPa}\text{.} [/latex] What is the speed of sound through the sample?

Suppose a bat uses sound echoes to locate its insect prey, 3.00 m away. (See (Figure) .) (a) Calculate the echo times for temperatures of [latex] 5.00\text{°}\text{C} [/latex] and [latex] 35.0\text{°}\text{C}\text{.} [/latex] (b) What percent uncertainty does this cause for the bat in locating the insect? (c) Discuss the significance of this uncertainty and whether it could cause difficulties for the bat. (In practice, the bat continues to use sound as it closes in, eliminating most of any difficulties imposed by this and other effects, such as motion of the prey.)

a. [latex] {t}_{5.00\text{°}\text{C}}=0.0180\,\text{s},\enspace{t}_{35.0\text{°}\text{C}}=0.0171\,\text{s} [/latex]; b. [latex] \text{% uncertainty}=5.00\text{%} [/latex]; c. This uncertainty could definitely cause difficulties for the bat, if it didn’t continue to use sound as it closed in on its prey. A 5% uncertainty could be the difference between catching the prey around the neck or around the chest, which means that it could miss grabbing its prey.

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17.2: Speed of Sound, Frequency, and Wavelength

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Learning Objectives

By the end of this section, you will be able to:

Define pitch.
Describe the relationship between the speed of sound, its frequency, and its wavelength.
Describe the effects on the speed of sound as it travels through various media.
Describe the effects of temperature on the speed of sound.

Sound, like all waves, travels at a certain speed and has the properties of frequency and wavelength. You can observe direct evidence of the speed of sound while watching a fireworks display. The flash of an explosion is seen well before its sound is heard, implying both that sound travels at a finite speed and that it is much slower than light. You can also directly sense the frequency of a sound. Perception of frequency is called pitch . The wavelength of sound is not directly sensed, but indirect evidence is found in the correlation of the size of musical instruments with their pitch. Small instruments, such as a piccolo, typically make high-pitch sounds, while large instruments, such as a tuba, typically make low-pitch sounds. High pitch means small wavelength, and the size of a musical instrument is directly related to the wavelengths of sound it produces. So a small instrument creates short-wavelength sounds. Similar arguments hold that a large instrument creates long-wavelength sounds.

A photograph of a fireworks display in the sky.

The relationship of the speed of sound, its frequency, and wavelength is the same as for all waves:

\[v_w = f\lambda,\]

where $v_w$ is the speed of sound, $f$ is its frequency, and $\lambda$ is its wavelength. The wavelength of a sound is the distance between adjacent identical parts of a wave—for example, between adjacent compressions as illustrated in Figure $\PageIndex{2}$. The frequency is the same as that of the source and is the number of waves that pass a point per unit time.

A picture of a vibrating tuning fork is shown. The sound wave compressions and rarefactions are shown to emanate from the fork on both the sides as semicircular arcs of alternate bold and dotted lines. The wavelength is marked as the distance between two successive bold arcs. The frequency of the vibrations is shown as f and velocity of the wave represented by v sub w.

Table $\PageIndex{1}$ makes it apparent that the speed of sound varies greatly in different media. The speed of sound in a medium is determined by a combination of the medium’s rigidity (or compressibility in gases) and its density. The more rigid (or less compressible) the medium, the faster the speed of sound. For materials that have similar rigidities, sound will travel faster through the one with the lower density because the sound energy is more easily transferred from particle to particle. The speed of sound in air is low, because air is compressible. Because liquids and solids are relatively rigid and very difficult to compress, the speed of sound in such media is generally greater than in gases.

The speed of sound is affected by temperature in a given medium. For air at sea level, the speed of sound is given by

\[v_w = (331 \, m/s)\sqrt{\dfrac{T}{273 \, K}},\]

where the temperature (denoted as $T$) is in units of kelvin. The speed of sound in gases is related to the average speed of particles in the gas, $v_{rms}$, and that

\[v_{rms} = \sqrt{\dfrac{3 \, kT}{m}},\]

where $k$ is the Boltzmann constant $(1.38 \times 10^{-23} \, J/K)$ and $m$ is the mass of each (identical) particle in the gas. So, it is reasonable that the speed of sound in air and other gases should depend on the square root of temperature. While not negligible, this is not a strong dependence. At $0^oC$, the speed of sound is 331 m/s, whereas at $20^oC$ it is 343 m/s, less than a 4% increase. Figure $\PageIndex{3}$ shows a use of the speed of sound by a bat to sense distances. Echoes are also used in medical imaging.

The picture is of a bat trying to catch its prey an insect using sound echoes. The incident sound and sound reflected from the bat are shown as semicircular arcs.

One of the more important properties of sound is that its speed is nearly independent of frequency. This independence is certainly true in open air for sounds in the audible range of 20 to 20,000 Hz. If this independence were not true, you would certainly notice it for music played by a marching band in a football stadium, for example. Suppose that high-frequency sounds traveled faster—then the farther you were from the band, the more the sound from the low-pitch instruments would lag that from the high-pitch ones. But the music from all instruments arrives in cadence independent of distance, and so all frequencies must travel at nearly the same speed. Recall that

\[v_w = f\lambda.\]

In a given medium under fixed conditions, $v_w$ is constant, so that there is a relationship between $f$ and $\lambda$; the higher the frequency, the smaller the wavelength. See Figure $\PageIndex{4}$ and consider the following example.

Picture of a speaker having a woofer and a tweeter. High frequency sound coming out of the woofer shown as small circles closely spaced. Low frequency sound coming out of tweeter are shown as larger circles distantly spaced.

Example $\PageIndex{1}$: Calculating Wavelengths: What Are the Wavelengths of Audible Sounds?

Calculate the wavelengths of sounds at the extremes of the audible range, 20 and 20,000 Hz, in $30.0^oC$ air. (Assume that the frequency values are accurate to two significant figures.)

To find wavelength from frequency, we can use $v_w = f\lambda$.

Identify knowns. The value for $v_w$, is given by \[v_w = (331 \, m/s)\sqrt{\dfrac{T}{273 \, K}}. \nonumber\]
Convert the temperature into kelvin and then enter the temperature into the equation \[v_w = (331 \, m/s)\sqrt{\dfrac{303 \, K}{273 \, K}} = 348.7 \, m/s. \nonumber\]
Solve the relationship between speed and wavelength for $\lambda$: \[\lambda = \dfrac{v_w}{f}. \nonumber \]
Enter the speed and the minimum frequency to give the maximum wavelength: \[\lambda_{max} = \dfrac{348.7 \, m/s}{20 \, Hz} = 17 \, m. \nonumber\]
Enter the speed and the maximum frequency to give the minimum wavelength: \[\lambda_{min} = \dfrac{348.7 \, m/s}{20,000 \, Hz} = 0.017 \, m = 1.7 \, cm. \nonumber\]

Because the product of $f$ multiplied by $\lambda$ equals a constant, the smaller $f$ is, the larger $\lambda$ must be, and vice versa.

The speed of sound can change when sound travels from one medium to another. However, the frequency usually remains the same because it is like a driven oscillation and has the frequency of the original source. If $v_w$ changes and $f$ remains the same, then the wavelength $\lambda$ must change. That is, because $v_w = f\lambda$, the higher the speed of a sound, the greater its wavelength for a given frequency.

MAKING CONNECTIONS: TAKE-HOME INVESTIGATION - VOICE AS A SOUND WAVE

Suspend a sheet of paper so that the top edge of the paper is fixed and the bottom edge is free to move. You could tape the top edge of the paper to the edge of a table. Gently blow near the edge of the bottom of the sheet and note how the sheet moves. Speak softly and then louder such that the sounds hit the edge of the bottom of the paper, and note how the sheet moves. Explain the effects.

Exercise $\PageIndex{1A}$

Imagine you observe two fireworks explode. You hear the explosion of one as soon as you see it. However, you see the other firework for several milliseconds before you hear the explosion. Explain why this is so.

Sound and light both travel at definite speeds. The speed of sound is slower than the speed of light. The first firework is probably very close by, so the speed difference is not noticeable. The second firework is farther away, so the light arrives at your eyes noticeably sooner than the sound wave arrives at your ears.

Exercise $\PageIndex{1B}$

You observe two musical instruments that you cannot identify. One plays high-pitch sounds and the other plays low-pitch sounds. How could you determine which is which without hearing either of them play?

Compare their sizes. High-pitch instruments are generally smaller than low-pitch instruments because they generate a smaller wavelength.

The relationship of the speed of sound $v_w$, its frequency $f$, and its wavelength $\lambda$ is given by $v_w = f\lambda,$ which is the same relationship given for all waves.
In air, the speed of sound is related to air temperature $T$ by $v_w = (331 \, m/s) \sqrt{\dfrac{T}{273 \, K}}.$ $v_w$ is the same for all frequencies and wavelengths.

17.2 Speed of Sound, Frequency, and Wavelength

Learning objectives.

By the end of this section, you will be able to:

Define pitch.
Describe the relationship between the speed of sound, its frequency, and its wavelength.
Describe the effects on the speed of sound as it travels through various media.
Describe the effects of temperature on the speed of sound.

Sound, like all waves, travels at a certain speed and has the properties of frequency and wavelength. You can observe direct evidence of the speed of sound while watching a fireworks display. The flash of an explosion is seen well before its sound is heard, implying both that sound travels at a finite speed and that it is much slower than light. You can also directly sense the frequency of a sound. Perception of frequency is called pitch . The wavelength of sound is not directly sensed, but indirect evidence is found in the correlation of the size of musical instruments with their pitch. Small instruments, such as a piccolo, typically make high-pitch sounds, while large instruments, such as a tuba, typically make low-pitch sounds. High pitch means small wavelength, and the size of a musical instrument is directly related to the wavelengths of sound it produces. So a small instrument creates short-wavelength sounds. Similar arguments hold that a large instrument creates long-wavelength sounds.

The relationship of the speed of sound, its frequency, and wavelength is the same as for all waves:

where v w v w is the speed of sound, f f is its frequency, and λ λ is its wavelength. The wavelength of a sound is the distance between adjacent identical parts of a wave—for example, between adjacent compressions as illustrated in Figure 17.8 . The frequency is the same as that of the source and is the number of waves that pass a point per unit time.

Table 17.1 makes it apparent that the speed of sound varies greatly in different media. The speed of sound in a medium is determined by a combination of the medium’s rigidity (or compressibility in gases) and its density. The more rigid (or less compressible) the medium, the faster the speed of sound. For materials that have similar rigidities, sound will travel faster through the one with the lower density because the sound energy is more easily transferred from particle to particle. The speed of sound in air is low, because air is compressible. Because liquids and solids are relatively rigid and very difficult to compress, the speed of sound in such media is generally greater than in gases.

Earthquakes, essentially sound waves in Earth’s crust, are an interesting example of how the speed of sound depends on the rigidity of the medium. Earthquakes have both longitudinal and transverse components, and these travel at different speeds. The bulk modulus of granite is greater than its shear modulus. For that reason, the speed of longitudinal or pressure waves (P-waves) in earthquakes in granite is significantly higher than the speed of transverse or shear waves (S-waves). Both components of earthquakes travel slower in less rigid material, such as sediments. P-waves have speeds of 4 to 7 km/s, and S-waves correspondingly range in speed from 2 to 5 km/s, both being faster in more rigid material. The P-wave gets progressively farther ahead of the S-wave as they travel through Earth’s crust. The time between the P- and S-waves is routinely used to determine the distance to their source, the epicenter of the earthquake. The time and nature of these wave differences also provides the evidence for the nature of Earth's core. Through careful analysis of seismographic records of large earthquakes whose waves could be clearly detected around the world, Richard Dixon Oldham established that waves passing through the center of the Earth behaved as if they were moving through a different medium: a liquid. Later on, Inge Lehmann used more precise observations (partly based on a better coordinated network of seismographs she helped set up) to better define the nature of the core: that it was a solid inner core surrounded by a liquid outer core.

The speed of sound is affected by temperature in a given medium. For air at sea level, the speed of sound is given by

where the temperature (denoted as T T ) is in units of kelvin. The speed of sound in gases is related to the average speed of particles in the gas, v rms v rms , and that

where k k is the Boltzmann constant ( 1.38 × 10 −23 J/K 1.38 × 10 −23 J/K ) and m m is the mass of each (identical) particle in the gas. So, it is reasonable that the speed of sound in air and other gases should depend on the square root of temperature. While not negligible, this is not a strong dependence. At 0ºC 0ºC , the speed of sound is 331 m/s, whereas at 20.0ºC 20.0ºC it is 343 m/s, less than a 4% increase. Figure 17.9 shows a use of the speed of sound by a bat to sense distances. Echoes are also used in medical imaging.

In a given medium under fixed conditions, v w v w is constant, so that there is a relationship between f f and λ λ ; the higher the frequency, the smaller the wavelength. See Figure 17.10 and consider the following example.

Example 17.1

Calculating wavelengths: what are the wavelengths of audible sounds.

Calculate the wavelengths of sounds at the extremes of the audible range, 20 and 20,000 Hz, in 30.0ºC 30.0ºC air. (Assume that the frequency values are accurate to two significant figures.)

To find wavelength from frequency, we can use v w = fλ v w = fλ .

Identify knowns. The value for v w v w , is given by v w = 331 m/s T 273 K . v w = 331 m/s T 273 K . 17.5
Convert the temperature into kelvin and then enter the temperature into the equation v w = 331 m/s 303 K 273 K = 348 . 7 m/s . v w = 331 m/s 303 K 273 K = 348 . 7 m/s . 17.6
Solve the relationship between speed and wavelength for λ λ : λ = v w f . λ = v w f . 17.7
Enter the speed and the minimum frequency to give the maximum wavelength: λ max = 348 . 7 m/s 20 Hz = 17 m . λ max = 348 . 7 m/s 20 Hz = 17 m . 17.8
Enter the speed and the maximum frequency to give the minimum wavelength: λ min = 348 . 7 m/s 20 , 000 Hz = 0 . 017 m = 1 . 7 cm . λ min = 348 . 7 m/s 20 , 000 Hz = 0 . 017 m = 1 . 7 cm . 17.9

Because the product of f f multiplied by λ λ equals a constant, the smaller f f is, the larger λ λ must be, and vice versa.

The speed of sound can change when sound travels from one medium to another. However, the frequency usually remains the same because it is like a driven oscillation and has the frequency of the original source. If v w v w changes and f f remains the same, then the wavelength λ λ must change. That is, because v w = fλ v w = fλ , the higher the speed of a sound, the greater its wavelength for a given frequency.

Making Connections: Take-Home Investigation—Voice as a Sound Wave

Check Your Understanding

Compare their sizes. High-pitch instruments are generally smaller than low-pitch instruments because they generate a smaller wavelength.

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90 Speed of Sound, Frequency, and Wavelength

[latexpage]

Learning Objectives

Define pitch.
Describe the relationship between the speed of sound, its frequency, and its wavelength.
Describe the effects on the speed of sound as it travels through various media.
Describe the effects of temperature on the speed of sound.

Sound, like all waves, travels at a certain speed and has the properties of frequency and wavelength. You can observe direct evidence of the speed of sound while watching a fireworks display. The flash of an explosion is seen well before its sound is heard, implying both that sound travels at a finite speed and that it is much slower than light. You can also directly sense the frequency of a sound. Perception of frequency is called pitch . The wavelength of sound is not directly sensed, but indirect evidence is found in the correlation of the size of musical instruments with their pitch. Small instruments, such as a piccolo, typically make high-pitch sounds, while large instruments, such as a tuba, typically make low-pitch sounds. High pitch means small wavelength, and the size of a musical instrument is directly related to the wavelengths of sound it produces. So a small instrument creates short-wavelength sounds. Similar arguments hold that a large instrument creates long-wavelength sounds.

The relationship of the speed of sound, its frequency, and wavelength is the same as for all waves:

where ${v}_{w}$ is the speed of sound, $f$ is its frequency, and $\lambda $ is its wavelength. The wavelength of a sound is the distance between adjacent identical parts of a wave—for example, between adjacent compressions as illustrated in (Figure) . The frequency is the same as that of the source and is the number of waves that pass a point per unit time.

(Figure) makes it apparent that the speed of sound varies greatly in different media. The speed of sound in a medium is determined by a combination of the medium’s rigidity (or compressibility in gases) and its density. The more rigid (or less compressible) the medium, the faster the speed of sound. This observation is analogous to the fact that the frequency of a simple harmonic motion is directly proportional to the stiffness of the oscillating object. The greater the density of a medium, the slower the speed of sound. This observation is analogous to the fact that the frequency of a simple harmonic motion is inversely proportional to the mass of the oscillating object. The speed of sound in air is low, because air is compressible. Because liquids and solids are relatively rigid and very difficult to compress, the speed of sound in such media is generally greater than in gases.

The speed of sound is affected by temperature in a given medium. For air at sea level, the speed of sound is given by

where the temperature (denoted as $T$) is in units of kelvin. The speed of sound in gases is related to the average speed of particles in the gas, ${v}_{\text{rms}}$, and that

where $k$ is the Boltzmann constant ($1.38×{\text{10}}^{\text{−23}}\phantom{\rule{0.25em}{0ex}}\text{J/K}$) and $m$ is the mass of each (identical) particle in the gas. So, it is reasonable that the speed of sound in air and other gases should depend on the square root of temperature. While not negligible, this is not a strong dependence. At $\text{0ºC}$, the speed of sound is 331 m/s, whereas at $\text{20.0ºC}$ it is 343 m/s, less than a 4% increase. (Figure) shows a use of the speed of sound by a bat to sense distances. Echoes are also used in medical imaging.

In a given medium under fixed conditions, ${v}_{\text{w}}$ is constant, so that there is a relationship between $f$ and $\lambda $ ; the higher the frequency, the smaller the wavelength. See (Figure) and consider the following example.

Calculate the wavelengths of sounds at the extremes of the audible range, 20 and 20,000 Hz, in $\text{30.0ºC}$ air. (Assume that the frequency values are accurate to two significant figures.)

To find wavelength from frequency, we can use ${v}_{\text{w}}=\mathrm{f\lambda }$.

Identify knowns. The value for ${v}_{\text{w}}$, is given by ${v}_{\text{w}}=\left(\text{331}\phantom{\rule{0.25em}{0ex}}\text{m/s}\right)\sqrt{\frac{T}{\text{273}\phantom{\rule{0.25em}{0ex}}\text{K}}}.$
Convert the temperature into kelvin and then enter the temperature into the equation ${v}_{\text{w}}=\left(\text{331}\phantom{\rule{0.25em}{0ex}}\text{m/s}\right)\sqrt{\frac{\text{303 K}}{\text{273}\phantom{\rule{0.25em}{0ex}}\text{K}}}=\text{348}\text{.}7\phantom{\rule{0.25em}{0ex}}\text{m/s}.$
Solve the relationship between speed and wavelength for $\lambda $ : $\lambda =\frac{{v}_{w}}{f}.$
Enter the speed and the minimum frequency to give the maximum wavelength: ${\lambda }_{\text{max}}=\frac{\text{348}\text{.}7\phantom{\rule{0.25em}{0ex}}\text{m/s}}{\text{20 Hz}}=\text{17}\phantom{\rule{0.25em}{0ex}}\text{m}.$
Enter the speed and the maximum frequency to give the minimum wavelength: ${\lambda }_{\text{min}}=\frac{\text{348}\text{.}7\phantom{\rule{0.25em}{0ex}}\text{m/s}}{\text{20}\text{,}\text{000 Hz}}=0\text{.}\text{017}\phantom{\rule{0.25em}{0ex}}\text{m}=1\text{.}\text{7 cm}.$

Because the product of $f$ multiplied by $\lambda $ equals a constant, the smaller $f$ is, the larger $\lambda $ must be, and vice versa.

The speed of sound can change when sound travels from one medium to another. However, the frequency usually remains the same because it is like a driven oscillation and has the frequency of the original source. If ${v}_{\text{w}}$ changes and $f$ remains the same, then the wavelength $\lambda $ must change. That is, because ${v}_{\text{w}}=\mathrm{f\lambda }$, the higher the speed of a sound, the greater its wavelength for a given frequency.

Compare their sizes. High-pitch instruments are generally smaller than low-pitch instruments because they generate a smaller wavelength.

Section Summary

The relationship of the speed of sound ${v}_{w}$, its frequency $f$, and its wavelength $\lambda $ is given by

which is the same relationship given for all waves.

In air, the speed of sound is related to air temperature $T$ by

${v}_{\text{w}}$ is the same for all frequencies and wavelengths.

Conceptual Questions

How do sound vibrations of atoms differ from thermal motion?

When sound passes from one medium to another where its propagation speed is different, does its frequency or wavelength change? Explain your answer briefly.

Problems & Exercises

When poked by a spear, an operatic soprano lets out a 1200-Hz shriek. What is its wavelength if the speed of sound is 345 m/s?

What frequency sound has a 0.10-m wavelength when the speed of sound is 340 m/s?

Calculate the speed of sound on a day when a 1500 Hz frequency has a wavelength of 0.221 m.

(a) What is the speed of sound in a medium where a 100-kHz frequency produces a 5.96-cm wavelength? (b) Which substance in (Figure) is this likely to be?

Show that the speed of sound in $\text{20.0ºC}$ air is 343 m/s, as claimed in the text.

Air temperature in the Sahara Desert can reach $\text{56.0ºC}$ (about $\text{134ºF}$). What is the speed of sound in air at that temperature?

Dolphins make sounds in air and water. What is the ratio of the wavelength of a sound in air to its wavelength in seawater? Assume air temperature is $\text{20.0ºC}$.

A sonar echo returns to a submarine 1.20 s after being emitted. What is the distance to the object creating the echo? (Assume that the submarine is in the ocean, not in fresh water.)

(b) Discuss the limits this time resolution imposes on the ability of the sonar system to detect the size and shape of the object creating the echo.

(b) This means that sonar is good for spotting and locating large objects, but it isn’t able to resolve smaller objects, or detect the detailed shapes of objects. Objects like ships or large pieces of airplanes can be found by sonar, while smaller pieces must be found by other means.

A physicist at a fireworks display times the lag between seeing an explosion and hearing its sound, and finds it to be 0.400 s. (a) How far away is the explosion if air temperature is $\text{24.0ºC}$ and if you neglect the time taken for light to reach the physicist? (b) Calculate the distance to the explosion taking the speed of light into account. Note that this distance is negligibly greater.

Suppose a bat uses sound echoes to locate its insect prey, 3.00 m away. (See (Figure) .) (a) Calculate the echo times for temperatures of $\text{5.00ºC}$ and $\text{35.0ºC}$. (b) What percent uncertainty does this cause for the bat in locating the insect? (c) Discuss the significance of this uncertainty and whether it could cause difficulties for the bat. (In practice, the bat continues to use sound as it closes in, eliminating most of any difficulties imposed by this and other effects, such as motion of the prey.)

(a) 18.0 ms, 17.1 ms

(c) This uncertainty could definitely cause difficulties for the bat, if it didn’t continue to use sound as it closed in on its prey. A 5% uncertainty could be the difference between catching the prey around the neck or around the chest, which means that it could miss grabbing its prey.

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17 Physics of Hearing

129 17.2 Speed of Sound, Frequency, and Wavelength

Define pitch.
Describe the relationship between the speed of sound, its frequency, and its wavelength.
Describe the effects on the speed of sound as it travels through various media.
Describe the effects of temperature on the speed of sound.

Sound, like all waves, travels at a certain speed and has the properties of frequency and wavelength. You can observe direct evidence of the speed of sound while watching a fireworks display. The flash of an explosion is seen well before its sound is heard, implying both that sound travels at a finite speed and that it is much slower than light. You can also directly sense the frequency of a sound. Perception of frequency is called pitch . The wavelength of sound is not directly sensed, but indirect evidence is found in the correlation of the size of musical instruments with their pitch. Small instruments, such as a piccolo, typically make high-pitch sounds, while large instruments, such as a tuba, typically make low-pitch sounds. High pitch means small wavelength, and the size of a musical instrument is directly related to the wavelengths of sound it produces. So a small instrument creates short-wavelength sounds. Similar arguments hold that a large instrument creates long-wavelength sounds.

The relationship of the speed of sound, its frequency, and wavelength is the same as for all waves:

where[latex]\boldsymbol{v_{\textbf{w}}}[/latex]is the speed of sound,[latex]\boldsymbol{f}[/latex]is its frequency, and[latex]\boldsymbol{\lambda}[/latex]is its wavelength. The wavelength of a sound is the distance between adjacent identical parts of a wave—for example, between adjacent compressions as illustrated in Figure 2 . The frequency is the same as that of the source and is the number of waves that pass a point per unit time.

Table 1 makes it apparent that the speed of sound varies greatly in different media. The speed of sound in a medium is determined by a combination of the medium’s rigidity (or compressibility in gases) and its density. The more rigid (or less compressible) the medium, the faster the speed of sound. This observation is analogous to the fact that the frequency of a simple harmonic motion is directly proportional to the stiffness of the oscillating object. The greater the density of a medium, the slower the speed of sound. This observation is analogous to the fact that the frequency of a simple harmonic motion is inversely proportional to the mass of the oscillating object. The speed of sound in air is low, because air is compressible. Because liquids and solids are relatively rigid and very difficult to compress, the speed of sound in such media is generally greater than in gases.

The speed of sound is affected by temperature in a given medium. For air at sea level, the speed of sound is given by

where the temperature (denoted as[latex]\boldsymbol{T})[/latex]is in units of kelvin. The speed of sound in gases is related to the average speed of particles in the gas,[latex]\boldsymbol{v_{\textbf{rms}}},[/latex]and that

where[latex]\boldsymbol{k}[/latex]is the Boltzmann constant[latex](\boldsymbol{1.38\times10^{-23}\textbf{ J/K}})[/latex]and[latex]\boldsymbol{m}[/latex]is the mass of each (identical) particle in the gas. So, it is reasonable that the speed of sound in air and other gases should depend on the square root of temperature. While not negligible, this is not a strong dependence. At[latex]\boldsymbol{0^{\circ}\textbf{C}},[/latex]the speed of sound is 331 m/s, whereas at[latex]\boldsymbol{20.0^{\circ}\textbf{C}}[/latex]it is 343 m/s, less than a 4% increase. Figure 3 shows a use of the speed of sound by a bat to sense distances. Echoes are also used in medical imaging.

In a given medium under fixed conditions,[latex]\boldsymbol{v_{\textbf{w}}}[/latex]is constant, so that there is a relationship between[latex]\boldsymbol{f}[/latex]and[latex]\boldsymbol{\lambda};[/latex]the higher the frequency, the smaller the wavelength. See Figure 4 and consider the following example.

Example 1: Calculating Wavelengths: What Are the Wavelengths of Audible Sounds?

Calculate the wavelengths of sounds at the extremes of the audible range, 20 and 20,000 Hz, in[latex]\boldsymbol{30.0^{\circ}\textbf{C}}[/latex]air. (Assume that the frequency values are accurate to two significant figures.)

To find wavelength from frequency, we can use[latex]\boldsymbol{v_{\textbf{w}}=f\lambda}.[/latex]

Identify knowns. The value for[latex]\boldsymbol{v_{\textbf{w}}},[/latex]is given by [latex]\boldsymbol{v_{\textbf{w}}=(331\textbf{ m/s})}[/latex][latex]\boldsymbol{\sqrt{\frac{T}{273\textbf{ K}}}}.[/latex]
Convert the temperature into kelvin and then enter the temperature into the equation [latex]\boldsymbol{v_{\textbf{w}}=(331\textbf{ m/s})}[/latex][latex]\boldsymbol{\sqrt{\frac{303\textbf{ K}}{273\textbf{ K}}}}[/latex][latex]\boldsymbol{=348.7\textbf{ m/s}}.[/latex]
Solve the relationship between speed and wavelength for[latex]\boldsymbol{\lambda}:[/latex] [latex]\boldsymbol{\lambda\:=}[/latex][latex]\boldsymbol{\frac{v_{\textbf{w}}}{f}}.[/latex]
Enter the speed and the minimum frequency to give the maximum wavelength: [latex]\boldsymbol{\lambda_{\textbf{max}}\:=}[/latex][latex]\boldsymbol{\frac{348.7\textbf{ m/s}}{20\textbf{ Hz}}}[/latex][latex]\boldsymbol{=17\textbf{ m}}.[/latex]
Enter the speed and the maximum frequency to give the minimum wavelength: [latex]\boldsymbol{\lambda_{\textbf{min}}=}[/latex][latex]\boldsymbol{\frac{348.7\textbf{ m/s}}{20,000\textbf{ Hz}}}[/latex][latex]\boldsymbol{=0.017\textbf{ m}=1.7\textbf{ cm}}.[/latex]

Because the product of[latex]\boldsymbol{f}[/latex]multiplied by[latex]\boldsymbol{\lambda}[/latex]equals a constant, the smaller[latex]\boldsymbol{f}[/latex]is, the larger[latex]\boldsymbol{\lambda}[/latex]must be, and vice versa.

The speed of sound can change when sound travels from one medium to another. However, the frequency usually remains the same because it is like a driven oscillation and has the frequency of the original source. If[latex]\boldsymbol{v_{\textbf{w}}}[/latex]changes and[latex]\boldsymbol{f}[/latex]remains the same, then the wavelength[latex]\boldsymbol{\lambda}[/latex]must change. That is, because[latex]\boldsymbol{v_{\textbf{w}}=f\lambda}[/latex]the higher the speed of a sound, the greater its wavelength for a given frequency.

Making Connections: Take-Home Investigation—Voice as a Sound Wave

Check Your Understanding 1

Check Your Understanding 2

Section Summary

The relationship of the speed of sound[latex]\boldsymbol{v_{\textbf{w}}},[/latex]its frequency[latex]\boldsymbol{f},[/latex]and its wavelength[latex]\boldsymbol{\lambda}[/latex]is given by

which is the same relationship given for all waves.

In air, the speed of sound is related to air temperature[latex]\boldsymbol{T}[/latex]by

[latex]\boldsymbol{v_{\textbf{w}}}[/latex]is the same for all frequencies and wavelengths.

Conceptual Questions

1: How do sound vibrations of atoms differ from thermal motion?

2: When sound passes from one medium to another where its propagation speed is different, does its frequency or wavelength change? Explain your answer briefly.

1: When poked by a spear, an operatic soprano lets out a 1200-Hz shriek. What is its wavelength if the speed of sound is 345 m/s?

2: What frequency sound has a 0.10-m wavelength when the speed of sound is 340 m/s?

3: Calculate the speed of sound on a day when a 1500 Hz frequency has a wavelength of 0.221 m.

4: (a) What is the speed of sound in a medium where a 100-kHz frequency produces a 5.96-cm wavelength? (b) Which substance in Table 1 is this likely to be?

5: Show that the speed of sound in[latex]\boldsymbol{20.0^{\circ}\textbf{C}}[/latex]air is 343 m/s, as claimed in the text.

6: Air temperature in the Sahara Desert can reach[latex]\boldsymbol{56.0^{\circ}\textbf{C}}[/latex](about[latex]\boldsymbol{134^{\circ}\textbf{F}}[/latex]). What is the speed of sound in air at that temperature?

7: Dolphins make sounds in air and water. What is the ratio of the wavelength of a sound in air to its wavelength in seawater? Assume air temperature is[latex]\boldsymbol{20.0^{\circ}\textbf{C}}.[/latex]

8: A sonar echo returns to a submarine 1.20 s after being emitted. What is the distance to the object creating the echo? (Assume that the submarine is in the ocean, not in fresh water.)

9: (a) If a submarine’s sonar can measure echo times with a precision of 0.0100 s, what is the smallest difference in distances it can detect? (Assume that the submarine is in the ocean, not in fresh water.)

(b) Discuss the limits this time resolution imposes on the ability of the sonar system to detect the size and shape of the object creating the echo.

10: A physicist at a fireworks display times the lag between seeing an explosion and hearing its sound, and finds it to be 0.400 s. (a) How far away is the explosion if air temperature is[latex]\boldsymbol{24.0^{\circ}\textbf{C}}[/latex]and if you neglect the time taken for light to reach the physicist? (b) Calculate the distance to the explosion taking the speed of light into account. Note that this distance is negligibly greater.

11: Suppose a bat uses sound echoes to locate its insect prey, 3.00 m away. (See Figure 3 .) (a) Calculate the echo times for temperatures of[latex]\boldsymbol{5.00^{\circ}\textbf{C}}[/latex]and[latex]\boldsymbol{35.0^{\circ}\textbf{C}}.[/latex](b) What percent uncertainty does this cause for the bat in locating the insect? (c) Discuss the significance of this uncertainty and whether it could cause difficulties for the bat. (In practice, the bat continues to use sound as it closes in, eliminating most of any difficulties imposed by this and other effects, such as motion of the prey.)

Compare their sizes. High-pitch instruments are generally smaller than low-pitch instruments because they generate a smaller wavelength.

Problems & Exercises

[latex]\begin{array}{lcl} \boldsymbol{v_{\textbf{w}}} & \boldsymbol{=} & \boldsymbol{(331\textbf{ m/s})\sqrt{\frac{T}{273\textbf{ K}}}=(331\textbf{ m/s})\sqrt{\frac{293\textbf{ K}}{273\textbf{ K}}}} \\ {} & \boldsymbol{=} & \boldsymbol{343\textbf{ m/s}} \end{array}[/latex]

(a) 18.0 ms, 17.1 ms

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Useful Info

46 Solving physics problems

Stay organized.

Over the last twenty years, I’ve watched lots of students struggle doing physics problems on homework sets and tests. I’ve seen them get wrong answers and not be able to track down their mistakes. Sometimes, it’s because the student is confused about the physics. More often, though, it’s because the student is trying to “jump to the answer”- trying random things in hopes of pumping out a right answer quickly. It works (sometimes), too- otherwise students wouldn’t do it. B ut it is a recipe for disaster for anything but the simplest “formula plug and chug” type of textbook exercises.

The key to solving science problems reliably and quickly is a systematic approach. Worked examples in physics books follow a common format- with good reason. The exact method varies, but the basic steps are universal. Versions of the template below have been published in education journals and are written into almost every physics textbook. Physicists use this approach because it works. Learn the method and use it!

At first, you might think this method is a waste of paper and/or time. Over time, you will develop your own streamlined version of this template- and that’s great. However, keep the essential elements in mind- 1) stay organized and 2) write all the important stuff down- and you can expect success.

Physics problem solving template

Identify the important physics concept in the problem. In this book, one sentence and one equation are usually enough.
Draw an appropriate diagram. In many problems, a diagram will help you remember and/or recognize important details.
List known and unknown quantities with letter names and units. This step is crucial. It is where you get organized. It is where you identify the info that will help you get to the answer (as well as irrelevant info put there to distract you). Get these details right, or expect trouble down the road.
Do the algebra. This means turning the equation(s) around so that the quantity you want to find is alone on the left hand side of the equation. Don’t plug in any numbers yet!
Do any necessary unit conversions and then plug in numbers. Do unit conversions (if needed). Replace the letters in equations with the correct numbers. Then pull out the calculator.
Could the answer possibly be correct? Does it have the right units? Is it in the right “ballpark”?
What did you learn from the problem? Pay attention to what just happened. Did you have any questions? Was there anything that made you stop and take notice?

Don’t believe me?

Check out these websites. Their advice has a lot in common with mine…

http://blog.cambridgecoaching.com/4-tricks-for-solving-any-physics-problem
http://www.smarterthanthat.com/physics/physics-dont-panic-10-steps-to-solving-most-physics-problems/
https://youtu.be/YocWuzi4JhY
https://youtu.be/ywZPAsM1FeU

Keep in mind that the example you are about to read is written by a textbook author for a physics student. When you write your own solutions, you will be writing for your own study purposes, or perhaps for a grader. What you produce will almost certainly involve a lot of shorthand- and that’s OK.

Example: To the bat cave

Bats figure out the distance to nearby objects by emitting short bursts of ultrasonic sound and “listening” for the echo. Fred, the house bat, emits a 50 kHz ultrasonic “chirp” that lasts for just 0.1 milliseconds and hears the echo 8.0 ms later. How far away is the obstacle that caused the echo?

Identify important physics concept : This problem is about how sound travels. The important physics to know is that sounds in air all travel at the same constant speed- no matter what. The important equation is

[latex]v = \frac{d} {t}[/latex]

Diagram: A diagram for this problem underscores an important feature in this problem: sound goes from the bat to the obstacle and back again before the bat hears the chirp.

List known and unknown quantities (with letter names and units):

This problem provides too much information . Knowing the physics helps sort the useful info from the useless. To solve the problem, all that’s really needed is 1) the speed of sound in air and 2) the amount of time it takes for the sound to travel from bat to obstacle. The “50 kHz” is irrelevant- all sound travels at the same speed, no matter what the frequency is. How long the sound lasts- the “0.1 milliseconds”- is also irrelevant. I do, however, need to know the speed of sound in air- Google says it’s 343 m/s at 20 °C (room temperature). Since the problem did not provide the temperature, I will assume the air in the cave is at 20 °C.

[latex]d = ?[/latex]

[latex]v=343 \: \frac{m}{s}[/latex]

The sound takes 8.0 ms for the round trip from bat to obstacle and back, so it takes sound half that time to go one way. (The problem asks for the distance from bat to obstacle- not distance from bat to obstacle and back).

[latex]t=4.0 \: ms[/latex]

Do the algebra: Solve [latex]v = \frac{d} {t}[/latex] for distance: [latex]d=vt[/latex]).

Do unit conversions (if needed) then plug in numbers: Units are not consistent here- speed is given in m/s and time is milliseconds- so unit conversion is needed. It’s probably easiest to convert the milliseconds to seconds:

[latex]4.0 \: ms =0.004 \: s[/latex]

Then plug the number into the equation:

[latex]d=vt=(343 \: m/s)(0.0040 \: s)=1.36 \: m[/latex]

Reflect on the answer:

Answer is pretty small- 1.36 meters is a little more than a yard?! This bat’s really close to something! Is that right? Yes, as it turns out. Google the speed of sound in feet per second, you get 1125 ft/sec- or about 1.1 ft per millisecond. (Sound engineers routinely approximate the speed of sound as 1 foot per millisecond for figuring time delays on stage).
Units work.

How it looks on a student paper

Obviously, as a student, you aren’t going to write out all the steps the same way a textbook author does- it’s just too time consuming. The picture below shows what a student solution for the bat-obstacle problem above might look in a student notebook or homework paper. Notice that the essential elements are all here- a sketch, a list of known and unknown quantities with units, algebra, numbers plugged in with units and a short evaluation of the answer.

Students who are a little more careful might also include additional stuff, like a list of the extraneous givens (the 50 kHz and 0.1 ms) and why each could be ignored and/or a note about the pitfalls in the problem (“don’t forget to divide by two for out and back”).

Share This Book

Physics Formulas

Speed Of Sound Formula

The formula of the speed of sound formula is expressed as

P = pressure

ρ = density

γ = Ratio of specific heat.

The sound wave with density o.o43 kg/m 3 and pressure of 3kPa having the temp 3 0 C travels in the air. Find out the speed of the sound?

Temperature T = 276 K

Density ρ = 0.043 kg/m 3

Pressure p = 3kPa = 3000 Pa

The ratio of specific heat in air = 1.4

The formula for speed of sound is

c = √[ γ×(P/ρ)]

c = √[1.4× (3000 / 0.043)]

Therefore, speed of sound = 312.52 m/s

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Waves, Sound and Light: Sound Waves

Is it a sound of music...or of speech? Scientists uncover how our brains try to tell the difference

Slow and steady waves sound like music while faster and irregular ones like speech.

Music and speech are among the most frequent types of sounds we hear. But how do we identify what we think are differences between the two?

An international team of researchers mapped out this process through a series of experiments -- yielding insights that offer a potential means to optimize therapeutic programs that use music to regain the ability to speak in addressing aphasia. This language disorder afflicts more than 1 in 300 Americans each year, including Wendy Williams and Bruce Willis.

"Although music and speech are different in many ways, ranging from pitch to timbre to sound texture, our results show that the auditory system uses strikingly simple acoustic parameters to distinguish music and speech," explains Andrew Chang, a postdoctoral fellow in New York University's Department of Psychology and the lead author of the paper, which appears in the journal PLOS Biology . "Overall, slower and steady sound clips of mere noise sound more like music while the faster and irregular clips sound more like speech."

Scientists gauge the rate of signals by precise units of measurement: Hertz (Hz). A larger number of Hz means a greater number of occurrences (or cycles) per second than a lower number. For instance, people typically walk at a pace of 1.5 to 2 steps per second, which is 1.5-2 Hz. The beat of Stevie Wonder's 1972 hit "Superstition" is approximately 1.6 Hz, while Anna Karina's 1967 smash "Roller Girl" clocks in at 2 Hz. Speech, in contrast, is typically two to three times faster than that at 4-5 Hz.

It has been well documented that a song's volume, or loudness, over time -- what's known as "amplitude modulation" -- is relatively steady at 1-2 Hz. By contrast, the amplitude modulation of speech is typically 4-5 Hz, meaning its volume changes frequently.

Despite the ubiquity and familiarity of music and speech, scientists previously lacked clear understanding of how we effortlessly and automatically identify a sound as music or speech.

To better understand this process in their PLOS Biology study, Chang and colleagues conducted a series of four experiments in which more than 300 participants listened to a series of audio segments of synthesized music- and speech-like noise of various amplitude modulation speeds and regularity.

The audio noise clips allowed only the detection of volume and speed. The participants were asked to judge whether these ambiguous noise clips, which they were told were noise-masked music or speech, sounded like music or speech. Observing the pattern of participants sorting hundreds of noise clips as either music or speech revealed how much each speed and/or regularity feature affected their judgment between music and speech. It is the auditory version of "seeing faces in the cloud," the scientists conclude: If there's a certain feature in the soundwave that matches listeners' idea of how music or speech should be, even a white noise clip can sound like music or speech. Examples of both music and speech may be downloaded from the research page.

The results showed that our auditory system uses surprisingly simple and basic acoustic parameters to distinguish music and speech: to participants, clips with slower rates (<2Hz) and more regular amplitude modulation sounded more like music, while clips with higher rates (~4Hz) and more irregular amplitude modulation sounded more like speech.

Knowing how the human brain differentiates between music and speech can potentially benefit people with auditory or language disorders such as aphasia, the authors note. Melodic intonation therapy, for instance, is a promising approach to train people with aphasia to sing what they want to say, using their intact "musical mechanisms" to bypass damaged speech mechanisms. Therefore, knowing what makes music and speech similar or distinct in the brain can help design more effective rehabilitation programs.

The paper's other authors were Xiangbin Teng of Chinese University of Hong Kong, M. Florencia Assaneo of National Autonomous University of Mexico (UNAM), and David Poeppel, a professor in NYU's Department of Psychology and managing director of the Ernst Strüngmann Institute for Neuroscience in Frankfurt, Germany.

The research was supported by a grant from the National Institute on Deafness and Other Communication Disorders, part of the National Institutes of Health (F32DC018205), and Leon Levy Scholarships in Neuroscience.

Language Acquisition
Learning Disorders
Child Development
Neuroscience
Learning disability
Mensa International
Competition
Experimental economics
Hearing impairment

Story Source:

Materials provided by New York University . Original written by James Devitt. Note: Content may be edited for style and length.

Journal Reference :

Andrew Chang, Xiangbin Teng, M. Florencia Assaneo, David Poeppel. The human auditory system uses amplitude modulation to distinguish music from speech . PLOS Biology , 2024; 22 (5): e3002631 DOI: 10.1371/journal.pbio.3002631

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Zildjian’s Alchem-E e-drums solve one of the biggest problems with electronic percussion

By Stan Horaczek

Posted on May 26, 2024 10:00 AM EDT

6 minute read

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Stan Horaczek

Electronic drumsets have come a long way since the early days of fake-sounding samples and clunky rubber pads. Mesh heads and advanced computing hardware have made e-drums totally viable for recording and performing. Electronic cymbals, however, are a different story. Fake cymbals typically feel nothing like their traditional metal counterparts, which can really cramp a player’s style. It’s fitting, then, that Zildjian—a company continuously making cymbals since the 1600s —has discovered a solution to the e-cymbal problem and integrated it into its new Alchem-E drum kits.

How do electronic drums work?

E-drums work similarly to most other electronic instruments. An electronic keyboard , for instance, replaces hammers striking metal wires with sensors that cue electronic signals when you apply pressure to the keys. Electronic drums typically contain a sensor in a rubber pad or a traditional drum with a special mesh head. A central computing device triggers when you hit those inputs and translates action into whatever sound you desire. That can be heard through an amplified speaker, or, more importantly, they also work with headphones so that you can get in real practice sessions while drastically reducing noise to the outside world. To an external observer, it simply sounds like a series of clicks and taps instead of crashes and splashes, cracks and thwacks.

Zildjian’s Alchem-E drums offer real maple shells (a wood typically found in high-end kits) with mesh heads. The sensors inside detect multiple variables to tell how hard you hit and where the stick landed on the head. Hit the rim and drum simultaneously, and you get an extra-loud rim shot, just like with acoustic drums. Behind the kit, it all feels very natural. The bouncy mesh heads allow for rolls and precise changes in dynamics. They feel like very high-end electronic drums because that’s what they are. I’m a semi-experienced drummer, and the learning curve felt almost non-existent.

What about electronic cymbals?

Zildjian e-drumset with sticks on the snare

This is where Zildjian really sets itself apart from other electronic drum makers. The E-series cymbals sport an array of laser-cut holes, which drastically reduces the amount of metal involved. This feature reduces the cymbals’ overall sound output by roughly 80 percent while also cutting the duration of the sound. The company has been offering this type of cymbal as a lower-volume practice-oriented option since 2017, but now it’s fulfilling its electronic potential.

A sensor clips into the holes in the cymbal to create a connection with the electronic brain (which Zildjian calls the E-Vault). Once attached, the sensor can register many different inputs, whether a hard hit on the edge for a crash, a regular hit on the bow, or a precise whack on the bell for a defined ping.

These are real cymbals, so the feel is impeccable. You’re applying wood (or nylon tips in some cases) to actual metal alloy, so it’s as if you’re playing on a normal kit. I was very impressed with the responsiveness, especially in the context of my very sloppy technique. I played everything from tight ride patterns to aggressive crashes and got what I expected each time.

I could spend hours switching through the huge library of cymbals just to see how they sound with different types of hits. An eight-inch capacitive touchscreen allows players to scroll through pre-programmed and customizable kits designed to emulate common musical genres. Want to play hair metal with big boomy toms and a reverb-laden snare drum? There’s a preset for that. Or maybe you want to play a small jazz kit with meticulously tuned toms and super-dry cymbals. And Zildjian can update the E-Vault down the road to include even more sampled cymbal sounds.

One particularly impressive feature: You can choke the cymbals just like the real thing. When you want a short, punchy crash that doesn’t ring out over time, you can simply grab it to stop it from resonating. That’s typically a no-go for electronic cymbals, but Zildjian seems to have mastered it. The sampled cymbal sounds come from real models pulled from the company’s vault. When you choke the cymbal, the sound doesn’t simply stop; rather, it makes a unique and abrupt sound that a real cymbal makes when you mute its vibrations. If you need an example, fire up “Master of Puppets” by Metallica like I did during my demo time with this kit.

OK, so they feel good, but what do they sound like?

All the feel in the world isn’t worth anything if it doesn’t make pleasant noise. Zildjian has gone to great lengths to ensure the output is worthy of your input. The E-Vault computing captures activity from all of the sensors and translates it into whatever kit suits your style.

Each sound is carefully sampled from real drums and cymbals. The results are extremely convincing. I played through headphones and an amplified speaker and was very impressed in both cases. Once you start messing around with selecting specific cymbals and drum sounds, you could easily spend hours tweaking to get your exact preferences set up.

The E-Vault (shown below) accepts up to six drums and six cymbals at a time, so you can add more inputs down the road if you want more options. You could add a small special effects cymbal that you’d never actually buy in an acoustic version or just map one of the cymbals to a dog bark sound. They are electric, after all.

Playing quietly

The Zildjian brain for its electronic drumset

As stated before, Zildjian’s perforated speakers provide an 80 percent reduction in overall sound output compared to a typical model. It’s hard to convey just how impressive the reduction is without hearing it in person. Or not hearing it, in this case. Even when playing with relatively normal technique, the overall sound output to the outside world is still apartment-friendly. You wouldn’t want to play in the middle of the night, but it’s overall less disturbing to those around you than watching an action movie on a typical soundbar.

Built-in Bluetooth makes the E-Vault compatible with computers and smartphones, so you can stream your favorite songs or practice tracks from an external device and play along. With headphones on , it almost feels like you’re playing with a real band.

The Alchem-E kit options

Zildjian designed Alchem-E for professionals or high-end amateurs, so the kits aren’t cheap. All of the kits come with a single bass drum, hardware, the E-Vault, a snare drum, toms, and cymbals. The least expensive model is the $4,500 Bronze EX kit , which includes shortened shells with no resonant heads for each of the toms. The bass drum is also shallower than a typical option. This is the most portable kit.

The $5,999 Alchem-E Gold kit offers full-length maple shells for the toms and the bass drum. Since they have resonant heads on the back, you can simply swap out the mesh heads for more traditional skins and use this as a standard acoustic drum set. It’s a typical four-piece configuration with one rack tom and one floor tom. The cymbal kit includes a standard ride, crash, and high-hat. The flagship Gold EX setup costs $6,999 and includes two rack toms in addition to the floor tom. The cymbal kit also gets two crashes, a ride, and a high-hat.

Yes, these kits are expensive, but they’re beautifully made and, at least in my first impressions, perform impressively. Plus, if you go with the Gold models, you’re essentially getting two kits since a simple head swap can turn them from e-drums into an acoustic set. Plus, no price is too high when it comes to pretending you’re Lars from Metallica without annoying all of your roommates.

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17.3: Speed of Sound
Figure 17.3.1 :The mass of a fluid in a volume is equal to the density times the volume, m = ρV = ρAx. The mass flow rate is the time derivative of the mass. Now consider a sound wave moving through a parcel of air. A parcel of air is a small volume of air with imaginary boundaries (Figure 17.3.5 ).
Sound Waves Problem Sets
Problem 24: An organ pipe which acts as a closed-end resonator has a length of 83 cm. Sound waves travel at 350 m/s through the air column of the pipe. a. Determine the fundamental frequency of the organ pipe. b. Determine the frequency of the next two harmonics of this closed-end organ pipe. Audio Guided Solution.
Physics Tutorial: The Speed of Sound
At normal atmospheric pressure, the temperature dependence of the speed of a sound wave through dry air is approximated by the following equation: v = 331 m/s + (0.6 m/s/C)•T. where T is the temperature of the air in degrees Celsius. Using this equation to determine the speed of a sound wave in air at a temperature of 20 degrees Celsius ...
14.1 Speed of Sound, Frequency, and Wavelength
The amplitude of a sound wave decreases with distance from its source, because the energy of the wave is spread over a larger and larger area. But some of the energy is also absorbed by objects, such as the eardrum in Figure 14.5, and some of the energy is converted to thermal energy in the air. Figure 14.4 shows a graph of gauge pressure versus distance from the vibrating string.
Speed of Sound in Physics
Speed of Sound in Physics. This entry was posted on June 17, 2023 by Anne Helmenstine (updated on June 22, 2023) The speed of sound in dry air at room temperature is 343 m/s or 1125 ft/s. In physics, the speed of sound is the distance traveled per unit of time by a sound wave through a medium. It is highest for stiff solids and lowest for gases.
17.2 Speed of Sound
17.5. In an ideal gas (see The Kinetic Theory of Gases ), the equation for the speed of sound is. v = γRTK M− −−−−−√, 17.6. where γ is the adiabatic index, R = 8.31J/mol ⋅ K is the gas constant, TK is the absolute temperature in kelvins, and M is the molar mass.
7.3: The Speed of Sound
The physics of sound waves is obviously a three-dimensional problem. However, we can learn a lot about sound by considering motion of air in only one-dimension. Consider, for example, standing waves in the air in a long narrow tube like an organ pipe, shown in cartoon form in Figure $ 7.3$.
Sound Waves Problem Sets
Sound Waves: Equation Overview. There are 23 ready-to-use problem sets on the topic of Sound Waves. The problems target your ability to determine wave quantities such as frequency, period, wavelength, intensity and speed from verbal descriptions and diagrams of physical situations pertaining to sound waves and resonance in strings and air columns.
17.2 Speed of Sound
While not negligible, this is not a strong dependence. At 0°C 0 °C, the speed of sound is 331 m/s, whereas at 20.0°C 20.0 °C, it is 343 m/s, less than a 4% 4 % increase. (Figure) shows how a bat uses the speed of sound to sense distances. Figure 17.6 A bat uses sound echoes to find its way about and to catch prey.
17.2: Speed of Sound, Frequency, and Wavelength
vw = fλ, where vw is the speed of sound, f is its frequency, and λ is its wavelength. The wavelength of a sound is the distance between adjacent identical parts of a wave—for example, between adjacent compressions as illustrated in Figure 17.2.2. The frequency is the same as that of the source and is the number of waves that pass a point ...
17.2 Speed of Sound, Frequency, and Wavelength
4.6 Problem-Solving Strategies; 4.7 Further Applications of Newton's Laws of Motion; 4.8 Extended Topic: The Four Basic Forces—An Introduction; Glossary; ... The speed of sound is affected by temperature in a given medium. For air at sea level, the speed of sound is given by. v w = 331 m/s T 273 K, v w = 331 m/s T 273 K,
Speed of Sound, Frequency, and Wavelength
The wavelength of a sound is the distance between adjacent identical parts of a wave—for example, between adjacent compressions as illustrated in (Figure). The frequency is the same as that of the source and is the number of waves that pass a point per unit time. A sound wave emanates from a source vibrating at a frequency \ (f\), propagates ...
17.2 Speed of Sound, Frequency, and Wavelength
The relationship of the speed of sound vw, v w, its frequency f, f, and its wavelength λ λ is given by. vw = f λ, v w = f λ, which is the same relationship given for all waves. In air, the speed of sound is related to air temperature T T by. vw = (331 m/s) v w = ( 331 m/s) √ T 273 K. T 273 K.
PDF Speed of Sound
overhead at an altitude of 15 km at Mach 2.5. Calculate the speed of the plane in km/h if the ground temperature is 23oC. 5. Calculate the wavelength in the following substances if the frequency is 1000s-1 and the speed of sound in the medium is given: a) Helium (1230 m/s) b) Hydrogen (1267 m/s) c) Steel (5130 m/s) d) Glass (4700 m/s) 6.
Speed of Sound in Solids, Liquids, and Gases
This physics video tutorial provides a basic introduction into the speed of sound in solids, liquids, and gases. It contains plenty of examples and practice...
PDF waves & Sound practice problems
PROBLEMS. 1. What is the speed of sound in air at (a) 27.0°C, (b) 100°C, (c) 200°C? 2. What is the intensity level in decibels of a sound wave of intensity (a) 10-6 W/m2 and (b) 10-5 W/m2? 3. What is the intensity of a sound whose intensity level is (a) 40 dB and (b) 100 dB? 4. The dB level at a distance of 2.0 m from a sound source is 100 dB.
Calculating the Speed of Sound in Air: Physics Problem Solving
Join us in this physics problem-solving session as we dive into the fascinating topic of wave motion. We'll explore how to calculate the speed of sound in ai... Common Core
Solving physics problems
To solve the problem, all that's really needed is 1) the speed of sound in air and 2) the amount of time it takes for the sound to travel from bat to obstacle. The "50 kHz" is irrelevant- all sound travels at the same speed, no matter what the frequency is. How long the sound lasts- the "0.1 milliseconds"- is also irrelevant.
PDF Wave Speed Equation Practice Problems
Microsoft Word - 614 - 94 - WS-Basic Wave Speed-answers.doc. Problems for you to try: Complete the following practice problems. You MUST show ALL the work outlined in the steps in the example problems. A wave with a frequency of 14 Hz has a wavelength of 3 meters.
Speed Of Sound Formula Physics
The speed of sound is defined as the travelled by the sound wave that propagates per unit time through an elastic medium. In a perpendicular medium, the speed of the sound depends on the elasticity and density of the medium. If the speed of the sound is greater then the elasticity is more and density is less. γ = Ratio of specific heat.
A Collection of Maths Problem Solving Questions:#397 (Speed
May and her mom stopped at a railroad crossing to let a train pass. As the train began to pass, May counted 6 cars in the first 10 seconds. It took the train...
Problem solving
The term problem solving has a slightly different meaning depending on the discipline. For instance, it is a mental process in psychology and a computerized process in computer science. There are two different types of problems: ill-defined and well-defined; different approaches are used for each. Well-defined problems have specific end goals ...
Transverse wave
Mathematical formulation. Mathematically, the simplest kind of transverse wave is a plane linearly polarized sinusoidal one. "Plane" here means that the direction of propagation is unchanging and the same over the whole medium; "linearly polarized" means that the direction of displacement too is unchanging and the same over the whole medium; and the magnitude of the displacement is a ...
Mesoscale Eddy Effects on Vertical Correlation of Sound Field ...
To solve the problem of array detection performance in environments with mesoscale eddies, this study utilizes the Gaussian eddy model to describe the sound speed structure disturbed by eddies. Through numerical simulations, the corresponding sound field is obtained, and the transmission loss influenced by the eddy is analyzed. Furthermore, to investigate the relation between the array gain ...
Is It a Sound of Music…or of Speech? Scientists Uncover How Our ...
It has been well documented that a song's volume, or loudness, over time—what's known as "amplitude modulation"—is relatively steady at 1-2 Hz. By contrast, the amplitude modulation of speech is typically 4-5 Hz, meaning its volume changes frequently. Despite the ubiquity and familiarity of music and speech, scientists previously lacked clear understanding of how we effortlessly ...
Sound Waves Problem Sets
Includes 7 problems. Problem Set SW4 - Wavelength, Frequency, and Speed 2. Use the v=f•λ equation to relate the speed of sound to the wavelength of sound and the frequency of sound. Several problems center around a diagram. Includes 6 problems. Problem Set SW5 - Speed, Tension, Density.
Is it a sound of music...or of speech? Scientists ...
"Overall, slower and steady sound clips of mere noise sound more like music while the faster and irregular clips sound more like speech." Scientists gauge the rate of signals by precise units of ...
Solved A projectile is fired from ground level at time t=0,
It has an initial speed v0 = 60.0 m/s . In this problem we are assuming that the ground is level.Check that the angle in your calculator is in "degrees". sin30o = 0.5Part A - Calculate the x- and y- components of the initial velocity in m/s, Keep
Zildjian's Alchem-E e-drums solve one of the biggest problems with
These pro-grade electronic drums provide authentic sound and feel for both acoustic and electronic playing. That includes the cymbals. By Stan Horaczek Posted on May 26, 2024 10:00 AM EDT 6 minute ...
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17.2 Speed of Sound

Speed of Sound in Various Media

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17.2: Speed of Sound, Frequency, and Wavelength

17.2 Speed of Sound, Frequency, and Wavelength

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129 17.2 Speed of Sound, Frequency, and Wavelength

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Making Connections: Take-Home Investigation—Voice as a Sound Wave

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