speed of sound problem solving pdf

14.1 Speed of Sound, Frequency, and Wavelength
Introduction
1.1 Physics: Definitions and Applications
1.2 The Scientific Methods
1.3 The Language of Physics: Physical Quantities and Units
Section Summary
Key Equations
Concept Items
Critical Thinking Items
Performance Task
Multiple Choice
Short Answer
Extended Response
2.1 Relative Motion, Distance, and Displacement
2.2 Speed and Velocity
2.3 Position vs. Time Graphs
2.4 Velocity vs. Time Graphs
3.1 Acceleration
3.2 Representing Acceleration with Equations and Graphs
4.2 Newton's First Law of Motion: Inertia
4.3 Newton's Second Law of Motion
4.4 Newton's Third Law of Motion
5.1 Vector Addition and Subtraction: Graphical Methods
5.2 Vector Addition and Subtraction: Analytical Methods
5.3 Projectile Motion
5.4 Inclined Planes
5.5 Simple Harmonic Motion
6.1 Angle of Rotation and Angular Velocity
6.2 Uniform Circular Motion
6.3 Rotational Motion
7.1 Kepler's Laws of Planetary Motion
7.2 Newton's Law of Universal Gravitation and Einstein's Theory of General Relativity
8.1 Linear Momentum, Force, and Impulse
8.2 Conservation of Momentum
8.3 Elastic and Inelastic Collisions
9.1 Work, Power, and the Work–Energy Theorem
9.2 Mechanical Energy and Conservation of Energy
9.3 Simple Machines
10.1 Postulates of Special Relativity
10.2 Consequences of Special Relativity
11.1 Temperature and Thermal Energy
11.2 Heat, Specific Heat, and Heat Transfer
11.3 Phase Change and Latent Heat
12.1 Zeroth Law of Thermodynamics: Thermal Equilibrium
12.2 First law of Thermodynamics: Thermal Energy and Work
12.3 Second Law of Thermodynamics: Entropy
12.4 Applications of Thermodynamics: Heat Engines, Heat Pumps, and Refrigerators
13.1 Types of Waves
13.2 Wave Properties: Speed, Amplitude, Frequency, and Period
13.3 Wave Interaction: Superposition and Interference
14.2 Sound Intensity and Sound Level
14.3 Doppler Effect and Sonic Booms
14.4 Sound Interference and Resonance
15.1 The Electromagnetic Spectrum
15.2 The Behavior of Electromagnetic Radiation
16.1 Reflection
16.2 Refraction
16.3 Lenses
17.1 Understanding Diffraction and Interference
17.2 Applications of Diffraction, Interference, and Coherence
18.1 Electrical Charges, Conservation of Charge, and Transfer of Charge
18.2 Coulomb's law
18.3 Electric Field
18.4 Electric Potential
18.5 Capacitors and Dielectrics
19.1 Ohm's law
19.2 Series Circuits
19.3 Parallel Circuits
19.4 Electric Power
20.1 Magnetic Fields, Field Lines, and Force
20.2 Motors, Generators, and Transformers
20.3 Electromagnetic Induction
21.1 Planck and Quantum Nature of Light
21.2 Einstein and the Photoelectric Effect
21.3 The Dual Nature of Light
22.1 The Structure of the Atom
22.2 Nuclear Forces and Radioactivity
22.3 Half Life and Radiometric Dating
22.4 Nuclear Fission and Fusion
22.5 Medical Applications of Radioactivity: Diagnostic Imaging and Radiation
23.1 The Four Fundamental Forces
23.2 Quarks
23.3 The Unification of Forces
A | Reference Tables

Section Learning Objectives

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

Teacher Support

The learning objectives in this section will help your students master the following standards:

In addition, the High School Physics Laboratory Manual addresses content in this section in the lab titled: Waves, as well as the following standards:

(B) investigate and analyze characteristics of waves, including velocity, frequency, amplitude, and wavelength, and calculate using the relationship between wave speed, frequency, and wavelength.

Section Key Terms

[BL] [OL] Review waves and types of waves—mechanical and non-mechanical, transverse and longitudinal, pulse and periodic. Review properties of waves—amplitude, period, frequency, velocity and their inter-relations.

Properties of Sound Waves

Sound is a wave. More specifically, sound is defined to be a disturbance of matter that is transmitted from its source outward. A disturbance is anything that is moved from its state of equilibrium. Some sound waves can be characterized as periodic waves, which means that the atoms that make up the matter experience simple harmonic motion .

A vibrating string produces a sound wave as illustrated in Figure 14.2 , Figure 14.3 , and Figure 14.4 . As the string oscillates back and forth, part of the string’s energy goes into compressing and expanding the surrounding air. This creates slightly higher and lower pressures. The higher pressure... regions are compressions, and the low pressure regions are rarefactions . The pressure disturbance moves through the air as longitudinal waves with the same frequency as the string. Some of the energy is lost in the form of thermal energy transferred to the air. You may recall from the chapter on waves that areas of compression and rarefaction in longitudinal waves (such as sound) are analogous to crests and troughs in transverse waves .

A string bulges to the right, creating compression ahead of it and rarefaction behind it.

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. From this figure, you can see that the compression of a longitudinal wave is analogous to the peak of a transverse wave, and the rarefaction of a longitudinal wave is analogous to the trough of a transverse wave. Just as a transverse wave alternates between peaks and troughs, a longitudinal wave alternates between compression and rarefaction.

A sound wave with areas of alternating compression and rarefaction enter an ear canal and eardrum.

The Speed of Sound

[BL] Review the fact that sound is a mechanical wave and requires a medium through which it is transmitted.

[OL] [AL] Ask students if they know the speed of sound and if not, ask them to take a guess. Ask them why the sound of thunder is heard much after the lightning is seen during storms. This phenomenon is also observed during a display of fireworks. Through this discussion, develop the concept that the speed of sound is finite and measurable and is much slower than that of light.

The speed of sound varies greatly depending upon the medium it is traveling through. 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. The greater the density of a medium, the slower the speed of sound. 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. Table 14.1 shows the speed of sound in various media. Since temperature affects density, the speed of sound varies with the temperature of the medium through which it’s traveling to some extent, especially for gases.

Misconception Alert

Students might be confused between rigidity and density and how they affect the speed of sound. The speed of sound is slower in denser media. Solids are denser than gases. However, they are also very rigid, and hence sound travels faster in solids. Stress on the fact that the speed of sound always depends on a combination of these two properties of any medium.

[BL] Note that in the table, the speed of sound in very rigid materials such as glass, aluminum, and steel ... is quite high, whereas the speed in rubber, which is considerably less rigid, is quite low.

The Relationship Between the Speed of Sound and the Frequency and Wavelength of a Sound Wave

The photograph shows fireworks exploding in the night sky.

Sound, like all waves, travels at certain speeds through different media and has the properties of frequency and wavelength . Sound travels much slower than light—you can observe this while watching a fireworks display (see Figure 14.6 ), since the flash of an explosion is seen before its sound is heard.

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

where v is the speed of sound (in units of m/s), f is its frequency (in units of hertz), and λ λ is its wavelength (in units of meters). Recall that wavelength is defined as the distance between adjacent identical parts of a wave. The wavelength of a sound, therefore, is the distance between adjacent identical parts of a sound wave. Just as the distance between adjacent crests in a transverse wave is one wavelength, the distance between adjacent compressions in a sound wave is also one wavelength, as shown in Figure 14.7 . The frequency of a sound wave is the same as that of the source. For example, a tuning fork vibrating at a given frequency would produce sound waves that oscillate at that same frequency. The frequency of a sound is the number of waves that pass a point per unit time.

A tuning fork vibrates back and forth, creating sound waves.

[BL] [OL] [AL] In musical instruments, shorter strings vibrate faster and hence produce sounds at higher pitches. Fret placements on instruments such as guitars, banjos, and mandolins, are mathematically determined to give the correct interval or change in pitch. When the string is pushed against the fret wire, the string is effectively shortened, changing its pitch. Ask students to experiment with strings of different lengths and observe how the pitch changes in each case.

One of the more important properties of sound is that its speed is nearly independent of frequency. If this were not the case, and high-frequency sounds traveled faster, for example, then the farther you were from a band in a football stadium, the more the sound from the low-pitch instruments would lag behind 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 = f λ v = f λ , and in a given medium under fixed temperature and humidity, v is constant. Therefore, the relationship between f and λ λ is inverse: The higher the frequency, the shorter the wavelength of a sound wave.

Teacher Demonstration

Hold a meter stick flat on a desktop, with about 80 cm sticking out over the edge of the desk. Make the meter stick vibrate by pulling the tip down and releasing, while holding the meter stick tight to the desktop. While it is vibrating, move the stick back onto the desktop, shortening the part that is sticking out. Students will see the shortening of the vibrating part of the meter stick, and hear the pitch or number of vibrations go up—an increase in frequency.

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 maintains the frequency of the original source. If v changes and f remains the same, then the wavelength λ λ must change. Since v = f λ v = f λ , the higher the speed of a sound, the greater its wavelength for a given frequency.

[AL] Ask students to predict what would happen if the speeds of sound in air varied by frequency.

Virtual Physics

This simulation lets you see sound waves. Adjust the frequency or amplitude (volume) and you can see and hear how the wave changes. Move the listener around and hear what she hears. Switch to the Two Source Interference tab or the Interference by Reflection tab to experiment with interference and reflection.

Tips For Success

Make sure to have audio enabled and set to Listener rather than Speaker, or else the sound will not vary as you move the listener around.

Voice as a Sound Wave

In this lab you will observe the effects of blowing and speaking into a piece of paper in order to compare and contrast different sound waves.

sheet of paper

Instructions

Grasp Check

Which sound wave property increases when you are speaking more loudly than softly?

Worked Example

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 conditions where sound travels at 348.7 m/s.

To find wavelength from frequency, we can use v = f λ v = f λ .

(1) Identify the knowns. The values for v and f are given.

(2) Solve the relationship between speed, frequency and wavelength for λ λ .

(3) Enter the speed and the minimum frequency to give the maximum wavelength.

(4) Enter the speed and the maximum frequency to give the minimum wavelength.

Because the product of f multiplied by λ λ equals a constant velocity in unchanging conditions, the smaller f is, the larger λ λ must be, and vice versa. Note that you can also easily rearrange the same formula to find frequency or velocity.

Practice Problems

Links To Physics

Echolocation.

A bat produces sound waves that bounce off of a fly and back to the bat.

Echolocation is the use of reflected sound waves to locate and identify objects. It is used by animals such as bats, dolphins and whales, and is also imitated by humans in SONAR—Sound Navigation and Ranging—and echolocation technology.

Bats, dolphins and whales use echolocation to navigate and find food in their environment. They locate an object (or obstacle) by emitting a sound and then sensing the reflected sound waves. Since the speed of sound in air is constant, the time it takes for the sound to travel to the object and back gives the animal a sense of the distance between itself and the object. This is called ranging . Figure 14.8 shows a bat using echolocation to sense distances.

Echolocating animals identify an object by comparing the relative intensity of the sound waves returning to each ear to figure out the angle at which the sound waves were reflected. This gives information about the direction, size and shape of the object. Since there is a slight distance in position between the two ears of an animal, the sound may return to one of the ears with a bit of a delay, which also provides information about the position of the object. For example, if a bear is directly to the right of a bat, the echo will return to the bat’s left ear later than to its right ear. If, however, the bear is directly ahead of the bat, the echo would return to both ears at the same time. For an animal without a sense of sight such as a bat, it is important to know where other animals are as well as what they are; their survival depends on it.

Principles of echolocation have been used to develop a variety of useful sensing technologies. SONAR, is used by submarines to detect objects underwater and measure water depth. Unlike animal echolocation, which relies on only one transmitter (a mouth) and two receivers (ears), manmade SONAR uses many transmitters and beams to get a more accurate reading of the environment. Radar technologies use the echo of radio waves to locate clouds and storm systems in weather forecasting, and to locate aircraft for air traffic control. Some new cars use echolocation technology to sense obstacles around the car, and warn the driver who may be about to hit something (or even to automatically parallel park). Echolocation technologies and training systems are being developed to help visually impaired people navigate their everyday environments.

Check Your Understanding

Use these questions to assess student achievement of the section’s Learning Objectives. If students are struggling with a specific objective, these questions will help identify which and direct students to the relevant content.

What sort of motion do the particles of a medium experience when a sound wave passes through it?

What does the speed of sound depend on?

What property of a gas would affect the speed of sound traveling through it?

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17.3: Speed of Sound

Learning Objectives

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 \(\PageIndex{1}\)). 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.

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

\[v = f \lambda, \label{17.3}\]

where \(v\) is the speed of the wave, \(f\) is its frequency, and \(\lambda\) 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 \(\PageIndex{2}\)). The frequency is the same as that of the source and is the number of waves that pass a point per unit time.

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

Speed of Sound in Various Media

Table \(\PageIndex{1}\) 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,

\[v = \sqrt{\frac{\text{elastic property}}{\text{inertial property}}} \ldotp\]

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

\[\frac{\partial^{2} y (x,t)}{\partial x^{2}} = \frac{1}{v^{2}} \frac{\partial^{2} y (x,t)}{\partial t^{2}} \ldotp\]

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

\[v = \sqrt{\frac{B}{\rho}} \ldotp \label{17.4}\]

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

\[v = \sqrt{\frac{Y}{\rho}} \ldotp \label{17.5}\]

In an ideal gas (see The Kinetic Theory of Gases ), the equation for the speed of sound is

\[v = \sqrt{\frac{\gamma RT_{K}}{M}}, \label{17.6}\]

where \(\gamma\) is the adiabatic index, R = 8.31 J/mol • K is the gas constant, T K 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

\[v = 331\; m/s \sqrt{1 + \frac{T_{C}}{273 °C}} = 331\; m/s \sqrt{\frac{T_{K}}{273\; K}} \label{17.7}\]

where the temperature in the first equation (denoted as T C ) is in degrees Celsius and the temperature in the second equation (denoted as T K ) is in kelvins. The speed of sound in gases is related to the average speed of particles in the gas,

\[v_{rms} = \sqrt{\frac{3k_{B}T}{m}}.\]

where \(k_B\) is the Boltzmann constant (1.38 x 10 −23 J/K) 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 \(v_{rms}\) 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 0°C , the speed of sound is 331 m/s, whereas at 20.0 °C, it is 343 m/s, less than a 4% increase. Figure \(\PageIndex{3}\) 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.

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 \(\PageIndex{4}\)). The mass in a small volume of length \(x\) of the pipe is equal to the density times the volume, or

\[m = \rho V = \rho Ax.\]

The mass flow rate is

\[\frac{dm}{dt} = \frac{d}{dt} (\rho V) = \frac{d}{dt} (\rho Ax) = \rho A \frac{dx}{dt} = \rho Av \ldotp\]

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,

\[\rho_{in} A_{in}v_{in} = \rho_{out} A_{out}v_{out}.\]

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.

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 \(\PageIndex{5}\)). The density, temperature, and velocity on one side of the volume of the fluid are given as \(\rho\), T, v, and on the other side are \(\rho\) + d\(\rho\), \(T + dT\), \(v + dv\).

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.

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

\[\rho Av = (\rho + d \rho)A(v + dv) \ldotp\]

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

\[\begin{split} \rho v & = (\rho + d \rho)(v + dv) \\ & = \rho v + \rho (dv) + (d \rho)v + (d \rho)(dv) \\ 0 & = \rho (dv) + (d \rho) v \\ \rho\; dv & = -v\; d \rho \ldotp \end{split}\]

The net force on the volume of fluid (Figure \(\PageIndex{6}\)) equals the sum of the forces on the left face and the right face:

\[\begin{split} F_{net} & = p\; dy\; dz - (p + dp)\; dy\; dz \ & = p\; dy\; dz\; - p\; dy\; dz - dp\; dy\; dz \\ & = -dp\; dy\; dz \\ ma & = -dp\; dy\; dz \ldotp \end{split}\]

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 \(\PageIndex{6}\):

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

\[\begin{split} ma & = -dp\; dy\; dz \\ a & = - \frac{dp\; dy\; dz}{m} = - \frac{dp\; dy\; dz}{\rho\; dx\; dy\; dz} = - \frac{dp}{\rho\; dx} \\ \frac{dv}{dt} & = - \frac{dp}{\rho\; dx} \\ dv & = - \frac{dp}{\rho dx} dt = - \frac{dp}{\rho} \frac{1}{v} \\ \rho v\; dv & = -dp \ldotp \end{split}\]

From the continuity equation \(\rho\) dv = −vd\(\rho\), we obtain

\[\begin{split} \rho v\; dv & = -dp \\ (-v\; d \rho)v & = -dp \\ v & = \sqrt{\frac{dp}{d \rho}} \ldotp \end{split}\]

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, \(pV^{\gamma} = \text{constant}\), where \(p\) is the pressure, \(V\) is the volume, and gamma (\(\gamma\)) is a constant that depends on the gas. For air, \(\gamma\) = 1.40. The density equals the number of moles times the molar mass divided by the volume, so the volume is equal to V = \(\frac{nM}{\rho}\). The number of moles and the molar mass are constant and can be absorbed into the constant p \(\left(\dfrac{1}{\rho}\right)^{\gamma}\) = constant. Taking the natural logarithm of both sides yields ln p − \(\gamma\) ln \(\rho\) = constant. Differentiating with respect to the density, the equation becomes

\[\begin{split} \ln p - \gamma \ln \rho & = constant \\ \frac{d}{d \rho} (\ln p - \gamma \ln \rho) & = \frac{d}{d \rho} (constant) \\ \frac{1}{p} \frac{dp}{d \rho} - \frac{\gamma}{\rho} & = 0 \\ \frac{dp}{d \rho} & = \frac{\gamma p}{\rho} \ldotp \end{split}\]

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

\[\begin{split} pV & = nRT = \frac{m}{M} RT \\ p & = \frac{m}{V} \frac{RT}{M} = \rho \frac{RT}{M} \ldotp \end{split}\]

Here M is the molar mass of air:

\[\frac{dp}{d \rho} = \frac{\gamma p}{\rho} = \frac{\gamma \left(\rho \frac{RT}{M}\right)}{\rho} = \frac{\gamma RT}{M} \ldotp\]

Since the speed of sound is equal to v = \(\sqrt{\frac{dp}{d \rho}}\), the speed is equal to

\[v = \sqrt{\frac{\gamma RT}{M}} \ldotp\]

Note that the velocity is faster at higher temperatures and slower for heavier gases. For air, \(\gamma\) = 1.4, M = 0.02897 kg/mol, and R = 8.31 J/mol • K. If the temperature is T C = 20 °C (T = 293 K), the speed of sound is v = 343 m/s. The equation for the speed of sound in air v = \(\sqrt{\frac{\gamma RT}{M}}\) can be simplified to give the equation for the speed of sound in air as a function of absolute temperature:

\[\begin{split} v & = \sqrt{\frac{\gamma RT}{M}} \\ & = \sqrt{\frac{\gamma RT}{M} \left(\dfrac{273\; K}{273\; K}\right)} = \sqrt{\frac{(273\; K) \gamma R}{M}} \sqrt{\frac{T}{273\; K}} \\ & \approx 331\; m/s \sqrt{\frac{T}{273\; K}} \ldotp \end{split}\]

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

\[v = f \lambda \ldotp\]

In a given medium under fixed conditions, \(v\) is constant, so there is a relationship between \(f\) and \(\lambda\); the higher the frequency, the smaller the wavelength (Figure \(\PageIndex{7}\)).

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.

Example \(\PageIndex{1}\): Calculating Wavelengths

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

To find wavelength from frequency, we can use \(v = f \lambda\).

Significance

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, 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 \(\lambda\) must change. That is, because \(v = f \lambda\), the higher the speed of a sound, the greater its wavelength for a given frequency.

Exercise \(\PageIndex{1}\)

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.

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 \(\PageIndex{8}\)).

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.

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.

Since the speed of a wave is defined as the distance that a point on a wave (such as a compression or a rarefaction) travels per unit of time, it is often expressed in units of meters/second (abbreviated m/s). In equation form, this is

The faster a sound wave travels, the more distance it will cover in the same period of time. If a sound wave were observed to travel a distance of 700 meters in 2 seconds, then the speed of the wave would be 350 m/s. A slower wave would cover less distance - perhaps 660 meters - in the same time period of 2 seconds and thus have a speed of 330 m/s. Faster waves cover more distance in the same period of time.

Factors Affecting Wave Speed

The speed of any wave depends upon the properties of the medium through which the wave is traveling. Typically there are two essential types of properties that affect wave speed - inertial properties and elastic properties. Elastic properties are those properties related to the tendency of a material to maintain its shape and not deform whenever a force or stress is applied to it. A material such as steel will experience a very small deformation of shape (and dimension) when a stress is applied to it. Steel is a rigid material with a high elasticity. On the other hand, a material such as a rubber band is highly flexible; when a force is applied to stretch the rubber band, it deforms or changes its shape readily. A small stress on the rubber band causes a large deformation. Steel is considered to be a stiff or rigid material, whereas a rubber band is considered a flexible material. At the particle level, a stiff or rigid material is characterized by atoms and/or molecules with strong attractions for each other. When a force is applied in an attempt to stretch or deform the material, its strong particle interactions prevent this deformation and help the material maintain its shape. Rigid materials such as steel are considered to have a high elasticity. (Elastic modulus is the technical term). The phase of matter has a tremendous impact upon the elastic properties of the medium. In general, solids have the strongest interactions between particles, followed by liquids and then gases. For this reason, longitudinal sound waves travel faster in solids than they do in liquids than they do in gases. Even though the inertial factor may favor gases, the elastic factor has a greater influence on the speed ( v ) of a wave, thus yielding this general pattern:

Inertial properties are those properties related to the material's tendency to be sluggish to changes in its state of motion. The density of a medium is an example of an inertial property . The greater the inertia (i.e., mass density) of individual particles of the medium, the less responsive they will be to the interactions between neighboring particles and the slower that the wave will be. As stated above, sound waves travel faster in solids than they do in liquids than they do in gases. However, within a single phase of matter, the inertial property of density tends to be the property that has a greatest impact upon the speed of sound. A sound wave will travel faster in a less dense material than a more dense material. Thus, a sound wave will travel nearly three times faster in Helium than it will in air. This is mostly due to the lower mass of Helium particles as compared to air particles.

The Speed of Sound in Air

The speed of a sound wave in air depends upon the properties of the air, mostly the temperature, and to a lesser degree, the humidity. Humidity is the result of water vapor being present in air. Like any liquid, water has a tendency to evaporate. As it does, particles of gaseous water become mixed in the air. This additional matter will affect the mass density of the air (an inertial property). The temperature will affect the strength of the particle interactions (an elastic property). At normal atmospheric pressure, the temperature dependence of the speed of a sound wave through dry air is approximated by the following equation:

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 yields the following solution.

v = 331 m/s + (0.6 m/s/C)•(20 C)

v = 331 m/s + 12 m/s

v = 343 m/s

(The above equation relating the speed of a sound wave in air to the temperature provides reasonably accurate speed values for temperatures between 0 and 100 Celsius. The equation itself does not have any theoretical basis; it is simply the result of inspecting temperature-speed data for this temperature range. Other equations do exist that are based upon theoretical reasoning and provide accurate data for all temperatures. Nonetheless, the equation above will be sufficient for our use as introductory Physics students.)

Look It Up!

Using wave speed to determine distances.

At normal atmospheric pressure and a temperature of 20 degrees Celsius, a sound wave will travel at approximately 343 m/s; this is approximately equal to 750 miles/hour. While this speed may seem fast by human standards (the fastest humans can sprint at approximately 11 m/s and highway speeds are approximately 30 m/s), the speed of a sound wave is slow in comparison to the speed of a light wave. Light travels through air at a speed of approximately 300 000 000 m/s; this is nearly 900 000 times the speed of sound. For this reason, humans can observe a detectable time delay between the thunder and the lightning during a storm. The arrival of the light wave from the location of the lightning strike occurs in so little time that it is essentially negligible. Yet the arrival of the sound wave from the location of the lightning strike occurs much later. The time delay between the arrival of the light wave (lightning) and the arrival of the sound wave (thunder) allows a person to approximate his/her distance from the storm location. For instance if the thunder is heard 3 seconds after the lightning is seen, then sound (whose speed is approximated as 345 m/s) has traveled a distance of

If this value is converted to miles (divide by 1600 m/1 mi), then the storm is a distance of 0.65 miles away.

Another phenomenon related to the perception of time delays between two events is an echo . A person can often perceive a time delay between the production of a sound and the arrival of a reflection of that sound off a distant barrier. If you have ever made a holler within a canyon, perhaps you have heard an echo of your holler off a distant canyon wall. The time delay between the holler and the echo corresponds to the time for the holler to travel the round-trip distance to the canyon wall and back. A measurement of this time would allow a person to estimate the one-way distance to the canyon wall. For instance if an echo is heard 1.40 seconds after making the holler , then the distance to the canyon wall can be found as follows:

The canyon wall is 242 meters away. You might have noticed that the time of 0.70 seconds is used in the equation. Since the time delay corresponds to the time for the holler to travel the round-trip distance to the canyon wall and back, the one-way distance to the canyon wall corresponds to one-half the time delay.

While an echo is of relatively minimal importance to humans, echolocation is an essential trick of the trade for bats. Being a nocturnal creature, bats must use sound waves to navigate and hunt. They produce short bursts of ultrasonic sound waves that reflect off objects in their surroundings and return. Their detection of the time delay between the sending and receiving of the pulses allows a bat to approximate the distance to surrounding objects. Some bats, known as Doppler bats, are capable of detecting the speed and direction of any moving objects by monitoring the changes in frequency of the reflected pulses. These bats are utilizing the physics of the Doppler effect discussed in an earlier unit (and also to be discussed later in Lesson 3 ). This method of echolocation enables a bat to navigate and to hunt.

The Wave Equation Revisited

Like any wave, a sound wave has a speed that is mathematically related to the frequency and the wavelength of the wave. As discussed in a previous unit , the mathematical relationship between speed, frequency and wavelength is given by the following equation.

Using the symbols v , λ , and f , the equation can be rewritten as

Check Your Understanding

1. An automatic focus camera is able to focus on objects by use of an ultrasonic sound wave. The camera sends out sound waves that reflect off distant objects and return to the camera. A sensor detects the time it takes for the waves to return and then determines the distance an object is from the camera. If a sound wave (speed = 340 m/s) returns to the camera 0.150 seconds after leaving the camera, how far away is the object?

Answer = 25.5 m

The speed of the sound wave is 340 m/s. The distance can be found using d = v • t resulting in an answer of 25.5 m. Use 0.075 seconds for the time since 0.150 seconds refers to the round-trip distance.

2. On a hot summer day, a pesky little mosquito produced its warning sound near your ear. The sound is produced by the beating of its wings at a rate of about 600 wing beats per second.

a. What is the frequency in Hertz of the sound wave? b. Assuming the sound wave moves with a velocity of 350 m/s, what is the wavelength of the wave?

Part a Answer: 600 Hz (given)

Part b Answer: 0.583 meters

3. Doubling the frequency of a wave source doubles the speed of the waves.

a. True b. False

Doubling the frequency will halve the wavelength; speed is unaffected by the alteration in the frequency. The speed of a wave depends upon the properties of the medium.

4. Playing middle C on the piano keyboard produces a sound with a frequency of 256 Hz. Assuming the speed of sound in air is 345 m/s, determine the wavelength of the sound corresponding to the note of middle C.

Answer: 1.35 meters (rounded)

Let λ = wavelength. Use v = f • λ where v = 345 m/s and f = 256 Hz. Rearrange the equation to the form of λ = v / f. Substitute and solve.

5. Most people can detect frequencies as high as 20 000 Hz. Assuming the speed of sound in air is 345 m/s, determine the wavelength of the sound corresponding to this upper range of audible hearing.

Answer: 0.0173 meters (rounded)

Let λ = wavelength. Use v = f • λ where v = 345 m/s and f = 20 000 Hz. Rearrange the equation to the form of λ = v / f. Substitute and solve.

6. An elephant produces a 10 Hz sound wave. Assuming the speed of sound in air is 345 m/s, determine the wavelength of this infrasonic sound wave.

Answer: 34.5 meters

Let λ = wavelength. Use v = f • λ where v = 345 m/s and f = 10 Hz. Rearrange the equation to the form of λ = v / f. Substitute and solve.

7. Determine the speed of sound on a cold winter day (T=3 degrees C).

Answer: 332.8 m/s

The speed of sound in air is dependent upon the temperature of air. The dependence is expressed by the equation:

v = 331 m/s + (0.6 m/s/C) • T

where T is the temperature in Celsius. Substitute and solve.

v = 331 m/s + (0.6 m/s/C) • 3 C v = 331 m/s + 1.8 m/s v = 332.8 m/s

8. Miles Tugo is camping in Glacier National Park. In the midst of a glacier canyon, he makes a loud holler. He hears an echo 1.22 seconds later. The air temperature is 20 degrees C. How far away are the canyon walls?

Answer = 209 m

The speed of the sound wave at this temperature is 343 m/s (using the equation described in the Tutorial). The distance can be found using d = v • t resulting in an answer of 343 m. Use 0.61 second for the time since 1.22 seconds refers to the round-trip distance.

9. Two sound waves are traveling through a container of unknown gas. Wave A has a wavelength of 1.2 m. Wave B has a wavelength of 3.6 m. The velocity of wave B must be __________ the velocity of wave A.

a. one-ninth b. one-third c. the same as d. three times larger than

The speed of a wave does not depend upon its wavelength, but rather upon the properties of the medium. The medium has not changed, so neither has the speed.

10. Two sound waves are traveling through a container of unknown gas. Wave A has a wavelength of 1.2 m. Wave B has a wavelength of 3.6 m. The frequency of wave B must be __________ the frequency of wave A.

Since Wave B has three times the wavelength of Wave A, it must have one-third the frequency. Frequency and wavelength are inversely related.

Interference and Beats

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Speed of Sound Formula

What is the Speed of Sound?

Sound is a vibration or noise which travels through any medium. It travels in energy waves from one molecule to another, and when it enters a person's ear, it can be easily perceived. When an object vibrates, for example, it passes energy to the surrounding particles, causing them to vibrate as well. Owing to the lack of particles to serve as a conduit, sound cannot move through the vacuum. It can only flow through a medium like air, water, or solid.

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The distance travelled per unit time by a sound wave travelling in an elastic medium is known as the speed of sound. The speed of sound in a given medium depends on the elasticity and density of that medium. The higher the sound speed, the greater is the elasticity and the lower is the mass. As a result, the speed of sound is the maximum in solids and the minimum in gases.

The speed of sound equation is expressed as,

Speed of sound = The square root of (the coefficient ratio of specific heats × the pressure of the gas / the density of the medium).

Speed of sound = \[\frac{\text{(the coefficient ratio of specific heats} \times {\text{ the pressure of the gas)}}}{\text{(the density of the medium)}}\]

c = \[\frac{\gamma \times P}{\rho}\]

c: Speed of sound

P: Pressure

ρ : Density

γ : Specific heat ratio

Speed of Sound in Air Formula

The speed of sound in air formula is,

γ = \[\sqrt{\frac{\gamma \times R \times T}{M}}\]

For air, γ = 1.4, R = 8.31 J/mol, and M = 0.02897 kg/mol.

Speed of Sound in Solid Formula

The sound speed for pressure waves in rigid materials like metals is sometimes given for "long rods" of the material, which are simpler to measure.

The speed of pure pressure waves may be simplified in rods with diameters less than a wavelength and the speed of sound in solid formula is given by:

C\[_{solid}\] = \[\frac{E}{\rho}\]

Where E stands for Young's modulus. This expression is identical to that for shear waves, with the exception that Young's modulus replaces the shear modulus. The speed of sound for pressure waves in long rods will always be somewhat slower than the same speed in the homogeneous 3-dimensional solids, and the ratio of the speeds in the two types of objects is determined by the material's Poisson's ratio.

Speed of Sound in Water Formula

The only non-zero stiffness in a fluid is due to volumetric deformation (fluid does not sustain shear forces).

In conclusion, the sound speed in a fluid (water) is given by

C\[_{fluid}\] = \[\frac{K}{\rho}\]

K = bulk modulus of the fluid

ρ = Density of fluid

Speed of Sound in Gas Formula

The speed of sound in the fluid is,

K = bulk modulus of the fluid.

ρ = Density of fluid.

By adiabatic compression and rarefaction, sound waves move through a gas (expansion). The speed of sound in gas formula is given by,

γ = \[\sqrt{\frac{\gamma \times R \times T}{M}}\]

γ = Adiabatic index

R = 8.314 J/mol - k universal gas constant

T = Absolute temperature

M = Molecular mass

Wavelength Sound Formula

In simple terms, the wavelength is the distance between two consecutive crests or two consecutive troughs of a wave.

In addition, many different things move in similar forms of waves, like water, strings, air (sound waves), the earth or ground, and light. Furthermore, the wavelength of the wave is represented by the Greek symbol lambda (λ). Furthermore, the wave's wavelength is equal to the wave's velocity divided by its frequency. We also use meters (m) to measure wavelength in units.

The wavelength formula sound is given by,

Wavelength = \[\frac{\text{Speed of sound}}{Frequency}\]

Solved Examples

Ex.1. The frequency of the middle C on the piano is 342 Hz. What will be the wavelength of the sound corresponding to the note of middle C if the speed of sound in air is assumed to be 445 m/s?

By using the wavelength formula sound we get,

v = f.λ

Rearranging the wavelength sound formula we get,

λ = \[\frac{445}{342}\]

λ = 1.3011 m/s

The wavelength is 1.3011 m/s.

Ex.2. The sound wave with density 0.037 Kg/m\[^{3}\] and pressure of 4 kPa having the temp 50 degrees Celsius travels in the air. Find out the speed of the sound.

Temperature T = 276 K

Density ρ = 0.037 Kg/m\[^{3}\]

Pressure p = 4kPa = 4000 Pa

The specific heat in air = 1.4

The speed of sound equation is given by,

c = \[\sqrt{\gamma \times \frac{P}{\rho}}\]

c = \[\sqrt{1.4 \times \frac{4000}{0.037}}\]

c = \[\sqrt{1.4 \times 108, 108.1081}\]

c = \[\sqrt{151351.3513}\]

c = 389.0390 m/s

FAQs on Speed of Sound Formula

Q.1) In Which Material Does Sound Travel the Fastest?

Answer: Solids. Sound waves move the slowest through gases, faster through liquids, and the fastest through solids.

Q.2) What is the Highest Sound Frequency?

Answer: The highest sound frequency is 20 kHz.

Frequencies between 20 Hz (lowest pitch) and 20 kHz are heard by the human ear (highest pitch). Although certain animals can hear any noises below 20 Hz. These are classified as infrasounds.

Q.3) Can Sound Waves Travel in a Vacuum?

Answer: Sound waves are the vibrations of particles travelling through a medium like air, water, or metal. As a result, they are unable to move across vacuum since there are no atoms or molecules to vibrate.

Q.4) What is the Relationship Between the Speed of Sound Temperature Formula?

Answer: Temperature is another factor that affects the speed of sound. Heat is a kind of kinetic energy, much like sound. Molecules with more energy vibrate quicker at higher temperatures, allowing sound waves to travel quicker. Sound travels at 346 meters per second in room temperature air.

5th Grade Physical Science Worksheet

Speed of sound.

How fast does sound travel? Help your child figure out the physics of sound waves with this informative worksheet. He'll learn about the speed of sound through different materials, and even try a tough math challenge.

View aligned standards

Speed of Sound Calculator

Where Ratio of Specific Heats ( γ ) = 1.4

Enter the unknown value as ‘ x ‘

Pressure ( P 0 ) = P a

Density ( ρ ) = k g m − 3

Speed of sound ( c ) = m s − 1

Speed of Sound Calculator is a free online tool that displays the speed of the sound. BYJU’S online speed of sound calculator tool makes the calculation faster and it displays the sound speed in a fraction of seconds.

How to Use the Speed of Sound Calculator?

The procedure to use the speed of sound calculator is as follows:

Step 1: Enter the pressure, density and x for the unknown value in the input field

Step 2: Now click the button “Calculate x” to get the sound speed

Step 3: Finally, the speed of the sound will be displayed in the output field

What is Meant by Speed of Sound?

In Physics, the speed of sound represents the speed of the sound waves . It is the dynamic propagation of the sound waves in the elastic medium. The formula to calculate the speed of sound is given by the formula

\(\begin{array}{l}v =\sqrt{\frac{\gamma P}{\rho }}\end{array} \)

v is the speed of sound

\(\begin{array}{l}\rho\end{array} \) is the density

P is the pressure

\(\begin{array}{l}\gamma\end{array} \) is the adiabatic expansion coefficient

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COMMENTS

14.1 Speed of Sound, Frequency, and Wavelength
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. The greater the density of a medium, the slower the speed of sound. The speed of sound in air is low, because air is compressible.
17.3: Speed of Sound
If the temperature is T C = 20 °C (T = 293 K), the speed of sound is v = 343 m/s. The equation for the speed of sound in air v = √γRT M can be simplified to give the equation for the speed of sound in air as a function of absolute temperature: v = √γRT M = √γRT M (273 K 273 K) = √(273 K)γR M √ T 273 K ≈ 331 m / s√ 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.
PDF Speed of Sound
sound: • Isothermal Speed of Sound, cT =(RT) 1 2 • Adiabatic Speed of Sound, cA =(γRT) 1 2 We note that in air cA is 18.3% larger than cT. It is appropriate to detail some examples of the magnitude of the speed of sound. In air with γ =1.4and R = 280 m2/s2 Ko, the adiabatic speed of sound at a temperature of T = 293oK is 339m/s.Interms
PDF 12-02,03
2. The speed of sound in air depends on the _____ of the air. At 0oC, the speed of sound in air is _____ m/s. For every degree above 0oC, the speed _____ by 0.6 m/s. For every degree below 0oC, the speed _____by 0.6 m/s. The equation is: 3. What is the speed of sound at 35oC? _____ 4.
Sound Waves Problem Sets
The speed ( v) at which sound travels through air is dependent upon the temperature of the air and seems to follow the equation v = 331 m/s + 0.6 m/s/°C * T where T is the Celsius temperature of the air. Determine the speed of sound … a. … on a cold day when the outdoor temperature is 4°C. b. … inside the school where the temperature is 24°C.
PDF Problem Set IX Solutions Fall 2006 Physics 200a
The speed of sound in water and air is 1450 and 330m/sm/s respectively. Sound ... Let us begin solving this problem by drawing the free body diagrams for both masses. Since we know that both masses are at rest that means that the forces on each mass must balance. Focusing our attention on the tension in the wire gives us these two equations
PDF Matthew Schwartz Lecture 21: The Doppler eﬀect
cally (less than the speed of sound). As the source approaches the speed of sound, you can see from the picture that the wave-fronts in the forward direction start bunching up. When the speed of sound is hit, they all come together. Remember for sound these are wavefronts of pressure. The large increase in pressure
PDF Wave Speed Equation Practice Problems
7. A wave has a wavelength of 125 meters is moving at a speed of 20 m/s. What is it's frequency? 8. A wave has a frequency of 900 Hz and a wavelength of 200 m. At what speed is this wave traveling? 9. A wave has a wavelength of 0.5 meters and a frequency of 120 Hz. What is the wave's speed? 10. Radio waves travel at a speed of 300,000,000 m/s.
PDF Distance, Time, Speed Practice Problems
DISTANCE, TIME, SPEED PRACTICE PROBLEMS YOU MUST SHOW YOUR WORK. ... 11. If you shout into the Grand Canyon, your voice travels at the speed of sound (340 m/s) to the bottom of the canyon and back, and you hear an echo. ... CHALLENGE PROBLEM Bill and Amy want to ride their bikes from their neighborhood to school which is 14.4 kilometers away ...
PDF Problem Solving Exercises
Problem Solving Exercises - Pearson Education
Physics Tutorial: The Speed of Sound
Using this equation to determine the speed of a sound wave in air at a temperature of 20 degrees Celsius yields the following solution. v = 331 m/s + (0.6 m/s/C)•T v = 331 m/s + (0.6 m/s/C)• (20 C) v = 331 m/s + 12 m/s v = 343 m/s
PDF Name: KEY Period: Speed /Frequency / Wavelength
9. If the limits of human hearing are 20 Hz. to 20,000 Hz, what are the sound wavelengths that are associated with both of these two extremes, assuming the speed of sound is 345 m/s. Frequency = 20 Hz : Wavelength = 1.725 1× 10 m 2 m Speed "c" (m/s) Frequency "ν" Wavelength (Hz) "λ" (m)
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NASA
Speed of Sound Formula
Find out the speed of the sound. Solution: Given, Temperature T = 276 K Density ρ = 0.037 Kg/m 3 Pressure p = 4kPa = 4000 Pa The specific heat in air = 1.4 The speed of sound equation is given by, c = γ × P ρ c = 1.4 × 4000 0.037 c = 1.4 × 108, 108.1081 c = 151351.3513 c = 389.0390 m/s Is this page helpful?
Speed of Sound Formula: Definition, Equations and Examples
Calculate the speed of the sound. Solution: As given here: Temperature, T = 2° C Density, ³ and, pressure P = 2k Pa i.e. P = 2000 Pa as we know that specific heat ratio in air is, Now the speed of sound formula is given by c = c = c= c = 286.97 Therefore, speed of sound = 286.97 m/s.
Speed of Sound (video)
In non-humid air at 20 degrees Celsius, the speed of sound is about 343 meters per second or 767 miles per hour. We can also watch the speed of sound of a repeating simple harmonic wave. The speed of the wave can again be determined by the speed of the compressed regions as they travel through the medium.
PDF Physics 1120: Waves Solutions
Solving for λ and T yields λ = 11.4 m and T = 0.100 s. We know that frequency is given by f = 1/T = 10.0 Hz. As well, the speed of the wave is given by v = λ/T = 114 m/s. To find the tension in the string, we take and rewrite it as . For the given values, Ftension = 260 N. The rate of energy flow, or energy per unit time, or power, is given ...
PDF Chapter 12:Physics of Ultrasound
Sound speed, acoustic impedance, and attenuation coefficient Acoustic properties from Zagzebski (1996) and Shung (2006) Material Sound speed (m/s) Air 330 Water 1480 Fat 1450-1460 Liver 1555-1570 Blood 1550-1560 Muscle 1550-1600 Skull bone 3360-4080 Sound speeds •highest : in solids •lowest : in gases Sound speed in soft tissues is
Sound questions (practice)
Choose 1 answer: Sound waves can propagate as longitudinal or transverse waves, depending on the transmitting medium. A. Sound waves can propagate as longitudinal or transverse waves, depending on the transmitting medium. Sound waves are transverse waves and they propagate perpendicular to the transmitting medium. B.
Speed of Sound
Entire Library Worksheets Fifth Grade Science Speed of Sound. 5th Grade Physical Science Worksheet Speed of Sound. How fast does sound travel? Help your child figure out the physics of sound waves with this informative worksheet. He'll learn about the speed of sound through different materials, and even try a tough math challenge.
Speed of Sound Calculator
The procedure to use the speed of sound calculator is as follows: Step 1: Enter the pressure, density and x for the unknown value in the input field. Step 2: Now click the button "Calculate x" to get the sound speed. Step 3: Finally, the speed of the sound will be displayed in the output field.