# Lesson 23: The Speed of Sound

**Notice**: Trying to get property 'post_author' of non-object in

**/home/forge/cilearning.ca/web/app/themes/buddyboss-theme/learndash/ld30/lesson.php**on line

**168**

Download here: Ontario Curriculum Expectations

Sound travels as a series of rarefactions and compressions making up a longitudinal wave. When you talk to a friend, the air particles that make up the rarefactions and compressions pass on their energy of vibration from the air particles in your larynx, to the air particles in your throat, to the air particles in your mouth, to the air particles in the air between the two of you, and finally, to the air particles in the ear of your friend. Of course, the air particles themselves move very little. It is the energy of vibration that is moving from one particle to the next.

As a general rule, as there are exceptions, the closer the particles are to each other, the more quickly the energy of vibration, and therefore the sound, can pass through the medium. As a result, sound usually passes most quickly through solids (where the particles are very close together) and most slowly through gases (where the particles are most spread out). The speed of sound through liquids is generally somewhere between that of solids and gases. These trends can be seen in the table below.

State | Medium | Speed of Sound |
---|---|---|

Solid | Glass | 5 200 m/s |

Steel | 6 000 m/s | |

Rubber | 1 600 m/s | |

Liquid | Sea Water | 1 533 m/s |

Water | 1 490 m/s | |

Gas | Hydrogen | 1 335 m/s |

Air | 346 m/s | |

Carbon Dioxide | 258 m/s |

View the long description of this table.

Sound can travel through warm air more quickly than through cool air. This is because when the air is warmer, the particles in the air have more energy and are moving faster. They can therefore transmit the sound vibrations more quickly. We can determine the speed of sound through air at a given temperature using the following equation:

The speed of sound through air at a pressure of 101 kPa is:

v_{sound} = 331 m/s + (c* _{air}*)(T)

where,

v_{sound} = the speed of sound through the air, measured in metres per second (m/s)

331 m/s is the speed of sound at 0 °C in dry air at sea level.

T = the temperature in degrees Celsius (°C)

C* _{air}* = 0.59 m/s·°C which is a constant that tells us that the speed of sound through air increases by 0.59 m/s for every degree Celsius.

#### Example 1

What is the speed of sound if the temperature is:

a) 20 °C

b) 0 °C

c) -10 °C**A)**

**Given:**

T = 20 °C

**Required:**

v_{sound} = ?

**Solution:**

v_{sound }= 331 m/s + (0.59 m/s·°C)(T)

v_{sound }= 331 m/s + (0.59 m/s·°C)( 20 °C)

v_{sound }= 343 m/s

The speed of sound through air at 20 °C is 343 m/s.**B)**

**Given:**

T = 0˚C

**Required:**

v_{sound} = ?

**Solution:**

v_{sound }= 331 m/s + (0.59 m/s·°C)(T)

v_{sound }= 331 m/s + (0.59 m/s·°C)( 0 °C)

v_{sound }= 331 m/s

The speed of sound through air at 0 °C is 331 m/s.

**C)**

**Given:**

T = -10 °C

**Required:**

v_{sound} = ?

Solution:

v_{sound }= 331 m/s + (0.59 m/s·°C)(T)

v_{sound }= 331 m/s + (0.59 m/s·°C)( -10 °C)

v_{sound }= 325 m/s

The speed of sound through air at -10 °C is 325 m/s.

#### Practice Questions

- What is the temperature if the speed of sound through air is measured to be 319 m/s?

**Answer**

Given:

v = 319 m/s

Required:

T = ?

Solution:

v = 331 m/s + (0.59 m/s·˚C)(T)

The temperature is -20.3 °C when the speed of sound through air is 319 m/s.

#### Echoes and Sonar

Sonar is an acronym for sound navigation and ranging. Sonar devices measure the distance to an object by emitting an ultrasonic pulse and timing how long it takes the pulse to return after hitting a distant object. By knowing the speed at which sound travels through the medium, the distance to the object can be determined.

#### Example 2

A boat is using sonar to determine the depth of a local lake. The speed of sound through the water is 1 490 m/s. It is found that it takes 0.34 s for the sound to return to the boat. How deep is the lake?

When solving echo and sonar questions, you must be careful to account for both the outward bound and returning distance covered by signal.

**Given:**

v = 1 490 m/s

Δt = 0.34 s. This is the total time for the signal to travel to the bottom of the lake and return to the boat. In 0.34 s, the signal has travelled twice the distance between the ship and the lake bottom. Therefore, we only need to consider the distance travelled in half of this time.

Δt_{oneway} = 0.17 s

**Required:**

Δd =?

**Solution:**

Δd = vΔt_{oneway}

Δd = (1 490 m/s)(0.17 s)

Δd = 253 m

The lake is 253 m deep.

#### Practice Questions

- While overlooking a canyon, you yell your name very loudly. If it is 20.0 °C, and you hear your name echoed back to you in 1.60 s, how wide is the canyon?

**Answer**

**Given:**

T = 20 °C

Δt_{return} = 1.60 s

Δt_{oneway} = 0.80 s

**Required:**

v = ?

Δd = ?

**Solution:**

v = 331 m/s + (0.59 m/s·°C)(T)

v = 331 m/s + (0.59 m/s·°C)( 20 °C)

v = 342.8 m/s

Δd = vΔt

Δd = (342.8 m/s)(0.80 s)

Δd = 274 m

The canyon is 274 m wide.