The speed of a sound wave in air depends upon the properties of the air, namely the temperature and the pressure. The pressure of air (like any gas) will effect the mass density of the air (an inertial property) and the temperature will effect the strength of the particle interactions (an elastic property). At normal atmospheric pressure, the temperature dependence of the speed of a sound wave through air is approximated by the following equation:
v = 331 m/s + (0.59 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 yields the following solution.

v = 331 m/s + (0.59 m/s/C)*T

v = 331 m/s + (0.59m/s/C)*20 °C

v = 331 m/s + 12 m/s

v = 343 m/s

At normal atmospheric pressure and a temperature of 20° Celsius, a sound wave will travel at approximately 343 m/s; this is approximately equal to 1200 km/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 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
distance = v * t = 345 m/s * 3 s = 1035 m

Another phenomenon related to the perception of time delays between two events is the phenomenon of echolation. 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 2.2 seconds after making the holler, then the distance to the canyon wall can be found as follows:
distance = v * t = 345 m/s * 1.1 s = 380 m

The canyon wall is 380 meters away. You might have noticed that the time of 1.1 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 the phenomenon of echolation is of relatively minimal importance to humans, it is an essential trick of the trade for bats. Being merely blind, bats must use sound waves to navigate and hunt. They produce short bursts of ultrasonic sound waves which reflect off 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 (see below). This method of echolation enables a bat to navigate and to hunt.

Like any wave, a sound wave has a speed which 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 be the following equation.
Speed = Wavelength * Frequency

Using the symbols v, , and f, the equation can be re-written as
v = f *lambda

The above equations are useful for solving mathematical problems related to the speed, frequency and wavelength relationship. However, one important misconception could be conveyed by the equation. Even though wave speed is calculated using the frequency and the wavelength, the wave speed is not dependent upon these quantities. An alteration in wavelength does not effect (i.e., change) wave speed. Rather, an alteration in wavelength effects the frequency in an inverse manner. A doubling of the wavelength results in a halving of the frequency; yet the wave speed is not changed. The speed of a sound wave depends on the properties of the medium through which it moves and the only way to change the speed is to change the properties of the medium.

A. Questions

1. a) Determine the speed of sound on a winter day in Mississauga (T=3° C).
b) Determine the speed of sound on a summer day in Mississauga (T=30° C).

2. An automatic focus camera is able to focus on objects by use of an ultrasonic sound wave. The camera sends out sound waves which 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 0.140 seconds after leaving the camera, how far away is the object?

3. Miles Tugo is camping in Glacier National Park. In the midst of a glacier canyon, he makes a loud holler. He hears an echo 2.0 seconds later. The air temperature is 20-degrees C. How far away are the canyon walls.

4. If you were to inhale a few breaths from a helium gas balloon, you would probably experience an amusing change in your voice. You would sound like Donald Duck or Alvin the Chipmunk. What is the cause of this curious high-pitched effect?

5. At 0° C, how long does it take sound to travel 5.00 km?

6. When jelly sets, it changes from a liquid to a solid . What would you expect to happen to the sound in the material as the jelly sets? Explain your reasoning.

7. A man drops a metal probe down a deep well drilling shaft that is 3920 meters deep. If the temperature is 25 C, which is the closest estimate of how long it takes to hear the echo after the probe is dropped?

The sudden change in pitch of a car horn as a car passes by (source motion) or in the pitch of a boom box on the sidewalk as you drive by in your car (observer motion) was first explained in 1842 by Christian Doppler. His Doppler Effect is the shift in frequency and wavelength of waves which results from a source moving with respect to the medium, a receiver moving with respect to the medium, or even a moving medium.

Stationary Sound Source

The movie at left shows a stationary sound source. Sound waves are produced at a constant frequency f0, and the wavefronts propagate symmetrically away from the source at a constant speed v, which is the speed of sound in the medium. The distance between wavefronts is the wavelength. All observers will hear the same frequency, which will be equal to the actual frequency of the source.

Source moving with vsource < vsound

In the movie at left the same sound source is radiating sound waves at a constant frequency in the same medium. However, now the sound source is moving to the right with a speed vs = 0.7 v. The wavefronts are produced with the same frequency as before. However, since the source is moving, the center of each new wavefront is now slightly displaced to the right. As a result, the wavefronts begin to bunch up on the right side (in front of) and spread further apart on the left side (behind) of the source. An observer in front of the source will hear a higher frequency f ´ > f0, and an observer behind the source will hear a lower frequency f ´ < f0.

B. Questions

1. State what happens to the apparent frequency of a sound source in each of the following situations:

a) The listener is stationary and the source is approaching.
b) The listener is stationary and the source is receding.
c) The source is stationary and the listener is approaching.
d) The source is stationary and the listener is receding.

Objects travelling at speeds less than the speed of sound in air have subsonic speeds. When the speed of an object equals the speed of sound in air at that locations, the speed is called Mach 1. The Mach number of a source of sound is the ratio of the speed of the source to the speed of sound in air at that location.

Mach number = speed of object/speed of sound.

Speeds greater than Mach 1 are supersonic. Speeds for supersonic aircraft, such as the Concorde and fighter aircraft, are given in terms of Mach number rather than kilometers per hour.

Source moving with vsource = vsound ( Mach 1 - breaking the sound barrier )

Now the source is moving at the speed of sound in the medium (vs = v, or Mach 1). The speed of sound in air at sea level is about 331m/s or about 1200 km/hour. The wavefronts in front of the source are now all bunched up at the same point. As a result, an observer in front of the source will detect nothing until the source arrives. The pressure front will be quite intense (a shock wave), due to all the wavefronts adding together, and will not be percieved as a pitch but as a "thump" of sound as the pressure wall passes by.

The figure below shows a bullet travelling at Mach 1.01. You can see the shock wave front just ahead of the bullet.

Source moving with vsource > vsound (Mach 1.4 - supersonic)

The sound source has now broken through the sound speed barrier, and is traveling at 1.4 times the speed of sound (Mach 1.4). Since the source is moving faster than the sound waves it creates, it actually leads the advancing wavefront. The sound source will pass by a stationary observer before the observer actually hears the sound it creates.

As you watch the animation, notice the clear formation of the Mach cone, the angle of which depends on the ratio of source speed to sound speed. It is this intense pressure front on the Mach cone that causes the shock wave known as a sonic boom as a supersonic aircraft passes overhead. The shock wave advances at the speed of sound v, and since it is built up from all of the combined wave fronts, the sound heard by an observer will be quite intense. A supersonic aircraft usually produces two sonic booms, one from the aircraft's nose and the other from its tail, resulting in a double thump.

Click here to see the Video of F-18

Click here for an explanation of how the condesation cloud formed.

A jet flying at an altitidue of 9000 m generates 128 dB sonic boom to the ground. The same jet, flying at an alitude of 30 m would generate a 263 dB sonic boom!. The pressure change from typical atmospheric pressure to the maximum pressure of the compressional wave of a sonic boom occurs over a time interval of 100ms for a fighter jet. Such a sudden, drastic pressure change can force windows to flex beyond their elastic limit and shatter!

The figure below shows a bullet travelling at Mach 2.45. The mach cone and shock wavefronts are very noticeable.

The Skinny on Sonic Booms!
Supersonic cars!

C. Questions

1. The speed of sound at sea level and 0° C is 331 m/s or approximately 1200 km/hour. At an altitude of 10 km, it is approximately 1060 km/h. What is the Mach number of an aircraft flying at an altitude of 10 km with a speed of 1800 km/h?
2. What is the Mach number of an aircraft travelling at sea level at 0° C with a speed of 920 km/h?
3. A military interceptor airplane can fly at Mach 2.0. What is its speed in kilometers per hour at sea level and at 0° C?
4. What is the sound barrier?
5. Why is it so difficult to break through the sound barrier?
6.. What is the sonic boom?
7. Is it true that only fast vehicles and aircraft can break the sound barrier? What else does?
8. Why does a condensation cloud sometimes form?

What does it take to fly faster than the speed of sound? Innovation! Go to the Concorde site and find out! Visit Concorde F.A.Q for some info on the Concorde accident in Paris and why they are not flown anymore.

1. Discuss how the each of the following has to be modified to fly at supersonic speeds!

a) airframe
b) engines
c) fuel
d) nose
e) speed,
f) wings

and ofcourse the airfare!