Introducing Sound Waves David William D. Ecoben III-Archimedes.

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Presentation transcript:

Introducing Sound Waves David William D. Ecoben III-Archimedes

The Nature of Sound  Sound travels in waves as it moves through the air or some other medium (substance)  Waves are produced by a vibrating object.  As a vibrating object moves outward, it compresses the surrounding medium, producing a region of compression called a condensation.  As the vibrating object then moves inward, the medium expands into the space formerly occupied by the object. This region of expansion is called a rarefaction.

 Sound waves consist of these condensations and rarefactions.  Sound waves must travel through a medium.  Sound is called a longitudinal wave because the vibrations are in the same direction as the motion and speed of the wave.  It travels at a speed of 331 m/s in air at 0 o C.  The speed increases, at a rate of 0.6 m/s per Celsius degree.  If something moves faster than sound, it is said to be supersonic. If it travels at the speed of sound, it travels at Mach 1; three times the speed of the sound is called Mach 3.

 Acoustics is the science of sound and of its effects on people.  Condensation is a region of compression in a sound wave.  Frequency of sound waves refers to the number of condensation or rarefactions produced by a vibrating object each second.  Decibel is the unit used to measure the intensity level of a sound. A 3000-hertz tone of zero decibels is the weakest sound that the normal human can hear.  Hertz is the unit used to measure frequency. One hertz equals one cycle (vibration) per second.  *Other terms used in the study of sound can be found in page 451.

 The nature of a particular sound can be described in terms of  Frequency and Pitch  Intensity and loudness and  Quality

Frequency and Pitch  Pitch is related to its frequency. Higher frequency means higher pitch.  Frequencies of sound waves humans can hear range from about 20-20,000 hertz or cycles per second.  The loudness of sound is related to the wave’s energy or intensity.  As the frequency of sound waves increases, the wavelength decreases. Wavelength is the distance between any point on one wave and the corresponding point on the next one.

The Doppler Effect on Waves  The apparent change in pitch produced by moving objects is called the Doppler Effect.  Sound waves undergo the Doppler Effect. When the source of the waves moves at a constant speed relative to you, the observer, the frequency of the wave as observed by you is shifted.

For sound, the shifted frequency f observed, in hertz, is given by:  f observed = f source x (v source - v observer ) (v source - v sound ) Where: F source = frequency in hertz at the source of the sound V source = the speed in m/s of the source of the sound V observer = the speed in m/s of the observer, v source – v observer = the relative speed in m/s of the source with respect to the observer, and V sound = the speed of sound in m/s.

Example 12.5 Emergency A fire truck siren emits a sinusoidal wave with a source frequency f source of 400 Hz. The speed of sound is 340 m/s. Find the wavelength of the siren while at rest. If the truck is moving with velocity v source = 40 m/s, find the wavelength of the siren going nearer to and then away from a stationary observer.

Given: Source frequency f source 400 Hz Velocity of source v source 40 m/s Velocity of the observer v observer 0 Find a. Wavelength if the source is at rest b. Wavelength if the source is nearing the observer c. Wavelength if the source is going away from the observer

 Solution:  A. When source is at rest, using the equation v = f λ  Solving for λ,  λ = v source   f source  λ = 340 m/s / 400 Hz  λ = O.85 m

 B. In front of the siren, since the object is getting farther, then the speed of sound relative to the observer is v – v s, so  λ = (v – v source ) / f source  = (340 m/s – 40 m/s) / 400 Hz  λ = 0.75 m

c. Behind the siren, since the object is going nearer, then the speed of sound relative to the observer is v + v s, so λ = (v + v source ) / f source = (340 m/s + 40 m/s) / 400 Hz λ = 0.95 m