Click here to purchase the entire book in PDF format. Frequency and Period Go back to the clarinet example. If we play a concert A, then it just so happens that the reed is opening and closing at a rate of 440 times per second. This therefore means that there are 440 cycles between a high and a low pressure coming out of the bell of the clarinet each second. We normally use the term Hertz (indicated Hz) to indicate the number of cycles per second in sound waves. Therefore 440 cycles per second is more commonly known as a frequency of 440 Hz. (In older books, you will see this called cycles per second or cps.) In order to find the frequency of a note one octave above this pitch, multiply by 2 (1 octave = twice the frequency). One octave below is one-half of the frequency. In order to find the frequency of a note one decade above this pitch, multiply by 1 (1 decade = ten times the frequency). One decade below is one-tenth of the frequency. Always remember that a complete cycle consists of a high and a low pressure. One cycle is measured from a point on the wave to the next identical point on the wave (i.e. the positive-going zero crossing to the next positive-going zero crossing or maximum to maximum...) If we know the frequency of a sound wave (i.e. 440 Hz), then we can calculate how long it takes a single cycle to exit the bell of the clarinet. If there are 440 cycles each second, then it takes 1/440th of a second to produce 1 cycle. The usual equation for calculating this amount of time (known as the period) is: $$T=\frac{1}{f}$$ (4.4) where $T$ is the period and $f$ is the frequency