Stationary Waves in Stretched String

TONE

Is a sound of regular frequency some times in called musical note.

ANTI NODE

Is the point on a wave with maximum amplitude.

  NODE

      Is the point on a wave with zero amplitude

FUNDAMENTAL FREQUENCY

Is the strongest available frequency present. Also called First Harmonic

                                                    OR

      Is the lowest frequency that a vibrating string or pipe can produce.

OVERTONES

      Are notes of a higher frequency which are produced with the fundamental note.

HARMONICS
Are frequencies which are multiples of the fundamental frequency from a musical instrument

In general,for a vibrating string

THE SONOMETER

A sonometer is an instrument used to study properties of waves, especially stationary waves. It is made up of a wire or string fixed on top of a box. The box increases the loudness of the sound produced by the wire

FACTORS AFFECTING THE FREQUENCY OF A VIBRATING STRING

 The frequency of a vibrating string depends on

                     i.            Length of the string

                   ii.            Tension of the string

                 iii.            Mass density of the string

LENGTH OF STRING

      The frequency of sound waves is inversely proportional to the length of the vibrating string.

¦ µ ¹/_L

TENSION OF STRING

The frequency of the sound is directly proportional to the square root of the tension of the string.

¦ µ ÖT

MASS DENSITY OF STRING (m)

The frequency of the sound is inversely proportional to the square root of the mass per unit length (mass density) of the string.

Mass density is the mass of the string per unit length.

                                             m = ^m/_l

The SI- unit of mass density is kilogram per unit metre (kg/m)

Example 1

Under constant tension, the note produced by a plucked string is 300 Hz when the length is 0.90 m.

a)       At what length is the frequency 200 Hz?

b)      What frequency is produced at a length of 0.30 m?

                                                             

                                   Solution

a)       ¦µ ¹/_L, hence, ¦L = constant.

                        \¦₁L₁ = ¦₂L₂

                    300(0.9) = 200(L₂)

                           L₂      = 300(0.9)

                                          200   

                                    = 3 X 0.9

                                           2

                                    = 2.7

                                        2

                                   = 1.35m

     \The length of the string = 1.35 m

  

b)              ¦₁L₂ = ¦₂L₂

           300(0.9) = ¦₂ (0.3)

                     ¦₂ = 300(0.9)

                                0.3

                          =   300(⁹/₃)

                          =   300(3)

                          =   900 Hz.

    \The frequency = 900 Hz