Stationary Waves in Pipes

A CLOSED PIPE

If a turning fork in sounded at the top of a tube with one end open and other and closed, the air in the tube resonates (vibrates freely) at a certain length of the tube.

In practical, the vibrations at the open end of a pipe extend into the free air, to a short distance called end correction. Hence L₁ + C =   ¼λ

In general, for any harmonics in a closed pipe,

Example 1

A turning fork of frequency 512 Hz is sound at the mouth of a tube closed at one end with a movable piston. It is found that resource occurs when the column of air is 18.0 cm along and again when the column is 51 cm long. Determine the velocity of sound in air.

Solution

L₁ = ^λ/₄     and         L₂ = ³/₄ λ

L₂ – L₁ = ³/₄ λ - ¼ λ
L₂ – L₁  =½λ

        (51 – 18) cm = ½ λ

         33 cm = ½ λ

   λ = 66 cm.

λ = 0.66 m

From      v = ¦λ

V = 512(0.66)

 = 338 m/s

\The velocity of sound in air = 338 m/s

Example 2

In a closed pipe; first resonance is at 23 cm and second resonance at 7.3 cm. Determine the wavelength and the end correction of the pipe.

Solution

L₁ + C = ¼ λ

                           0.23 + C = ¼ λ - - - - - - - - - - - (i)

L₂ + C = ³/₄ λ

                          0.73+ C = ³/₄ λ - - - - - - - - - - (ii)

From;

0.23+ C = ¼ λ

0.23+ C = ¼ (1)

0.23+ C = 0.25

C = 0.25 – 0.23

C = (0.25 – 0.23) m

= 0.02m or 2cm

\The end correction = 0.02 m.

B: OPEN PIPES

Open pipes have both ends open.

 

Example

A turning fork of frequency 250Hz is used produce resonance in an open pipe. Given that the velocity sound in air is 350m/s, find the length of the tube which gives.

a)       The first resonance

b)      The third (3^rd) resonance

RESONANCE

”Resonance is said to occur wherever a particular body or system is set in oscillation at its own natural frequency as a result of impulse received from some other system which is vibrating with the same frequency”

BEATS

Is the rise or fall in loudness of sound when two sources of sound of nearly equal frequencies produce sound together.

                                                                                Or

Is the difference between two frequencies of sound waves.

Beat frequency = ¦₂ - ¦₁ or ¦₁ - ¦₂

Example

A 256 Hz turning fork produces sound at the same time with a 249 Hz fork. What is the beat frequency?

Solution

Beat frequency = ¦₂ - ¦₁

                                        = 256 Hz – 249 Hz

                    = 7 Hz

\ The beat frequency = 7 Hz.