Standing waves

Cause : resonance.

Standing transverse waves

Form: sine .

Occurs in stringed instruments

Amplitude varies along the string. They all the vibrate with the same frequency.

Nodes  (no vibrations)  and  Antinodes (amplitude maximal value)

Nodes at the fixed ends and between the antinodes.

Between two nodes vibrate the points with equal phase

 Δφ = 0 ( points between two nodes) or 0.5 (points on either side of a node)

There is also:

v = f λ     

  • v speed of the travelling wave in the string
  • λ length of one sine

Standing waves in a string fixed at two ends 

Example

A string with length of 60 cm is fixed at both sides

The fundamental has a frequency of  400 Hz.

Calculate the speed of the wave in the string.

The distance  from node to node  N-N  = ½ λ     So  0.5 λ1= 0.60 m         λ1 = 1.20 m

λf is the wavelength of the fundamental  mode

v = f1  λ= (400)( 1.20) = 480 m/s

Example

Find now the frequency of the second harmonic

v  = f2 λ2

λ2 = 0.60 m      

480 = f2 0.60         f2= 800 Hz

The  frequency of the third harmonic   is 1200 Hz       ( 1.5 λ3 = 0.60 m

                                                                                                            λ3=  0.40 m)

Standing longitudinal waves

Wind instruments and  organ pipes

The same type of calculations

An antinode is created at the mouth of the pipe!

Example

Find the frequency of the fundamental of an open organ pipe with a length of 2.0 m

at a temperature of 20 ⁰C  (speed of the sound is 343 m/s)

Open organ pipe : the open top is also an antinode

Distance from antinode to antinode A-A = ½  λf 

So  λf = (2)( 2.0) = 4.0 m

v = 343 m/s

v = ff  λf

343 = ff . 4.0

ff = 85.8 Hz

Example

Find now the frequency of the fundamental of a stopped organ pipe with the same length

Fundamental:     

Stoppedorgan pipe : the closed top created a node

length pipe  (N-A) = ¼ λ       So  λ= (4)( 2.0) = 8.0 m

v = ff  λf       343 = ff   8.0      ff= 42.9 Hz    ( 2 x lower then the fof an open pipe)

Example

Calculate the frequency of the second harmonic of a stopped organ pipe with the same length of 2.0 m at the same temperature of 20 ⁰C

length organ pipe = 2.0 m

length = ¾ λ 1     λ1= 2.67 m

f1 =  128.5 Hz

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