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
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 λ1 = (400)( 1.20) = 480 m/s
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!
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
Find now the frequency of the fundamental of a stopped organ pipe with the same length
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 ff of an open pipe)
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