13.3. harmonics a vibrating string will produce standing waves whose frequencies depend upon the...
TRANSCRIPT
HarmonicsA vibrating
string will produce standing waves whose frequencies depend upon the length of the string.
Harmonics Video 2:34
In the lowest frequency of vibration, one wavelength will equal twice the length of string and its called the fundamental frequency (f1).
For f1, 1λ = 2LOne wavelength = 2*length of string
Fundamental Frequency Half of a wavelength
HarmonicsA Harmonic series is a series of frequencies
that include the fundamental frequency and multiples of that frequency.
1st harmonic = f1
2nd harmonic = f2 = 2*f1
3rd harmonic = f3 = 3*f1
Etc…
HarmonicsThe second harmonic is the next possible
standing wave for the same string length.This shows an increase in frequency, and a
decrease in wavelength.f2=2f1 λ2 = L
Second Harmonic = 2*fundamental frequency
HarmonicsAs the harmonic increases the frequency
increases and wavelength decreases.Ex:f3= 3f1 λ3 = 2/3λ1
f4= 4f1 λ4= ½ λ1
Standing Waves, Fixed at Both Ends Animation
Formula for other harmonicsHarmonic Series of standing waves fn = n* V n=1, 2, 3…
2L
Frequency = harmonic number x (speed of waves on the string)
(2)*(length of vibrating string)
Standing Waves in an Air ColumnIf both ends of a pipe are open, all harmonics
are present and the ends act as antinodes. This is the exact opposite of a vibrating string,
but the waves act the same so we can still use the same formula to calculate frequencies.
fn = n* V n=1, 2, 3… 2L
Frequency = harmonic number x (speed of waves on the string)
(2)*length of vibrating air column)
Standing Waves in an Air ColumnIf one end of the pipe is closed, only odd harmonics
are present (1, 3, 5, etc).This changes the formula:
fn = n* V n=1, 3, 5… 4L
Frequency = harmonic number*(speed of waves on the string)
(4)*length of vibrating air column)
ExampleWhat are the first three harmonics in a 2.45 m long
pipe that is open at both ends? Given that the speed of sound in air is 345 m/s.
L= 2.45 m v= 345 m/sfn = n*v/2L
1st harmonic: f1= 1*(345 m/s)/(2*2.45 m) = 70.4 Hz
2nd harmonic: f2= 2*(345 m/s)/(2*2.45 m) = 141 Hz
3rd harmonic: f3= 3*(345 m/s)/(2*2.45 m) = 211 Hz
ExampleWhat are the first three harmonics of this pipe when
one end of the pipe is closed? Given that the speed of sound in air is 345 m/s.
L= 2.45 m v= 345 m/sfn = n*v/4L
1st harmonic: f1= 1*(345 m/s)/(4*2.45 m) = 35.2 Hz
3rd harmonic: f3= 3*(345 m/s)/(4*2.45 m) = 106 Hz
5th harmonic: f5= 5*(345 m/s)/(4*2.45 m) = 176 Hz
Why do different instruments sound different?Timbre is the quality of a steady musical
sound that is the result of a mixture of harmonics present at different intensities.
This is why a clarinet and a trumpet can play the same pitch but they sound different.
Harmonics Applet
BeatWhen two waves of
slightly different frequencies travel in the same direction they interfere. This causes a listener to hear an alternation between loudness and softness and is called beat.