13.3

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.

BeatFormation of Beats Applet

The frequency difference between two sounds can be found by the number of beats per second.