Pressure waves in open pipe

Pressure waves in pipeclosed at one end

Musical Sounds• Consider a hollow pipe open at

both ends• a wave reflects even if the end is

open =>free ‘end’ =>anti-node

Fundamental or first harmonic f1 = v/ = v/2L

In general, n=2L/n n=1,2,3,…

fn = v/ n =nv/2L

for L=.4m, v=343m/s, f1=429Hz

Note: pressure has anode but displaementan anti-node

Musical Sounds• Consider a pipe with one

end closed• waves reflect at both ends

but there is a node at the closed end and an anti-node at the open end

Fundamental has /4 = L f1 = v/ = v/4L

In general, n = 4L/n but n is odd!

fn = v/ n = nv/4L n=1,3,5,...

Lower frequency as L increases

Lower than both open

Problem • Organ pipe A has both ends open and a

fundamental frequency of 300 Hz

• The 3rd harmonic of pipe B (one open end) has the same frequency as the second harmonic of pipe A

• How long is a) pipe A ? b) pipe B ?if the speed of sound is 343 m/s

Problem

• fundamental of A has LA=/2=v/2f =(343m/s)/2(300Hz) =.572 m

• 2nd harmonic has LA==.572m f=v/ =343/.572=600Hz

3rd harmonic of pipe B has n=3 v= f=(4 LB/3)600 =343 m/s LB = 343/800 = .429 m

Musical Sounds• Actual wave form produced by an instrument is a

superposition of various harmonics

Complexwave

Fourier Analysis• The principle of superposition can be used to

understand an arbitrary wave form

• Jean Baptiste Fourier (1786-1830) showed that an arbitrary wave form can be written as a sum of a large number of sinusoidal waves with carefully chosen amplitudes and frequencies

• e.g. y(0,t)= -(1/) sin(t)-(1/2) sin(2t) -(1/3) sin(3t)-(1/4) sin(4t)-...

y(x,t)=ym sin(kx- t)

y(0,t)=-ym sin(t)Decomposition into sinusoidal wavesis analogous to vector components r = x i + y j + z k

-(1/) sin(t) -(1/2) sin(2t)

-(1/) sin(t)-(1/2) sin(2t)

-(1/) sin(t)-(1/2) sin(2t) -(1/3) sin(3t) -(1/4) sin(4t) -(1/5) sin(5t) -(1/6) sin(6t)

T=2/2T=2/