Pressure waves in open pipe Pressure waves in pipe closed 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 f 1 = v/ = v/2L In general, n =2L/n n=1,2,3,… f n = v/ n =nv/2L for L=.4m, v=343m/s, f 1 =429Hz Note: pressure has a node but displaement an 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 f 1 = v/ = v/4L In general, n = 4L/n but n is odd! f n = 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 L A = /2=v/2f =(343m/s)/2(300Hz) =.572 m 2nd harmonic has L A = =.572m f=v/ =343/.572=600Hz 3rd harmonic of pipe B has n=3 v= f=(4 L B /3)600 =343 m/s L B = 343/800 =.429 m

Musical Sounds Actual wave form produced by an instrument is a superposition of various harmonics

Complex wave

Fourier Analysis The principle of superposition can be used to understand an arbitrary wave form Jean Baptiste Fourier ( ) 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(2  t) -(1/3  ) sin(3  t)-(1/4  ) sin(4  t)-... y(x,t)=y m sin(kx-  t) y(0,t)=-y m sin(  t) Decomposition into sinusoidal waves is analogous to vector components r = x i + y j + z k

-(1/  ) sin(  t) -(1/2  ) sin(2  t) -(1/  ) sin(  t)-(1/2  ) sin(2  t) -(1/  ) sin(  t)-(1/2  ) sin(2  t) -(1/3  ) sin(3  t) -(1/4  ) sin(4  t) -(1/5  ) sin(5  t) -(1/6  ) sin(6  t) T=2  /2  T=2  / 