Question

# Find the fundamental frequency and the frequency of the first three overtones of a pipe 90.0...

Find the fundamental frequency and the frequency of the first three overtones of a pipe 90.0 cm long, if the pipe is open at both ends.

ffund,fov1,fov2,fov3 =   Hz

Find the fundamental frequency and the frequency of the first three overtones of a pipe 90.0 cm long, if the pipe is closed at one end.

ffund,fov1,fov2,fov3 =

Hz

If the pipe is open at both ends, what is the number of the highest harmonic that may be heard by a person who can hear frequencies from 20.0 Hz to 2.00×104 Hz ?

n =

If the pipe is closed at one end, what is the number of the highest harmonic that may be heard by a person who can hear frequencies from 20.0 Hz to 2.00×104 Hz ?

n =

a)

in case of open pipe, frequency fn=n*v/2l'

fundamental frequency, f1=1*343/(2*90*10^-2)

f1=190.55 Hz

for first overtone or second harmonic,(n=2) f2=2*f1

f2=2*190.55

f2=381.1 Hz

for second overtone or third harmonic,(n=3) f3=3*f1

f3=3*190.55

f3=571.67 Hz

for third overtone or fourth harmonic,(n=4) f4=4*f1

f4=4*190.55

f4=762.2 Hz

b)

in case of open pipe, frequency fn=((2n-1)*v/4l

fundamental frequency, f1=(2*1-1)*343/(4*90*10^-2)

f1=95.3 Hz

for first overtone or second harmonic,(n=2) f2=(2*2-1)*343/(4*90*10^-2)

f2=285.8 Hz

for second overtone or third harmonic,(n=3) f3=(2*3-1)*343/(4*90*10^-2)

f3=476.4 Hz

for third overtone or fourth harmonic,(n=4) f4=(2*4-1)*343/(4*90*10^-2)

f4=666.9 Hz

c)

f1=20 HZ

fn=n*f1

2*10^4=n*20

===> n=1000

no of harmonic, n=1000

d)

f1=20 Hz

fn=(2n-1)*f1

2*10^4=(2*n-1)*20

==> n=500.5

no of harmonic, n=500

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