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.
Please enter your answer as four numbers, separated with commas.


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.
Please enter your answer as four numbers, separated with commas.


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 2.00×10^{4} 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 2.00×10^{4} Hz ?


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=((2n1)*v/4l
fundamental frequency, f1=(2*11)*343/(4*90*10^2)
f1=95.3 Hz
for first overtone or second harmonic,(n=2)
f2=(2*21)*343/(4*90*10^2)
f2=285.8 Hz
for second overtone or third harmonic,(n=3)
f3=(2*31)*343/(4*90*10^2)
f3=476.4 Hz
for third overtone or fourth harmonic,(n=4) f4=(2*41)*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=(2n1)*f1
2*10^4=(2*n1)*20
==> n=500.5
no of harmonic, n=500
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