Question

What frequencies will a 1.65 m long tube produce in the audible range (20 Hz - 20,000 Hz) at 17.0°C for the following cases?

(a) the tube is closed at one end
lowest frequency
Hz
second lowest frequency
Hz
highest frequency (rounded to the nearest Hz)
Hz

(b) the tube is open at both ends
lowest frequency
Hz
second lowest frequency
Hz
highest frequency (rounded to the nearest Hz)
Hz

Answer 1

1) for tube closed at one end

the first ocsillation will occur when lambda = 4L

the second oscilation will occur when lambda = (4/3)*L

similarly ,the 5th and 7 th harmonics will ocuur at (4/5)*L and (4/7)*L respectively

now, given length of tube =1.65 m

the speed of sound at given temperatutre = 341.77 m/s

a) now,

lowest frequency will be at first harmonics,

so,lamda=4L

=4*1.65

=6.6m

b)second lowest frequency will be at lambda= (4/3) *L

= 2.2 m

so frequency = speed/lambda=341.77/2.2 = 155.35 Hz

c) the highest frequency would be near to 20000 Hz

2) in this case

for first harmonics, lambda=2L

2nd harmonics will be lambda=(2/2)*L

3nd harmonics will be lambda=(2/3)*L

and so in

a) the lowest frequency will be in first hrmonics

Lambda= 2L = 3.3 m

so velocity = speed /lambda=341.77/3.3 =103.56 HZ

b) second lowest frequency =

lambda =L

so,

frequency = speed / lambda= 341.77/1.65

=207.133 Hz

c) highest frequency will be near higher audible range