a) 8.58 m
For an open-open tube, the fundamental frequency is given by
[tex]f=\frac{v}{2L}[/tex]
where
f is the fundamental frequency (the lowest frequency)
v is the speed of sound
L is the length of the tube
In this problem, we have
f = 20 Hz
v = 343 m/s (speed of sound in air)
Solving the equation for L, we find the shortest length of the tube:
[tex]L=\frac{v}{2f}=\frac{343 m/s}{2(20 Hz)}=8.58 m[/tex]
(b) 4.29 m
For an open-closed tube, the fundamental frequency is instead given by
[tex]f=\frac{v}{4L}[/tex]
Where in this problem, we have
f = 20 Hz
v = 343 m/s (speed of sound in air)
Solving the equation for L, we find the shortest length of the tube:
[tex]L=\frac{v}{4f}=\frac{343 m/s}{4(20 Hz)}=4.29 m[/tex]
