Question

The lowest frequency in the audible range is 20 Hz.a)What is the length of the shortest open-open tube needed to produce this frequency?b)What is the length of the shortest open-closed tube needed to produce this frequency?

Answer 1

a) 8.58 m

For an open-open tube, the fundamental frequency is given by

`f=(v)/(2L)`

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:

`L=(v)/(2f)=(343 m/s)/(2(20 Hz))=8.58 m`

(b) 4.29 m

For an open-closed tube, the fundamental frequency is instead given by

`f=(v)/(4L)`

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:

`L=(v)/(4f)=(343 m/s)/(4(20 Hz))=4.29 m`