Question

the to nearest harmonics of a tube closed at one end and open at other and are to 20 Hz and to 60 Hz what is the fundamental frequency of the system?​

Answer 1

Answer:

The given tube is closed at one end and open at the other. The fundamental frequency =υ/4l=n and other haromonics have frequencies 3n,5n,7n ets. ... This must be the fundamental frequency of the system. The frequency of differrency harmonies would be 20Hz,60Hz,100Hz,140Hz,180Hz,220Hzand260Hz

Answer 2

S O L U T I O N

Thinking Process : Frequency of nth harmonic in a closed end tube will be

F = ( 2n - 1 )v/ 4l [ n= 1,2,3... ]

Also, only odd harmonics exists in a closed end tube. Now, given two nearest harmonics are of frequency 200 Hz and 600 Hz.

∴ ( 2n-1 ) v/4l = 220 Hz _[i]

Next harmonic occurs at

( 2n+1 )v / 4l = 260 Hz _[ii]

On subtracting Eq [i] from Eq [ii] we get,

• { (2n + 1 ) - ( 2n - 1 ) } v / 4l = 260 - 220

• 2 [v/4l] = 40

• v/4l = 20 Hz

∴ The fundamental frequency of the system is 20 Hz