Question

A uniform narrow tube 1.70 m long is open at both ends. It resonates at two successive harmonics of frequencies 275 Hz and 330 Hz. What is (a) the fundamental frequency, and (b) the speed of sound in the gas in the tube

Answer 1

a. 55Hz

b. 187m/s

Step-by-step explanation:

Step one:

For an open-open tube, the harmonic frequencies follow the following rule:

`F_n= (nv)/(2L)\\\\F_n= 275= (nv)/(2L)= (nv)/(2*1.7)`

`F_(n+1)= 330=((n+1)v)/(2L) =( (n+1)v)/(2*1.7) \\\\F_(n+1) - F_n=((n+1)v)/(2*1.7) - (nv)/(2*1.7)\\\\F_(n+1) - F_n= ( (n+1)v- nv)/(3.4)`

`330-275= 55= (v)/(3.4)`

`v= 3.4*55\\\\v=187 m/s`

Step two:

The fundamental frequency is when n=1

`F_n= (1*187)/(2*1.7)\\\\F_n= (187)/(3.2) \\\\F_n= 55 Hz`