Question

An open pipe, 0.29 m long, vibrates in the second overtone with a frequency of 1,227 Hz. In this situation, the fundamental frequency of the pipe, in SI units, is closest to:

Answer 1

f = 409 Hz

Step-by-step explanation:

We have,

Length of the open organ pipe, l = 0.29 m

Frequency of vibration of second overtone,
`f_2 = 1227 Hz`

It is required to find the fundamental frequency of the pipe. For the open organ pipe, the frequency of second overtone is given by :

`f_2=(3v)/(2l)`

v is speed of sound

Let f is the fundamental frequency. It is given by :

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

The relation between f and f₂ can be written as :

`f_2=3f\\\\f=(f_2)/(3)\\\\f=(1227)/(3)\\\\f=409\ Hz`

So, the fundamental frequency of the pipe is 409 Hz.