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

Answer:

f = 409 Hz

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 :

$f_2=\dfrac{3v}{2l}$

v is speed of sound

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

$f=\dfrac{v}{2l}$

The relation between f and f₂ can be written as :

$f_2=3f\\\\f=\dfrac{f_2}{3}\\\\f=\dfrac{1227}{3}\\\\f=409\ Hz$

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